AP Chemistry. Introduction to Solubility Equilibria. Slide 1 / 91 Slide 2 / 91. Slide 3 / 91. Slide 4 / 91. Slide 5 / 91.
|
|
- John Campbell
- 6 years ago
- Views:
Transcription
1 Slide 1 / 91 Slide 2 / 91 P hemistry queous Equilibria II: Ksp & Solubility Products Slide 3 / 91 Slide 4 / 91 Table of ontents: K sp & Solubility Products Introduction to Solubility Equilibria alculating K sp from the Solubility alculating Solubility from Ksp lick on the topic to go to that section Introduction to Solubility Equilibria Factors ffecting Solubility Precipitation Reactions and Separation of Ions Return to the Table of ontents Slide 5 / 91 Introduction to Solubility Equilibria Slide 6 / 91 Introduction to Solubility Equilibria Ionic compounds dissociate into their ions to different degrees when placed in water and reach equilibrium with the non-dissociated solid phase when the solution is saturated. saturated solution of ao 3(s) a 2+ a 2+ O 3 O 3 nswer Many shells are made of relatively insoluble calcium carbonate, so the shells are not at huge risk of dissolving in the ocean. ao 3(s) alcium carbonate is a relatively insoluble ionic salt. Would the picture look different for a soluble ionic salt such as Na 2O 3? Which solution would be the better electrolyte?
2 Slide 7 / 91 Introduction to Solubility Equilibria onsider the equilibrium that exists in a saturated solution of ao 3 in water: ao 3(s) a 2+ (aq) + O 3 (aq) Unlike acid-base equilibria which are homogenous, solubility equilibria are heterogeneous, there is always a solid in the reaction. Slide 8 / 91 Introduction to Solubility Equilibria The equilibrium constant expression for this equilibrium is K sp = [a 2+ ] [O 3 2 ] where the equilibrium constant, K sp, is called the solubility product. There is never any denominator in K sp expressions because pure solids are not included in any equilibrium expressions. Slide 9 / 91 Solubility Equilibrium The degree to which an ionic compound dissociates in water can be determined by measuring it's "K sp" or solubility product equilibrium constant. ao 3(s) --> a 2+ (aq) + O 3 (aq) 25 = 5.0 x 10-9 MgO 3(s) --> Mg 2+ (aq) + O 3 (aq) 25 = 6.8 x 10-6 In both cases above, the equilibrium lies far to the left, meaning relatively few aqueous ions would be present in solution. nswer Slide 10 / 91 1 Which Ksp expression is correct for gl? [g + ]/[l - ] [g + ][l - ] [g 2+ ] 2 [l ] 2 [g + ] 2 [l - ] 2 E None of the above. Which saturated solution above would have the higher conductivity and why? Slide 11 / 91 2 Given the reaction at equilibrium: Zn(OH) 2 (s) Zn 2+ (aq) + 2OH - (aq) what is the expression for the solubility product constant, K sp, for this reaction? K sp = [Zn 2+ ][OH - ] 2 / [Zn(OH) 2 ] K sp = [Zn(OH) 2 ] / [Zn 2+ ][2OH - ] Slide 12 / 91 3 Which Ksp expression is correct for Fe 3(PO 4) 2? [Fe 2+ ] 3 [PO 3-4 ] 2 [Fe 2+ ] 3 /[PO 3-4 ] 2 [Fe 3+ ] 2 3- [PO 4 ] 2 [Fe 2+ ] 2 /[PO 3-4 ] 2 E None of the above. K sp = [Zn 2+ ][2OH - ] K sp= [Zn 2+ ][OH - ] 2
3 Slide 13 / 91 4 When 30 grams of Nal are mixed into 100 ml of distilled water all of the solid Nal dissolves. The solution must be saturated and the K sp for the Nal must be very high. True False Slide 14 / 91 5 The conductivity of a saturated solution of g 2O 3 would be expected to be less than the conductivity of a saturated solution of ao 3. Justify your answer. True False Slide 15 / 91 Solubility The term solubility represents the maximum amount of solute that can be dissolved in a certain volume before any precipitate is observed. The solubility of a substance can be given in terms of Example #1 Slide 16 / 91 Solubility onsider the slightly soluble compound barium oxalate, a 2O 4. The solubility of a 2O 4 is 1.3 x 10-3 mol/l. grams per liter g/l The ratio of cations to anions is 1:1. or in terms of moles per liter mol/l The latter is sometimes referred to as molar solubility. For any slightly soluble salt the molar solubility always refers to the ion with the lower molar ratio. This means that 1.3 x 10-3 moles of a 2+ can dissolve in one liter. lso, 1.3 x 10-3 moles of 2O 4 can dissolve in one liter. What is the maximum amount (in grams) of a 2O 4 that could dissolve in 2.5 L (before a solid precipitate or solid settlement occurs)? Example #1 Slide 17 / 91 Solubility What is the maximum amount (in grams) of a 2O 4 that could dissolve in 2.5 L (before a precipitate occurs)? The solubility of a 2O 4 is 1.3 x 10-3 mol/l. a 2O 4 (s) --> a 2+ (aq) + 2O 4 (aq) 1.3 x 10-3 mol a 2O x 10-3 g 2.5L x = a 2O 4 1 liter 3.25 x 10-3 g x 1 mole = 0.73g a 2O 4 a 2O g 0.73g is the maximum amount of a 2O 4 that could dissolve in 2.5 L before a precipitate forms. Example #2 Slide 18 / 91 Solubility onsider the slightly soluble compound lead chloride, Pbl 2. The solubility of Pbl 2 is mol/l. The ratio of cations to anions is 1:2. This means that moles of Pb 2+ can dissolve in one liter. Twice as much, or 2(0.016) = moles of l - can dissolve in one liter.
