Ch. 14/15: Acid-Base Equilibria Sections 14.6, 14.7, 15.1, 15.2
|
|
- Myra Stone
- 6 years ago
- Views:
Transcription
1 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 License 4.0. ChemWiki (CC BY-NC-SA 3.0): Unless otherwise noted, the StatWiki is licensed under a Creative Commons Attribution- Noncommercial-Share Alike 3.0 United States License. Permissions beyond the scope of this license may be available at copyright@ucdavis.edu. The mineral fluorite (CaF 2 ) is deposited through a precipitation process. Note that pure fluorite is colorless, and that the color in this sample is due to the presence of other metals in the crystal. Chemistry: OpenStax 1 Principles of General Chemistry (CC BY-NC-SA 3.0): 2 Neutralization Reactions Buffers and the Common Ion Effect Write the molecular, ionic, and net ionic equations for the following reactions: 1) HCl(aq) + NaOH(aq) H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq) Na + (aq) + Cl - (aq) + H 2 O(l) H + (aq) + OH - (aq) H 2 O(l) or H 3 O + (aq) + OH - (aq) 2H 2 O(l) 2) HF(aq) + NaOH(aq) What happens to HF in solution? HF(aq) + Na + (aq) + OH - (aq) Na + (aq) + F - (aq) + H 2 O(l) HF(aq) + OH - (aq) F - (aq) + H 2 O(l) What will happen when we add sodium acetate to a solution of acetic acid? A system at equilibrium will shift in response to being stressed. The addition of a reactant or a product can be an applied stress. CH 3 COOH(aq) + H 2 O(l) NaCH 3 COO(aq) H 2O Equilibrium is driven toward reactants H 3 O + (aq) + CH 3 COO (aq) Na + (aq) + CH 3 COO - (aq) addition A solution that contains a weak acid and its conjugate base (or a weak base and its conjugate acid) is a buffer (Common Ions) Group Work Buffer solutions resist changes in ph when small amounts of strong acid or base are added. They do this because they already contain both a weak acid and a weak base that don t neutralize each other. When a small amount of strong acid is added, the buffer s base will neutralize the added acid. When a small amount of strong base is added, the buffer s acid will neutralize the base. acid conjugate base CH 3 COOH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO (aq) reacts with added base reacts with added acid Identify the solutions below that would make good buffer solutions (what criteria need to be met?): HF and NaF ammonia and ammonium bromide KOH and KF CH 3 COOH and LiCH 3 COO nitric acid and sodium nitrate chlorous acid and potassium chlorite NaOH and NaCl HCl and NaCH 3 COO think about what these would do in solution! 1
2 Calculate the ph of a 0.50 M CH 3 COOH solution (K a = 1.8x10-5 ). ph = 2.52 What will happen to ph after we add 0.50 M NaCH 3 COO? The initial equation for a buffer solution looks like the dissociation of the weak acid or weak base in water. CH 3 COOH (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq) The ph of a buffer solution can be calculated using two initial concentrations in an ICE table ( ), or we can use the Henderson- Hasselbalch equation which is derived from the K a equation. ph = (Common Ions) 1) Add strong acid (e.g., HCl) HCl (aq) + NaCH 3 COO(aq) CH 3 COOH (aq) + NaCl(aq) Draw the product beaker of: Reactant beaker 1 = 2 moles of buffer solution (can use HA/NaA) Reactant beaker 2 = 1 mole of HCl 2) Add strong base (e.g., NaOH): NaOH(aq) + CH 3 COOH(aq) NaCH 3 COO(aq) + H 2 O(l) Draw the product beaker of: Reactant beaker 1 = 2 moles of buffer solution (can use HA/NaA) Reactant beaker 2 = 1 mole of NaOH Buffer Solution plus Strong Acid Buffer Solution plus Strong Base Calculate the ph of 75.0 ml of a 0.50 M acetic acid (K a = 1.8x10-5 ) and 0.50 M sodium acetate solution. ph = pk a + log (base / acid) = log (0.50 / 0.50) ph = 4.74 The volume doesn t affect the calculation of the Henderson- Hasselbalch equation. However, we ll need volumes when we calculate ph after adding strong acid or base. Because volumes are often measured in ml, it is helpful to use a new unit, mmol (millimole). Figure 14.18: This diagram shows the buffer action of these reactions. M = mmol / ml mmol = M x ml 2