4 Example #3 Slide 19 / 91 Solubility onsider the slightly soluble compound silver sulfate, g 2SO 4. The solubility of g 2SO 4 is mol/l. The ratio of cations to anions is 2:1. This means that moles of SO 4 can dissolve in one liter. Slide 20 / 91 Solubility Remember that molar solubility refers to the ion with the lower mole ratio. It does not always refer to the cation, although in most cases it does. ompound a 2O 4 Pbl 2 Molar Solubility of [ation] ompound 1.3 x 10-3 mol 1.3 x 10-3 mol [nion] 1.3 x 10-3 mol mol/l mol/l mol/l Twice as much, or 2(0.015) = moles of g + can dissolve in one liter. g 2SO mol/l mol/l mol/l Slide 21 / 91 Slide 22 / 91 6 If the solubility of barium carbonate, ao 3 is 7.1 x 10-5 M, this means that a maximum of barium ions, a 2+ ions can be dissolved per liter of solution. 7 If the solubility of barium carbonate, ao 3 is 7.1 x 10-5 M, this means that a maximum of carbonate ions, O 3 ions can be dissolved per liter of solution. 7.1 x 10-5 moles half of that 7.1 x 10-5 moles half of that twice as much twice as much one-third as much one-third as much E one-fourth as much E one-fourth as much Slide 23 / 91 Slide 24 / 91 8 If the solubility of g 2rO 4 is 6.5 x 10-5 M, this means that a maximum of silver ions, g +, can be dissolved per liter of solution. E 6.5 x 10-5 moles twice 6.5 x 10-5 moles half 6.5 x 10-5 moles one-fourth 6.5 x 10-5 moles four times 6.5 x 10-5 moles alculating K sp from the Solubility Return to the Table of ontents
5 Slide 25 / 91 alculating K sp from the Solubility Sample Problem The molar solubility of lead (II) bromide, Pbr 2 is 1.0 x 10-2 at 25 o. alculate the solubility product, K sp, for this compound. The molar solubility always refers to the ion of the lower molar ratio, therefore [Pb 2+ ] = 1.0 x 10-2 mol/l and [r - ] = 2.0 x 10-2 mol/l Substitute the molar concentrations into the K sp expression and solve. K sp = [Pb 2+ ][r - ] 2 = (1.0 x 10-2 )(2.0 x 10-2 ) 2 = 4.0 x 10-6 Slide 26 / 91 9 For the slightly soluble salt, os, the molar solubility is 5 x 10-5 M. alculate the K sp for this compound. 5 x x x x E 2.5 x 10-9 Slide 27 / For the slightly soluble salt, af 2, the molar solubility is 3 x 10-4 M. alculate the solubility-product constant for this compound. Slide 28 / For the slightly soluble salt, La(IO 3) 3, the molar solubility is 1 x 10-4 M. alculate K sp. 3 x x x x x x x x E 1.08 x E 1 x Slide 29 / For the slightly soluble compound, a 3(PO 4) 2, the molar solubility is 3 x 10-8 moles per liter. alculate the K sp for this compound. Slide 30 / The concentration of hydroxide ions in a saturated solution of l(oh) 3 is 1.58x What is the K sp of l(oh) 3? 9.00 x x x x E 3.0 x 10-20
6 Slide 31 / 91 Slide 32 / What is the Ksp of Fe(OH) 3(s) if a saturated solution of it has a ph of 11.3? 2.0 x alculating Solubility from the K sp 1.6 x x x 10-8 Return to the Table of ontents E 5.4 x Slide 33 / 91 Slide 34 / 91 alculating Solubility from the K sp alculating Solubility from the K sp Example: What is the molar solubility of a saturated aqueous solution of ao 3? = 5.0 x 10-9 ) ao 3(s) --> a 2+ (aq) + O 3 (aq) Ksp = 5.0 x 10-9 = [a 2+ ][O 3 ] Since neither ion concentration is known, we will substitute "x" for the [a 2+ ] and "x" for the [O 3 ]. 5.0 x 10-9 = (x)(x) = x 2 Example: What is the molar solubility of a saturated aqueous solution of PbI 2? = 1.39 x 10-8 ) PbI 2(s) --> Pb 2+ (aq) + 2I - (aq) Ksp = 1.39 x 10-8 = [Pb 2+ ][I - ] 2 Since neither ion concentration is known, we will substitute "x" for the [Pb 2+ ] and "2x" for the [I - ] x 10-8 = (x)(2x) 2 = 4x 3 "x" = [a 2+ ] = [O 3 ] = 7.07 x 10-5 M Since 1 a 2+ or 1 O 3 are required for 1 ao 3, the molar solubility of the ao 3(s) = 7.07 x 10-5 M. "x" = [Pb 2+ ] = 1.51 x 10-3 M Since 1 Pb 2+ required 1 PbI 2, the molar solubility of the PbI 2(s) = 1.51 x 10-3 M. Slide 35 / alculate the concentration of silver ion when the solubility product constant of gi is 1 x Slide 36 / alculate the molar solubility of PbF 2 that has a K sp at 25 = 3.6 x Students type their answers here 0.5 (1 x ) 2 (1 x ) (1 x ) 2 (1 x )
7 Slide 37 / The K sp of a compound of formula 3 is 1.8 x What is the molar solubility of the compound? Slide 38 / The K sp of a compound of formula 3 is 1.8 x The molar mass is 280g/mol. What is the solubility? Slide 39 / 91 Slide 40 / Which of the following ionic salts would have the highest molar solubility? NiO 3(s) Ksp = 6.61 x 10-9 MnO 3(s) Ksp = 1.82 x Factors ffecting Solubility ZnO 3(s) Ksp = 1.45 x g 2rO 4(s) Ksp = 9.00 x E ll have the same molar solubility Return to the Table of ontents Slide 41 / 91 ommon Ion Effect Slide 42 / 91 ommon Ion Effect onsider a saturate solution of barium sulfate: aso 4(s) a 2+ (aq) + SO 4 (aq) If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease. So adding any soluble salt containing either a 2+ or SO 4 ions will decrease the solubility of barium sulfate. Sample Problem alculate the solubility of af 2 in grams per liter in a) pure water b) a 0.15 M KF solution c) a M a(no 3) 2 solution The solubility product for calcium fluoride, af 2 is 3.9 x 10-11
8 Slide 43 / 91 ommon Ion Effect Slide 44 / 91 ommon Ion Effect a) pure water af 2(s) a 2+ (aq) + 2F - (aq) b) a 0.15 M KF solution Remember KF, a strong electrolyte, is completely ionized and the major source of F- ions. [F - ] =0.15M If we assume x as the dissociation then, a 2+ ions = x and [F - ] = 2x K sp = [a 2+ ] [F - ] 2 = (x)(2x) 2 Note The solubility product for calcium fluoride, af 2 is 3.9 x [ F - ] = 0.15M Ksp = [a 2+ ] [F - ] 2 = (x)(0.15) 2 Note K sp = 3.9 x = 4x 3 Ksp = 3.9 x = x So x = 2.13 x 10-4 mol/l x (78 g/mol af 2) So x = mol/l Solubility is g/l Solubility is = x (78 g/mol af 2) = g/l Slide 45 / 91 Slide 46 / 91 ommon Ion Effect ommon Ion Effect alculate the solubility of af 2 in grams per liter in c) a M a(no 3) 2 solution [a 2+ ] = 0.08M Recall from the ommon-ion Effect that adding a strong electrolyte to a weakly soluble solution with a common ion will decrease the solubility of the weak electrolyte. The solubility product for calcium fluoride,af 2 is 3.9 x af 2 (s) a 2+ (aq) + 2 F - (aq) ompare the solubilities from the previous Sample Problem af 2 (s) a 2+ (aq) + 2 F - (aq) Ksp = [a 2+ ] [F-] 2 = (0.080)(x) 2 af 2 dissolved with: Solubility of af2 Ksp = 3.9 x = 0.080x 2 pure water g/l So x = 2.2 x 10-5 mol/l * (78 g/mol af 2)/ 2 Solubility is g/l M KF 1.35x10-7 g/l M a(no 3) g/l These results support Le hatelier's Principle that increasing a product concentration will shift equilibrium to the left. Slide 47 / 91 Slide 48 / 91 # 20 What is the molar solubility of a saturated solution of g 2rO 4? K sp at 25 is = 1.2 x x x x x 10-7 E 2.2 x What is the molar solubility of a saturated solution of g 2rO 4 in 0.100M K 2rO 4? K sp at 25 is = 1.2 x x x x x 10-7 E 1.7 x 10-6
9 Slide 49 / What is the molar solubility of a saturated solution of g 2rO 4 in 0.200M gl? K sp at 25 is = 1.2 x x x x x 10-7 E 6.7 x 10-6 Slide 50 / 91 hanges in ph The solubility of almost any ionic compound is affected by changes in ph. onsider dissociation equation for magnesium hydroxide: Mg(OH) (s) # Mg 2+ (aq) + 2OH - (aq) What do you expect will happen to the equilibrium if the ph of this system is lowered by adding a strong acid? Will one of the substances in the equilibrium interact with the strong acid? Would the Mg(OH) 2 be more or less soluble? (Think Le hâtelier s Principle.) Slide 51 / 91 hanges in ph hanges in ph can also affect the solubility of salts that contain the conjugate base of a weak acid. onsider the dissociation of the salt calcium fluoride: af 2 (s) a 2+ (aq) + 2F - (aq) Slide 52 / 91 hanges in ph Sample Problem alculate the molar solubility of Mn(OH) 2 in a) in a solution buffered at ph=9.5 b) in a solution buffered at ph=8.0 c) pure water The solubility product for Mn(OH) 2 at 25 is 1.6 x What do you expect will happen to the equilibrium if the ph of this system is lowered by adding a strong acid? Will one of the substances in the equilibrium interact with the strong acid? Would the af 2 be more or less soluble? Slide 53 / 91 In a solution buffered at ph=9.5, the [H + ] = 3.2 x 10-10, the [OH - ] = 3.2x The solubility product for Mn(OH) 2, is 1.6 x [ OH - ] = 3.2 x 10-5 M hanges in ph Sample Problem alculate the molar solubility of Mn(OH) 2 in a) in a solution buffered at ph = x = [Mn 2+ ] [OH - ] 2 = (x)(3.2 x 10-5 ) 2 x = 1.6 x /(3.2 x 10-5 ) 2 = 1.56 x 10-4 mol/l Slide 54 / 91 In a solution buffered at ph=8.0, the [H + ] = 1 x 10-8, the [OH - ] = 1x The solubility product for Mn(OH) 2, is 1.6 x [ OH - ] = 1 x 10-6 M 1.6 x = [Mn 2+ ] [OH - ] 2 = (x)(1 x 10-6 ) 2 x = 1.6 x / (1 x 10-6 ) 2 = So x = 0.16 mol/l hanges in ph Sample Problem alculate the molar solubility of Mn(OH) 2 in b) in a solution buffered at ph = 8.0 Note
10 Slide 55 / 91 In pure water the ph=7.0, the [H + ] = 1 x 10-7, the [OH - ] = 1x The solubility product for Mn(OH) 2, is 1.6 x [ OH - ] = 1 x 10-7 M 1.6 x = [Mn 2+ ] [OH - ] 2 = (x)(1 x 10-7 ) 2 x = 1.6 x / (1 x 10-7 ) 2 = So x = 16 mol/liter hanges in ph Sample Problem alculate the molar solubility of Mn(OH) 2 in c) in pure water Note Slide 56 / 91 hanges in ph If a substance has a basic anion, it will be more soluble in an acidic solution. If a substance has an acidic cation, it will be more soluble in basic solutions. We will discuss in a little while the affect of ph changes on substances that are amphoteric. o you remember what it means when a substance is amphoteric? Slide 57 / Given the system at equilibrium gl (s) g + (aq) + l - (aq) When 0.01 M Hl is added to the sytem, the point of equilibrium will shift to the. Slide 58 / Which of the following substances are more soluble in acidic solution than in basic solution? Select all that apply. Pbl 2 right and the concentration of g + will decrese right and the concentration of g + will increase left and the concentration of g + will decrease left and the concentration of g + will increase ao 3 gi Fe(OH) 3 E MgF 2 Slide 59 / What is the solubility of Zn(OH) 2 in a solution that is buffered at ph = 8.5? K sp = 3.0 x Students type their answers here Slide 60 / Will the solubility of Zn(OH) 2 in a solution that is buffered at ph = 11.0 be greater than in a solution buffered at 8.5? Explain. Yes No
11 Slide 61 / The molar solubility of NH 4l increases as ph. Slide 62 / The molar solubility of Na 2O 3 increases as ph. increases decreases is unaffected by changes in ph increases decreases is unaffected by changes in ph Slide 63 / 91 omplex Ions Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent. The formation of complex ions particularly with transitional metals can dramatically affect the solubility of a metal salt. For example, the addition of excess ammonia to gl will cause the gl to dissolve. This process is the sum of two reactions resulting in: gl (s) + 2NH 3(aq) g(nh 3) 2 + (aq) + l - (aq) dded NH 3 reacts with g + forming g(nh 3) 2+. dding enough NH 3 results in the complete dissolution of gl. Slide 64 / 91 mphoterism Some metal oxides and hydroxides are soluble in strongly acidic and in strongly basic solutions because they can act either as acids or bases. These substances are said to be amphotheric. Examples of such these substances are oxides and hydroxides of l 3+, Zn 2+, and Sn 2+. They dissolve in acidic solutions because their anion is protonated by the added H + and is pulled from solution causing a shift in the equilibrium to the right. For example: l(oh) 3(s) # l 3+ (aq) + 3 OH - (aq) Slide 65 / 91 mphoterism # Slide 66 / Which of the following factors affect solubility? However these oxides and hydroxides also dissolve in strongly basic solutions. This is because they form complex ions containing several typically four hydroxides bound to the metal ion. luminum hydroxide reacts with OH - to form a complex ion in the following reaction: l(oh) 3(s) + OH - (aq) # l(oh) 4 - (aq) s a result of the formation of the complex ion, l(oh) 4 -, aluminum hydroxide is more soluble. Many metal hydroxides only react in strongly acidic solutions. a(oh) 2, Fe(OH) 2 and Fe(OH) 3 are only more soluble in acidic solution they are not amphoteric. E ph Formation of omplex Ions ommon-ion Effect and,, and
12 Slide 67 / 91 Slide 68 / 91 Precipitation Reactions and Separation of Ions Precipitation Reactions and Separation of Ions Return to the Table of ontents o you remember the solubility rules? They were useful before when we were trying to qualitatively determine if a given reaction would produce a precipitate. They will be useful now for the same reason however now we are going to add a quantitative component that we will discuss soon. In general, soluble salts were: ny salt made with a Group I metal is soluble. ll salts containing nitrate ion are soluble. ll salts containing ammonium ion are soluble. o you remember what metal cations tended to be insoluble? g +, Pb 2+, and Hg 2+ Slide 69 / What is the name of the solid precipitate that is formed when a solution of sodium chloride is mixed with a solution of silver nitrate? Slide 70 / What is the name of the solid precipitate that is formed when a solution of potassium carbonate is mixed with a solution of calcium bromide? sodium silver potassium bromide sodium nitrate calcium carbonate chloride nitrate potassium calcium silver chloride carbonate bromide E Not enough information E Not enough information Slide 71 / 91 Slide 72 / What is the name of the solid precipitate that is formed when a solution of lead (IV) nitrate is mixed with a solution of magnesium sulfate? 33 The K sp for Zn(OH) 2 is 5.0 x Will a precipitate form in a solution whose solubility is 8.0x10-2 mol/l Zn(OH) 2? PbSO 4 Pb(SO 4 ) 2 Pb 2 SO 4 Mg(NO 3 ) 2 E Not enough information E yes, because Q sp < K sp yes, because Q sp > K sp no, because Q sp = K sp no, because Q sp < K sp no, because Q sp > K sp
13 Slide 73 / 91 Slide 74 / The K sp for zinc carbonate is 1 x If equivalent amounts 0.2M sodium carbonate and 0.1M zinc nitrate are mixed, what happens? zinc carbonate precipitate forms, since Q>K. zinc carbonate precipitate forms, since Q<K. Separation of Ions When metals are found in natural they are usually found as metal ores. The metal contained in these ores are in the form of insoluble salts. To make extraction even more difficult the ores often contain several metal salts. In order to separate out the metals, one can use differences in solubilities of salts to separate ions in a mixture. sodium nitrate precipitate forms, since Q>K. No precipitate forms, since Q=K. Slide 75 / 91 Separation of Ions Imagine, you have a test tube that contains g +, Pb 2+ and u 2+ ions and you want to selectively remove each ion and place them into separate test tubes. What reagent could you add to the test tube that will form a precipitate with one or move of the cations and leave the others in solution? You can use your knowledge of the solubility rules or Ksp values for various metal salts to help you accomplish this goal. Slide 76 / 91 Separation of Ions You should remember that g + and Pb 2+ readily form insoluble salts and that u 2+ does not form insoluble salts as readily. Looking at some solubility product values, you will find the following: Salt Ksp g 2S 6 x PbS 3 x us 6 x gl 1.8 x Pbl x 10-5 You will notice that ul 2 is not to be found. This means ul 2 is a soluble salt! Slide 77 / 91 Separation of Ions dding l - should precipitate the g + and Pb 2+ ions but not the u 2+ ions. We can remove g + and Pb 2+ from the test tube. Now, how can we separate the g + and Pb 2+ ions? Salt Ksp g 2S 6 x PbS 3 x us 6 x gl 1.8 x Pbl x 10-5 o you notice the significant difference between the K sp values for g 2S and PbS? Maybe we can precipitate one of the salts out before the other if we control the concentration of S added. Which salt g 2S and PbS should precipitate first when we begin to add S? Slide 78 / 91 Separation of Ions If we have 0.100M concentrations of g + and Pb 2+ and we begin to add 0.200M K 2S the g 2S should precipitate first. For g 2S: K sp = 6 x = [g + ] 2 [S ] = (0.100) 2 (x) x = [S ] = 6 x M. If this concentration of S is added g 2S will precipitate. For PbS: Ksp = 3 x = [Pb 2+ ][S ] = 0.100(x) x = [S ] = 3 x M. greater amount of S is needed to precipitate the PbS. Therefore, g 2S will precipitate first.