3 with Strong Acid Calculate the ph of 75.0 ml of a 0.50 M acetic acid and 0.50 M sodium acetate solution after 5.0 ml of 1.0 M HCl is added. The added acid (H 3 O + ) reacts with the conjugate base (acetate ion). Since a strong acid is added, the reaction goes to completion. Use a change table to determine how many mmoles remain after reaction. Then use Henderson-Hasselbalch equation to calculate ph. NaCH 3 COO(aq) HCl(aq) + NaCl(aq) CH 3 COOH(aq) with Strong Base Calculate the ph of 75.0 ml of a 0.50 M acetic acid and 0.50 M sodium acetate solution after 5.0 ml of 1.0 M NaOH is added. The added base (OH - ) reacts with the acid (acetic). Since a strong base is added, the reaction goes to completion. Use a change table to determine how many mmoles remain after reaction. Then use Henderson-Hasselbalch equation to calculate ph. CH 3 COOH(aq) NaOH(aq) + H 2 O(l) NaCH 3 COO(aq) ph = log (32.5 / 42.5) = 4.63 ph = log (42.5 / 32.5) = 4.86 Example Practice Calculations Calculate the ph of a 50.0 ml solution made by combining 0.20 M HCOOH with 0.50 M NaHCOO. K a = 1.8x10-4 ph = 4.14 Practice: What is ph when ml of 0.50 M HCl is added? ph = 3.87 Work in class: What is ph when ml of 2.0 M HCl is added? ph = 0.36 Practice: What is ph when ml of 0.50 M NaOH is added? ph = 4.52 Problem Practice Calculations The ph of a 50.0 ml buffer solution made by combining 0.20 M HCOOH and 0.50 M NaHCOO is K a = 1.8x10-4 What is ph when ml of 2.0 M HCl is added? NaHCOO(aq) HCl(aq) + NaCl(aq) HCOOH(aq) 25 mmol 60 mmol mmol - 25 mmol - 25 mmol mmol 0 mol 35 mmol - 35 mmol [H 3 O + ] = 35 mmol / ( ) ml = M ph = 0.36 [H 3 O + ] from HCOOH = M (relatively negligible) What is the ph of a 1.00 L solution that is 0.75 M NH 3 and 0.25 M NH 4 Cl? K b = 1.8 x 10-5 K b K a pk a OR K b pk b pk a ph = pk a + log ([NH 3 ]/[NH 4+ ]) = log (0.75/0.25) ph = 9.73 To make a buffer with a specific ph: 1) Pick a weak acid whose pk a is close to the desired ph (within one unit of desired ph). 2) Substitute the ph and pk a into the Henderson- Hasselbalch equation to calculate the [base]/[acid] ratio. What is the ph of this solution after 0.01 mol HBr is added to it? ph = log (( ) / ( )) = ph =
4 Buffer Capacity Buffers have a certain range in which they can maintain a constant ph. Outside of this range, we say the buffer s capacity has been exceeded and we see large changes in ph. The best capacity for a buffer is when the conjugate acid and base concentrations are within a factor of 10 (the best capacity is when the acid and base concentrations are equal). The ph of a buffer cannot be more than one ph unit different than the pk a of the weak acid it contains. Example: 1) If we want to make a buffer with a ph of 5.00, which acid solution should we start with? Ascorbic acid: K a = 8.0x10-5 Acetic acid: K a = 1.8x10-5 Hydrofluoric acid: K a = 7.1x10-4 Citric acid: K a = 3.5 x ) Acetic acid gives a pk a of ph pk a = log (base/acid) = (base / acid) = = 1.80 : 1 (the base should be 1.8x more concentrated than the acid) ICE vs Change Tables 14.7 Strong Acid-Strong Base Titrations Arrows ICE table Change Table The reaction between the strong acid HCl and the strong base NaOH can be represented by: HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) When to use Weak acid or base in water At least one strong reactant Units molarity moles How to find x Labels Solve for x at equilibrium Initial, Change, Equilibrium Find x by determining the L.R. Initial, Change, Final Review terminology: Titrant solution in buret, being added Analyte unknown solution Equivalence point stoichiometric equivalents End point lightest color change Calculate the volume of base needed to reach the equivalence point when ml of 0.10 M HCl is titrated with 0.10 M NaOH. Titration Beaker Drawings Draw 4 beakers with strong acid. HCl + NaOH NaCl + H 2 O 1) 2 moles strong acid, 0 moles base 2) 2 moles strong acid, 1 mole base 3) 2 moles strong acid, 2 moles base 4) 2 moles strong acid, 3 moles base Titration Beaker Drawings Draw a WEAK acid reacting with a strong base. CH 3 COOH + NaOH NaCH 3 COO + H 2 O 1) 2 moles weak acid, 0 moles base 2) 2 moles weak acid, 1 mole base 3) 2 moles weak acid, 2 moles base 4) 2 moles weak acid, 3 moles base 4