14 Slide 79 / 91 Separation of Ions Problems Most of the problems in this section ask you whether a precipitate will form after mixing certain solutions together. General Problem-Solving Strategy Step 1 - etermine which of the products is the precipitate. Write the K sp expression for this compound. Step 2 - alculate the cation concentration of this slightly soluble compound. Step 3 - alculate the anion concentration of this slightly soluble compound. Step 4 - Substitute the values into the reaction quotient (Q) expression. Recall that this is the same expression as K. Step 5 - ompare Q to K to determine whether a precipitate will form. Sample Problem Slide 81 / 91 Separation of Ions Problems Will a precipitate form if you mix 50.0 ml of 0.20 M barium chloride, al 2, and 50.0 ml of 0.30 M sodium sulfate, Na 2SO 4? Step 1 - etermine which of the products is the precipitate. Write the K sp expression for this compound. aso 4 (s) a 2+ (aq) + SO 4 (aq) Step 2 - alculate the cation concentration of this slightly soluble compound. M 1V 1 =M 2V 2 M 2 = (M 1V 1) / V 2 M 2= (0.20M*50.0mL) / 100 ml M 2 = 0.10 M al 2 [a 2+ ] = 0.10 M Slide 80 / 91 Separation of Ions Problems In order for a precipitate to form the equilibrium that exists between the solution and the insoluble salt must reside on the left. We can determine to which side the equilibrium will shift using Q, the Reaction Quotient. If Q = Ksp If Q > Ksp If Q < Ksp then you have an exactly perfect saturated solution with not one speck of undissolved solid. then YES you will observe a precipitate; the number of cations and anions exceeds the solubility then NO precipitate will form; there are so few cations and anions that they all remain dissolved In a solution, If Q = K sp, the system is at equilibrium and the solution is saturated. If Q > K sp, the salt will precipitate until Q = K sp. If Q < K sp, more solid can dissolve until Q = K sp. Slide 82 / 91 Separation of Ions Problems Sample Problem - nswers (con't) Will a precipitate form if you mix 50.0 ml of 0.20 M barium chloride, al 2, and 50.0 ml of 0.30 M sodium sulfate, Na 2SO 4? Step 3 - alculate the anion concentration of this slightly soluble compound. M 1V 1 =M 2V 2 M 2 = (M 1V 1) / V 2 M 2= (0.30M*50.0mL) / 100 ml M 2 = 0.15 M Na 2SO 4 [SO 4 ] = 0.15 M Step 4 - Substitute the values into the reaction quotient (Q) expression. Recall that this is the same expression as K. Q = [a 2+ ] [SO 4 ] = (0.10) (0.15) = Slide 83 / 91 Separation of Ions Problems Sample Problem - nswers (con't) Will a precipitate form if you mix 50.0 ml of 0.20 M barium chloride, al 2, and 50.0 ml of 0.30 M sodium sulfate, Na 2SO 4? Step 5 - ompare Q to K to determine whether a precipitate will form. The K sp for barium sulfate is 1 x Therefore, since Q > K, there will be a precipitate formed when you mix equal amounts of 0.20 M al 2, and 0.30 M Na 2SO 4. Slide 84 / 91 Separation of Ions Problems In summary, to selectively precipitate metal ions from a solution that contains a number of metal ions you should use the solubility rules and K sp values to determine an experimental strategy. The solubility rules may lead you to the identity of an anion that will result in separation of certain metal ions however, at other times the quantity of the added anion will be instrumental in the separation given that metal salts have different degrees of solubility as seen in their K sp values.
15 Slide 85 / solution contains 2.0 x 10-5 M barium ions and 1.8 x 10-4 M lead (II) ions. If Na 2rO 4 is added, which will precipitate first from solution? The K sp for aro 4 is 2.1 x and the K sp for PbrO 4 is 2.8 x aro 4 PbrO 4 They will precipitate at the same time. It's impossible to determine with the information provided. Slide 86 / solution contains 2.0 x 10-4 M g + and 2.0 x 10-4 M Pb 2+. If Nal is added, will gl (K sp = 1.8 x ) or PbI 2 (K sp = 1.7 x 10-5 ) precipitate first? gl Pbl 2 They will precipitate at the same time. It is impossible to determine with the information provided. Slide 87 / solution contains 2.0 x 10-4 M g + and 1.7 x 10-3 M Pb 2+. If Nal is added. What concentration of l - is needed to Students type their answers here begin precipitation. gl (K sp = 1.8 x ) and PbI 2 (K sp = 1.7 x 10-5 ) Slide 88 / solution contains 2.0 x 10-4 M g + and 1.7 x 10-3 M Pb 2+. If Nal is added. What will be the concentration of the Students type their answers here first ion to precipitate when the second ion begins to precipitate? gl (Ksp = 1.8 x ) and PbI 2 (Ksp = 1.7 x 10-5 ) Slide 89 / Will o(oh) 2 precipitate from solution if the ph of a 0.002M solution of o(no 3) 2 is adjusted to 8.4? K sp for o(oh) 2 is 2.5 x Yes No Slide 90 / Will a precipitate form if you mix 25.0 ml of M calcium chloride, and 50.0 ml of M lithium chromate? The K sp for calcium chromate is 4.5 x Students type their answers here
16 Slide 91 / 91
AP Chemistry. Introduction to Solubility Equilibria. Slide 1 / 91 Slide 2 / 91. Slide 3 / 91. Slide 4 / 91. Slide 5 / 91.
Slide 1 / 91 Slide 2 / 91 P hemistry queous quilibria II: Ksp & Solubility Products Slide 3 / 91 Slide 4 / 91 Table of ontents: K sp & Solubility Products Introduction to Solubility quilibria alculating
More informationSolubility Products. Solubility Products. Solubility Products. Solubility Products. Slide 2 / 57. Slide 1 / 57. Slide 3 / 57.
Slide 1 / 57 Slide 2 / 57 Products queous equilibria II Products onsider the equilibrium that exists in a saturated solution of aso 4 in water: aso 4 (s) a 2+ (aq) + SO 4 2- (aq) Slide 3 / 57 Products
More informationAP Chemistry Table of Contents: Ksp & Solubility Products Click on the topic to go to that section
Slide 1 / 91 Slide 2 / 91 AP Chemistry Aqueous Equilibria II: Ksp & Solubility Products Table of Contents: K sp & Solubility Products Slide 3 / 91 Click on the topic to go to that section Introduction
More informationSolubility and Complex-ion Equilibria
Solubility and Complex-ion Equilibria Contents and Concepts Solubility Equilibria 1. The Solubility Product Constant 2. Solubility and the Common-Ion Effect 3. Precipitation Calculations 4. Effect of ph
More informationChapter 19. Solubility and Simultaneous Equilibria p
Chapter 19 Solubility and Simultaneous Equilibria p. 832 857 Solubility Product ) The product of molar concentrations of the constituent ions, each raised ot the power of its stoichiometric coefficients
More informationSolubility and Complex-ion Equilibria
Solubility and Complex-ion Equilibria Solubility Equilibria Many natural processes depend on the precipitation or dissolving of a slightly soluble salt. In the next section, we look at the equilibria of
More informationUnit 3: Solubility Equilibrium
Unit 3: Chem 11 Review Preparation for Chem 11 Review Preparation for It is expected that the student understands the concept of: 1. Strong electrolytes, 2. Weak electrolytes and 3. Nonelectrolytes. CHEM
More informationUnit 3: Solubility Equilibrium
Unit 3: Chem 11 Review Preparation for Chem 11 Review Preparation for It is expected that the student understands the concept of: 1. Strong electrolytes, 2. Weak electrolytes and 3. Nonelectrolytes. CHEM
More informationSolubility Equilibria
Solubility Equilibria Heretofore, we have investigated gas pressure, solution, acidbase equilibriums. Another important equilibrium that is used in the chemistry lab is that of solubility equilibrium.
More informationChapter 17. Additional Aspects of Equilibrium
Chapter 17. Additional Aspects of Equilibrium Sample Exercise 17.1 (p. 726) What is the ph of a 0.30 M solution of acetic acid? Be sure to use a RICE table, even though you may not need it. (2.63) What
More informationSaturated vs. Unsaturated
Solubility Equilibria in Aqueous Systems K sp (Equilibria of Slightly Soluble Salts, Ionic Compounds) Factors that Affect Solubility (Common Ion Effect, AcidBase Chemistry) Applications of Ionic Equilibria
More information1. Forming a Precipitate 2. Solubility Product Constant (One Source of Ions)
Chemistry 12 Solubility Equilibrium II Name: Date: Block: 1. Forming a Precipitate 2. Solubility Product Constant (One Source of Ions) Forming a Precipitate Example: A solution may contain the ions Ca
More informationModified Dr. Cheng-Yu Lai
Ch16 Aqueous Ionic Equilibrium Solubility and Complex Ion Equilibria Lead (II) iodide precipitates when potassium iodide is mixed with lead (II) nitrate Modified Dr. Cheng-Yu Lai Solubility-product constant
More informationChapter 17: Additional Aspects of Aqueous equilibria. Common-ion effect
Chapter 17: Additional Aspects of Aqueous equilibria Learning goals and key skills: Describe the common ion effect. Explain how a buffer functions. Calculate the ph of a buffer solution. Calculate the
More informationChapter 15 Additional Aspects of
Chemistry, The Central Science Chapter 15 Additional Aspects of Buffers: Solution that resists change in ph when a small amount of acid or base is added or when the solution is diluted. A buffer solution
More informationChapter 17. Additional Aspects of Aqueous Equilibria. Lecture Presentation. James F. Kirby Quinnipiac University Hamden, CT
Lecture Presentation Chapter 17 Additional Aspects of James F. Kirby Quinnipiac University Hamden, CT Effect of Acetate on the Acetic Acid Equilibrium Acetic acid is a weak acid: CH 3 COOH(aq) H + (aq)
More informationChapter 17. Additional Aspects of Equilibrium
Chapter 17. Additional Aspects of Equilibrium 17.1 The Common Ion Effect The dissociation of a weak electrolyte is decreased by the addition of a strong electrolyte that has an ion in common with the weak
More informationAP Chemistry. Slide 1 / 39. Slide 2 / 39. Slide 3 / 39. Equilibrium Part C : Solubility Equilibrium. Table of Contents
Slide 1 / 39 AP Chemistry Slide 2 / 39 Equilibrium Part C : Solubility Equilibrium 2014-10-29 www.njctl.org Table of Contents click on the topic to go to that section Slide 3 / 39 Molar Solubility Calculating
More informationChapter 17. Additional Aspects of Aqueous Equilibria 蘇正寬 Pearson Education, Inc.
Chapter 17 Additional Aspects of Aqueous Equilibria 蘇正寬 chengkuan@mail.ntou.edu.tw Additional Aspects of Aqueous Equilibria 17.1 The Common-Ion Effect 17.2 Buffers 17.3 Acid Base Titrations 17.4 Solubility
More informationName AP CHEM / / Chapter 15 Outline Applications of Aqueous Equilibria
Name AP CHEM / / Chapter 15 Outline Applications of Aqueous Equilibria Solutions of Acids or Bases Containing a Common Ion A common ion often refers to an ion that is added by two or more species. For
More informationSOLUBILITY REVIEW QUESTIONS
Solubility Problem Set 1 SOLUBILITY REVIEW QUESTIONS 1. What is the solubility of calcium sulphate in M, g/l, and g/100 ml? 2. What is the solubility of silver chromate? In a saturated solution of silver
More informationLearning Objectives. Solubility and Complex-ion Equilibria. Contents and Concepts. 3. Precipitation Calculations. 4. Effect of ph on Solubility
Solubility and Comple-ion Equilibria. Solubility and the Common-Ion Effect a. Eplain how the solubility of a salt is affected by another salt that has the same cation or anion. (common ion) b. Calculate
More informationTry this one Calculate the ph of a solution containing M nitrous acid (Ka = 4.5 E -4) and 0.10 M potassium nitrite.