5 Acid-Base Titrations Acid-Base Titrations: HCl + NaOH 4 key points on a titration curve At the beginning (before titrant is added) 0.00 ml NaOH Before the equivalence point ml NaOH At the equivalence point ml NaOH After the equivalence point ml NaOH Figure ) 0.00 ml NaOH HCl(aq) + NaOH(aq) H 2 O(l) NaCl(aq) 2.50 mmol 0 mmol mmol - 0 mmol mmol 2.50 mmol 0 mmol mmol Only strong acid in solution [HCl] = [H 3 O + ] ph = -log (0.10) = 1.00 Acid-Base Titrations: HCl + NaOH 2) ml NaOH Test Question Example: What is the ph of ml of 0.10 M HCl titrated with ml of 0.10 M NaOH? HCl(aq) + NaOH(aq) H 2 O(l) NaCl(aq) 2.50 mmol 1.00 mmol mmol mmol mmol 1.50 mmol 0 mmol mmol NaCl is neutral doesn t affect ph [HCl] = [H 3 O + ] = 1.50 mmol / (25+10 ml) = M ph = 1.37 Acid-Base Titrations: HCl + NaOH 3) ml NaOH added At the equivalence point, what is in solution? HCl(aq) + NaOH(aq) H 2 O(l) NaCl(aq) 2.50 mmol 2.50 mmol mmol mmol mmol 0 mmol 0 mmol mmol mmol acid = mmol base; only neutral salt in solution ph = 7.00 (always 7 for strong acid-strong base!) Acid-Base Titrations: HCl + NaOH 4) ml NaOH added HCl(aq) + NaOH(aq) H 2 O(l) NaCl(aq) 2.50 mmol 3.00 mmol mmol mmol mmol 0 mmol 0.50 mmol mmol Base is in excess. [NaOH] = [OH - ] = 0.50 mmol / (25+30 ml) = M ph = 14 (-log ) = Problem Acid-Base Titrations: HCl + NaOH Table 14.4 Check ph values in titrations Volume of 0.10 M NaOH ph value with 0.10 M HCl
6 Acid-Base Indicators Figure 14.22: Different indicators can be used depending on where the ph at the equivalence point is expected to be. Acid-Base Titrations Comparing Curves SA SB Titration WA SB Titration Weak Acid-Strong Base Titrations Consider the neutralization between acetic acid and sodium hydroxide: CH 3 COOH(aq) + NaOH(aq) NaCH 3 COO (aq) + H 2 O(l) The acetate ion that results from this neutralization undergoes hydrolysis: CH 3 COO (aq) + H 2 O(l)(aq) CH 3 COOH(aq) + OH (aq) Calculate the volume of 0.10 M NaOH needed to reach the equivalence point in a titration of ml of 0.10 M acetic acid. K a = 1.8x regions in titration curve (same as SA-SB titration): 1) Before adding base 0.00 ml base 2) Before equivalence point ml base 3) At equivalence point: ml base 4) After equivalence point ml base 1) 0 ml base. HA(aq) + NaOH(aq) H 2 O(l) NaA(aq) 2.50 mmol 0.00 mmol mmol mmol mmol 2.50 mmol 0 mmol mmol This is just weak acid. How do we find the ph of a weak acid? K a = x 2 / (0.10 M x) = 1.8x10-5 x = [H 3 O + ] [H 3 O + ] = M ph = ) ml base. This is the buffer zone (both weak acid and its conjugate base are present). Set up a change table to find mmoles excess acid and mmoles cong. base produced. All strong base is converted to conjugate acid (strong base is completely consumed). HA(aq) + NaOH(aq) H 2 O(l) NaA(aq) 2.50 mmol 1.00 mmol mmol mmol mmol 1.50 mmol 0 mmol mmol ph = pk a + log ([A - ] / [HA]) = log (1.00 / 1.50) ph = 4.57 Calculate ph after adding ml of NaOH. 6