Chapter 17 Applying equilibrium 17.1 The Common Ion Effect When the salt with the anion of a is added to that acid, it reverses the dissociation of the acid. Lowers the of the acid. The same principle
More informationChap 17 Additional Aspects of Aqueous Equilibria. Hsu Fu Yin
Chap 17 Additional Aspects of Aqueous Equilibria Hsu Fu Yin 1 17.1 The Common-Ion Effect Acetic acid is a weak acid: CH 3 COOH(aq) H + (aq) + CH 3 COO (aq) Sodium acetate is a strong electrolyte: NaCH
More informationSOLUBILITY EQUILIBRIA (THE SOLUBILITY PRODUCT)
SOLUBILITY EQUILIBRIA (THE SOLUBILITY PRODUCT) Saturated solutions of salts are another type of chemical equilibria. Slightly soluble salts establish a dynamic equilibrium with the hydrated cations and
More informationChapter 17 Additional Aspects of
Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 17 Additional Aspects of John D. Bookstaver St. Charles Community College Cottleville,
More informationAqueous Equilibria: Part II- Solubility Product
Aqueous Equilibria: Part II- Solubility Product PSI AP Chemistry Name-------------------------- I Solubility Product, K sp MC #63-103 a) Writing K sp expression b) Solving for K sp c) Solving for (molar)
More informationOperational Skills. Operational Skills. The Common Ion Effect. A Problem To Consider. A Problem To Consider APPLICATIONS OF AQUEOUS EQUILIBRIA
APPLICATIONS OF AQUEOUS EQUILIBRIA Operational Skills Calculating the common-ion effect on acid ionization Calculating the ph of a buffer from given volumes of solution Calculating the ph of a solution
More informationAqueous Equilibria, Part 2 AP Chemistry Lecture Outline
Aqueous Equilibria, Part 2 AP Chemistry Lecture Outline Name: The Common-Ion Effect Suppose we have a weak acid and a soluble salt of that acid. CH 3 COOH NaCH 3 COO CH 3 COOH CH 3 COO + H + Since NaCH
More informationSolubility Equilibria. Dissolving a salt... Chem 30S Review Solubility Rules. Solubility Equilibrium: Dissociation = Crystalization
Chem 30S Review Solubility Rules Solubility Equilibria Salts are generally more soluble in HOT water(gases are more soluble in COLD water) Alkali Metal salts are very soluble in water. NaCl, KOH, Li 3
More informationChapter 17. Additional Aspects of Equilibrium
Chapter 17. Additional Aspects of Equilibrium 17.1 The Common Ion Effect The dissociation of a weak electrolyte is decreased by the addition of a strong electrolyte that has an ion in common with the weak
More informationAcid-Base Equilibria and Solubility Equilibria
Acid-Base Equilibria and Solubility Equilibria Acid-Base Equilibria and Solubility Equilibria Homogeneous versus Heterogeneous Solution Equilibria (17.1) Buffer Solutions (17.2) A Closer Look at Acid-Base
More informationSolubility Equilibrium
2016 Ksp note.notebook Solubility Equilibrium Learning Goals: to understand what happens when a compound dissolves in water to calculate the extent of dissolution...the molar solubility to calculate the
More informationChapter 3: Solution Chemistry (For best results when printing these notes, use the pdf version of this file)
Chapter 3: Solution Chemistry (For best results when printing these notes, use the pdf version of this file) Section 3.1: Solubility Rules (For Ionic Compounds in Water) Section 3.1.1: Introduction Solubility
More informationBuffers/Titration Aqueous Equilibria - I
Slide 1 / 113 Slide 2 / 113 uffers/titration queous Equilibria - I Review hydrolysis of salts Slide 3 / 113 1 The hydrolysis of a salt of a weak base and a strong acid should give a solution that is. Slide
More informationChapter 4. Aqueous Reactions and Solution Stoichiometry
Sample Exercise 4.1 (p. 127) The diagram below represents an aqueous solution of one of the following compounds: MgCl 2, KCl, or K 2 SO 4. Which solution does it best represent? Practice Exercise 1 (4.1)
More informationChemistry 12 Solubility Equilibrium I. Name: Date: Block: 1. Solutions Vocab & Calculations 2. Predicting Solubility 3.
Chemistry 12 Solubility Equilibrium I Name: Date: Block: 1. Solutions Vocab & Calculations 2. Predicting Solubility 3. Writing Equations Solutions Vocab & Calculations What is a solution? A homogenous
More informationLecture #12 Complex Ions and Solubility
Lecture #12 Complex Ions and Solubility Stepwise exchange of NH 3 for H 2 O in M(H 2 O) 4 2+ M(H 2 O) 2 (NH 3 ) 2 2+ M(H 2 O) 4 2+ M(NH 3 ) 4 2+ M(H 2 O) 3 (NH 3 ) 2+ M(H 2 O)(NH 3 ) 3 2+ Formation Constants
More informationChapter 4: Types of Chemical Reactions and Solution Stoichiometry
Chapter 4: Types of Chemical Reactions and Solution Stoichiometry 4.1 Water, the Common Solvent 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes 4.3 The Composition of Solutions (MOLARITY!)
More information22. What is the maximum concentration of carbonate ions that will precipitate BaCO 3 but not MgCO 3 from a solution that is 2.
PX312-1718 1. What is the solubility product expression for Th(IO 3 ) 4? A) K sp = [Th 4+ ][4IO 3 ] 4 B) K sp = [Th 4+ ][IO 3 ] C) K sp = [Th][IO 3 ] 4 D) K sp = [Th 4+ ][IO 3 ] 4 E) K sp = [Th 4+ ][IO
More informationSolution Stoichiometry
Chapter 8 Solution Stoichiometry Note to teacher: You will notice that there are two different formats for the Sample Problems in the student textbook. Where appropriate, the Sample Problem contains the
More informationCHEM J-6 June 2014
CHEM1102 2014-J-6 June 2014 A solution is prepared that contains sodium chloride and sodium chromate (both 0.10 M). When a concentrated solution of silver nitrate is added slowly, white AgCl(s) begins
More informationAP Chemistry. CHAPTER 17- Buffers and Ksp 17.1 The Common Ion Effect Buffered Solutions. Composition and Action of Buffered Solutions
AP Chemistry CHAPTER 17- Buffers and Ksp 17.1 The Common Ion Effect The dissociation of a weak electrolyte is decreased by the addition of a strong electrolyte that has an ion in common with the weak electrolyte.
More informationEQUIVALENCE POINT. 8.8 millimoles is also the amount of acid left, and the added base gets converted to acetate ion!
184 Another interesting point: The halfway point phenolphthalein color change buffer region EQUIVALENCE POINT What's special about it? It's the point where we have added half the required base to reach
More informationIonic Equilibria in Aqueous Systems. Dr.ssa Rossana Galassi
Ionic Equilibria in Aqueous Systems Dr.ssa Rossana Galassi 320 4381420 rossana.galassi@unicam.it Ionic Equilibria in Aqueous Systems 19.1 Equilibria of Acid-Base Buffer Systems 19.2 Acid-Base Titration
More information5. Pb(IO 3) BaCO 3 8. (NH 4) 2SO 3
Chemistry 11 Solution Chemistry II Name: Date: Block: 1. Ions in Solutions 2. Solubility Table 3. Separating Ions Ions in Solutions Ionization Equation - Represents the salt breaking apart into ions. Practice:
More information... so we need to find out the NEW concentrations of each species in the system.
171 Take 100. ml of the previous buffer (0.050 M tris / 0.075 M tris-hcl), and add 5.0 ml of 0.10 M HCl. What is the ph of the mixture? The HCl should react with basic component of the buffer (tris), and
More informationCHAPTER 4 TYPES OF CHEMICAL REACTIONS & SOLUTION STOICHIOMETRY
Advanced Chemistry Name Hour Advanced Chemistry Approximate Timeline Students are expected to keep up with class work when absent. CHAPTER 4 TYPES OF CHEMICAL REACTIONS & SOLUTION STOICHIOMETRY Day Plans
More informationUnit 6 ~ Learning Guide Name:
Unit 6 ~ Learning Guide Name: Instructions: Using a pencil, complete the following notes as you work through the related lessons. Show ALL work as is explained in the lessons. You are required to have
More information2/4/2016. Chapter 15. Chemistry: Atoms First Julia Burdge & Jason Overby. Acid-Base Equilibria and Solubility Equilibria The Common Ion Effect
Chemistry: Atoms First Julia Burdge & Jason Overby 17 Acid-Base Equilibria and Solubility Equilibria Chapter 15 Acid-Base Equilibria and Solubility Equilibria Kent L. McCorkle Cosumnes River College Sacramento,
More informationChapter 19 Solubility and Complex Ion Equilibria
Chapter 19 Solubility and Complex Ion Equilibria "if you are not part of the solution, then you are part of the precipitate" - all solutions of salts exist as a balance between the dissolved cations and
More informationReview 7: Solubility Equilibria
Review 7: Solubility Equilibria Objectives: 1. Be able to write dissociation equations for ionic compounds dissolving in water. 2. Given Ksp, be able to determine the solubility of a substance in both
More informationCH 4 AP. Reactions in Aqueous Solutions
CH 4 AP Reactions in Aqueous Solutions Water Aqueous means dissolved in H 2 O Moderates the Earth s temperature because of high specific heat H-bonds cause strong cohesive and adhesive properties Polar,
More informationThe solvent is the dissolving agent -- i.e., the most abundant component of the solution
SOLUTIONS Definitions A solution is a system in which one or more substances are homogeneously mixed or dissolved in another substance homogeneous mixture -- uniform appearance -- similar properties throughout
More informationThe solubility of insoluble substances can be decreased by the presence of a common ion. AgCl will be our example.
COMMON ION EFFECT WORKED PROBLEMS The solubility of insoluble substances can be decreased by the presence of a common ion. AgCl will be our example. Present in silver chloride are silver ions (Ag + ) and
More informationMay 09, Ksp.notebook. Ksp = [Li + ] [F + ] Find the Ksp for the above reaction.
example: Constant Product K sp Solubility Product Constant Some compounds dissolve in water Some compounds dissolve better than others The more that a compound can dissolve, the more soluble the compound
More informationChapter 16. The Danger of Antifreeze. Buffers. Aqueous Equilibrium
hapter 16 Aqueous Equilibrium The Danger of Antifreeze Each year, thousands of pets and wildlife die from consuming antifreeze Most brands of antifreeze contain ethylene glycol sweet taste and initial
More informationAP Chemistry Unit #4. Types of Chemical Reactions & Solution Stoichiometry
AP Chemistry Unit #4 Chapter 4 Zumdahl & Zumdahl Types of Chemical Reactions & Solution Stoichiometry Students should be able to: Predict to some extent whether a substance will be a strong electrolyte,
More informationQuick Review. - Chemical equations - Types of chemical reactions - Balancing chemical equations - Stoichiometry - Limiting reactant/reagent
Quick Review - Chemical equations - Types of chemical reactions - Balancing chemical equations - Stoichiometry - Limiting reactant/reagent Water H 2 O Is water an ionic or a covalent compound? Covalent,
More informationSolubility Equilibrium When a substance dissolves an equilibrium results between the precipitate and the dissolved ions. The solution becomes
Solubility Equilibrium When a substance dissolves an equilibrium results between the precipitate and the dissolved ions. The solution becomes saturated. The particles dissolving equals the particles precipitating.
More informationAdditional Aspects of Aqueous Equilibria David A. Katz Department of Chemistry Pima Community College
Additional Aspects of Aqueous Equilibria David A. Katz Department of Chemistry Pima Community College The Common Ion Effect Consider a solution of acetic acid: HC 2 H 3 O 2(aq) + H 2 O (l) H 3 O + (aq)
More informationUnit 4a: Solution Stoichiometry Last revised: October 19, 2011 If you are not part of the solution you are the precipitate.
1 Unit 4a: Solution Stoichiometry Last revised: October 19, 2011 If you are not part of the solution you are the precipitate. You should be able to: Vocabulary of water solubility Differentiate between
More informationChapter 17: Solubility Equilibria
Previous Chapter Table of Contents Next Chapter Chapter 17: Solubility Equilibria Sections 17.1-17.2: Solubility Equilibria and the K sp Table In this chapter, we consider the equilibrium associated with
More informationChapter 16. Solubility and Complex Ion Equilibria
Chapter 16 Solubility and Complex Ion Equilibria Section 16.1 Solubility Equilibria and the Solubility Product Solubility Equilibria Solubility product (K sp ) equilibrium constant; has only one value
More informationUNIT III: SOLUBILITY EQUILIBRIUM YEAR END REVIEW (Chemistry 12)
I. Multiple Choice UNIT III: SOLUBILITY EQUILIBRIUM YEAR END REVIEW (Chemistry 12) 1) Which one of the following would form an ionic solution when dissolved in water? A. I 2 C. Ca(NO 3 ) 2 B. CH 3 OH D.
More informationEquilibri acido-base ed equilibri di solubilità. Capitolo 16
Equilibri acido-base ed equilibri di solubilità Capitolo 16 The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance.
More informationAP Chemistry Honors Unit Chemistry #4 2 Unit 3. Types of Chemical Reactions & Solution Stoichiometry
HO AP Chemistry Honors Unit Chemistry #4 2 Unit 3 Chapter 4 Zumdahl & Zumdahl Types of Chemical Reactions & Solution Stoichiometry Students should be able to:! Predict to some extent whether a substance
More information2nd Semester Exam Review. C. K eq = [N 2][H 2 ]
Name: ate: 1. Which pair of formulas represents the empirical formula and the molecular formula of a compound?. H 2 O, 4 H 6 O 4. HO, 6 H 12 O 6 8. Given the reaction at equilibrium: N 2 (g) + 3H 2 (g)
More information173 Buffer calculation: Tris buffer - Tris(hydroxymethyl)-aminomethane. tris base
173 Buffer calculation: Tris buffer - Tris(hydroxymethyl)-aminomethane tris base tris-hcl (conjugate acid of tris base) Calculate the ph of a buffer made from 50 ml of 0.10M tris and 50 ml of 0.15M tris-hcl.
More informationWe CAN have molecular solutions (ex. sugar in water) but we will be only working with ionic solutions for this unit.
Solubility Equilibrium The Basics (should be mostly review) Solubility is defined as the maximum amount of a substance which can be dissolved in a given solute at a given temperature. The solubility of
More informationSolubility Rules and Net Ionic Equations
Solubility Rules and Net Ionic Equations Why? Solubility of a salt depends upon the type of ions in the salt. Some salts are soluble in water and others are not. When two soluble salts are mixed together
More informationChapter 16: Applications of Aqueous Equilibrium Part 3. Solubilities of Ionic Compounds and K sp
Chapter 16: Applications of Aqueous Equilibrium Part 3 Solubilities of Ionic Compounds and K sp You ve already learned that not all ionic compounds are water soluble. You memorized the solubility rules
More informationNCEA Chemistry 2.2 Identify Ions AS 91162
NCEA Chemistry 2.2 Identify Ions AS 91162 What is this NCEA Achievement Standard? When a student achieves a standard, they gain a number of credits. Students must achieve a certain number of credits to
More informationAcid - Base Equilibria 3
Acid - Base Equilibria 3 Reading: Ch 15 sections 8 9 Ch 16 sections 1 7 * = important homework question Homework: Chapter 15: 97, 103, 107, Chapter 16: 29*, 33*, 35, 37*, 39*, 41, 43*, 49, 55, 57, 61,
More informationChapter 4 Three Major Classes of Chemical Reactions
Chapter 4 Three Major Classes of Chemical Reactions Solution Stoichiometry Many reactions (biochemical, marine, etc.) take place in solution. We need to be able to express the number of moles of particles
More informationChapter 15 - Applications of Aqueous Equilibria
Neutralization: Strong Acid-Strong Base Chapter 15 - Applications of Aqueous Equilibria Molecular: HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) SA-SB rxn goes to completion (one-way ) Write ionic and net ionic
More informationIonic Equilibria in Aqueous Systems
Ionic Equilibria in Aqueous Systems Chapter Nineteen AP Chemistry There are buffers in our blood that keep the ph of our blood at a constant level. The foods that we eat are often acidic or basic. This
More informationPart A Answer all questions in this part.
Part A Directions (1-24): For each statement or question, record on your separate answer sheet the number of the word or expression that, of those given, best completes the statement or answers the question.
More information5. [7 points] What is the mass of gallons (a fifth) of pure ethanol (density = g/cm 3 )? [1 gallon = Liters]
1 of 6 10/20/2009 3:55 AM Avogadro s Number, N A = 6.022 10 23 1. [7 points] Given the following mathematical expression: (15.11115.0)/(2.154 10 3 ) How many significant figures should the answer contain?
More informationSolubility and Complex Ion. Equilibria
Solubility and Complex Ion a mineral formed by marine organisms through biological precipitation CALCITE Equilibria CaCO 3(s) Ca 2+ (aq) + CO 3 2- (aq) K = K sp = [Ca 2+ ][CO 3 2- ] = 2.8 x 10-9 K sp =
More informationAcid-Base Equilibria and Solubility Equilibria Chapter 17
PowerPoint Lecture Presentation by J. David Robertson University of Missouri Acid-Base Equilibria and Solubility Equilibria Chapter 17 The common ion effect is the shift in equilibrium caused by the addition
More informationDuring photosynthesis, plants convert carbon dioxide and water into glucose (C 6 H 12 O 6 ) according to the reaction:
Example 4.1 Stoichiometry During photosynthesis, plants convert carbon dioxide and water into glucose (C 6 H 12 O 6 ) according to the reaction: Suppose that a particular plant consumes 37.8 g of CO 2
More information1. Hydrochloric acid is mixed with aqueous sodium bicarbonate Molecular Equation
NAME Hr Chapter 4 Aqueous Reactions and Solution Chemistry Practice A (Part 1 = Obj. 1-3) (Part 2 = Obj. 4-6) Objective 1: Electrolytes, Acids, and Bases a. Indicate whether each of the following is strong,
More informationinsoluble partial very soluble (< 0.1 g/100ml) solubility (> 1 g/100ml) Factors Affecting Solubility in Water
Aqueous Solutions Solubility is a relative term since all solutes will have some solubility in water. Insoluble substances simply have extremely low solubility. The solubility rules are a general set of
More informationChapter 4; Reactions in Aqueous Solutions. Chapter 4; Reactions in Aqueous Solutions. V. Molarity VI. Acid-Base Titrations VII. Dilution of Solutions
Chapter 4; Reactions in Aqueous Solutions I. Electrolytes vs. NonElectrolytes II. Precipitation Reaction a) Solubility Rules III. Reactions of Acids a) Neutralization b) Acid and Carbonate c) Acid and
More informationDepartment of Chemistry University of Texas at Austin
Physical Equilibria Unit Activity Solubility II KEY Today we will practice the skill of THINKING LIKE A CHEMIST while considering the concept of solubility. Platinum stars will be on the line. Consider
More informationIonic_Bonding_&_Ionic_Compounds_Presentation_v_1.1.notebook. October 26, Chemical Bonds. Notebook 4 Ionic Compounds and Ionic Bonding
Ionic_onding_&_Ionic_ompounds_Presentation_v_1.1.notebook hemical onds Notebook 4 Ionic ompounds and Ionic onding There are three basic types of bonds: Ionic The electrostatic attraction between ions ovalent
More informationCh. 14/15: Acid-Base Equilibria Sections 14.6, 14.7, 15.1, 15.2
Ch. 14/15: Acid-Base Equilibria Sections 14.6, 14.7, 15.1, 15.2 Creative Commons License Images and tables in this file have been used from the following sources: OpenStax: Creative Commons Attribution
More informationSolubility Equilibria. Even substances that are considered "insoluble" dissolve to a small extent.
Solubility Equilibria Even substances that are considered "insoluble" dissolve to a small extent. When a solution contains the maximum amount of dissolved material, it is saturated. 1 2 The undissolved
More information116 PLTL Activity sheet / Solubility Equilibrium Set 11
Predicting Solubility Solubility problems are equilibrium problems. The reactant in a solubility equilibrium is a slightly soluble salt and the equilibrium constant for the reaction is the solubility product
More informationAP Chemistry Semester 1 Practice Problems
AP Chemistry Semester 1 Practice Problems 1. Adipic Acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula? a) C 3 H 5 O 2 b) C 3 H 3 O 4 c) C 2 HO 3 d) C 2 H 5 O 4 e) C 3
More informationChapter 4 Chemical Formulas, Reactions, Redox and Solutions
Terms to Know: Solubility Solute Solvent Solution Chapter 4 the amount of substance that dissolves in a given volume of solvent at a given temperature. a substance dissolved in a liquid to form a solution
More informationFinal Exam Review Questions You will be given a Periodic Table, Activity Series, and a Common Ions Chart CP CHEMISTRY
Final Exam Review Questions You will be given a Periodic Table, Activity Series, and a Common Ions Chart CP CHEMISTRY Part A True-False State whether each statement is true or false. If false, correct
More informationName period AP Unit 8: equilibrium
Name period AP Unit 8: equilibrium 1. What is equilibrium? Rate of the forward reaction equals the rate of the reverse reaction 2. How can you tell when equilibrium has been reached? The concentrations
More informationSchool of Chemistry, Howard College Campus University of KwaZulu-Natal CHEMICAL ENGINEERING CHEMISTRY 2 (CHEM171)
School of Chemistry, Howard College Campus University of KwaZulu-Natal CHEMICAL ENGINEERING CHEMISTRY 2 (CHEM171) Lecture Notes 1 st Series: Solution Chemistry of Salts SALTS Preparation Note, an acid
More informationConsider a normal weak acid equilibrium: Which direction will the reaction shift if more A is added? What happens to the % ionization of HA?
ch16blank Page 1 Chapter 16: Aqueous ionic equilibrium Topics in this chapter: 1. Buffers 2. Titrations and ph curves 3. Solubility equilibria Buffersresist changes to the ph of a solution. Consider a
More informationChemistry 102 Chapter 17 COMMON ION EFFECT
COMMON ION EFFECT Common ion effect is the shift in equilibrium caused by the addition of an ion that takes part in the equilibrium. For example, consider the effect of adding HCl to a solution of acetic
More informationE09. Exp 09 - Solubility. Solubility. Using Q. Solubility Equilibrium. This Weeks Experiment. Factors Effecting Solubility.
E09 Exp 09 - Solubility Solubility Solvation The reaction coefficient Precipitating Insoluble Substances Comparing Q to Ksp Solubility Equilibrium Solubility Product, Ksp Relating Molar Solubility Factors
More informationChapter 4 Notes Types of Chemical Reactions and Solutions Stoichiometry A Summary
Chapter 4 Notes Types of Chemical Reactions and Solutions Stoichiometry A Summary 4.1 Water, the Common Solvent A. Structure of water 1. Oxygen s electronegativity is high (3.5) and hydrogen s is low (2.1)
More informationStoichiometry: Chemical Calculations. Chemistry is concerned with the properties and the interchange of matter by reaction i.e. structure and change.
Chemistry is concerned with the properties and the interchange of matter by reaction i.e. structure and change. In order to do this, we need to be able to talk about numbers of atoms. The key concept is
More information