7 3) ml base. HA(aq) + NaOH(aq) H 2 O(l) NaA(aq) 2.50 mmol 2.50 mmol mmol mmol mmol 0 mmol 0 mmol mmol At equivalence point: only salt in solution Basic salt determines ph NaCH 3 COO Na + (aq) + CH 3 COO - (aq) Acetate ion hydrolyzes to shift ph 3) ml base, continued. At equivalence point: Find ph of salt. We ve done this. Need to find concentration of salt: 2.50 mmoles / (25+25 ml) = 0.05 M CH 3 COO (aq) + H 2 O(l)(aq) CH 3 COOH(aq) + OH (aq) Need K b because OH - is produced K b = K w / K a = 1.0x10-14 / 1.8x10-5 = x10-10 K b = x 2 / (0.05 x); x = [OH - ] = x10-6 M poh = ph = ) ml base. HA(aq) + NaOH(aq) H 2 O(l) NaA(aq) 2.50 mmol 3.00 mmol mmol mmol mmol 0 mmol 0.50 mmol mmol After the equivalence point: [NaOH] = [OH - ] = 0.50 mmol / (25+30 ml) = M [A - ] is a minor contributor and doesn t need to be calculated [OH - ] = x10-6 M (doesn t change [OH - ]). ph = Problem Acid-Base Titrations: CH 3 COOH + NaOH Table 14.4 Check ph values in titrations Volume of 0.10 M NaOH ph value with 0.10 M CH 3 COOH Practice Calculations 1) Calculate the ph of a 1.0 M solution of acetic acid (K a = 1.8 x 10-5 ). ph = ) Calculate the ph of the above solution after 0.50 moles sodium acetate is added to it. Volume = 1.00 L. ph = ) Calculate the ph of the above solution (#2) after 0.10 mole of NaOH is added. ph = ) Calculate the ph of the solution in #2 after 1.2 moles of NaOH are added. ph = What is the ph of a solution made by combining ml of 0.10 M HCl with ml of 0.15 M NaOH? ph = What is the ph of a solution made by combining ml of 0.12 M ascorbic acid (K a = 7.9 x 10-5 ) with ml of 0.10 M NaOH? ph = 4.20 Determine what substance is in the flask and what substance is the titrant. CC-BY-NA-3.0: acid-base-titrations.html 7
8 Determine what substance is in the flask and what substance is the titrant. Determine what substance is in the flask and what substance is the titrant. CC-BY-NA-3.0: acid-base-titrations.html CC-BY-NA-3.0: acid-base-titrations.html Determine what substance is in the flask and what substance is the titrant. Acid-Base Titrations Comparing Curves CC-BY-NA-3.0: acid-base-titrations.html CC-BY-NA-3.0: acid-base-titrations.html Polyprotic Acid Titration: H 3 PO 4 with NaOH One eq. pt. for each proton. 3 eq. pts. and 3 half eq. pts Insoluble Salts (K sp ) Some combinations of ions in solution form precipitates. Use the solubility rules from Chapter 4/Periodic Table for the following: NaNO 3, PbCl 2, NH 4 Cl, Ag 2 S Even insoluble salts dissolve to a small extent. The solid (undissolved) salt and its dissociated ions establish equilibrium in solution. Soluble salt: > 0.10 M aqueous solution dissolves at room temperature (> 1 g dissolves in 100 ml H 2 O) Insoluble salt: < M aqueous solution at room temperature (< 1 g dissolves in 100 ml H 2 O) The value of the equilibrium constant, K sp, (solubility product) is determined at the saturation point. 8
9 Solubility Equilibria K sp is the solubility product constant. Its value indicates how soluble or insoluble a salt will be in water. See Appendix J (Solubility Products) for K sp values. AgBr(s) Ag + (aq) + Br (aq) K sp = [Ag + ][Br ] = 5.0 x PbCl 2 (s) Pb 2+ (aq) + 2Cl (aq) K sp = [Pb 2+ ][Cl ] 2 = 1.6 x 10-5 What does a large K sp value indicate? A small value? Write the dissolution equation and K sp expression for calcium phosphate. Review nomenclature!!!!! (Chapter 2) Figure 15.2: Silver chloride is a sparingly soluble ionic solid. When it is added to water, it dissolves slightly and produces a mixture consisting of a very dilute solution of Ag + and Cl ions in equilibrium with undissolved silver chloride. Calculations Using K sp We can solve for concentrations (or gram solubility) given K sp values or solve for K sp given concentrations (or gram solubility). Appendix J (Solubility Products) has a more complete list of K sp values! Substance Ppt color K sp value AgCl white 1.6x10-10 AgBr off-white 5.0x10-13 AgI yellow 1.5x10-16 PbS black 7x10-29 PbCl 2 white 1.6x10-5 Fe(OH) 3 rust red 4x10-38 CaCO 3 white 8.7x10-9 BaSO 4 white 2.3x10-8 The K sp of silver bromide is 5.0 x Calculate the molar solubility and solubility (or gram solubility). AgBr(s) Ag + (aq) + Br (aq) Initial concentration (M) 0 0 Change in concentration (M) +x +x Equilibrium concentration (M) x x K sp = [Ag + ][Br ] = 5.0 x = x 2 x = 7.1 x 10 7 M (molar sol.) Gram solubility (g/l): use molar mass of salt x10-7 mol/l x g/mol = 1.3x10-4 g/l (gram sol.) Example 15.1 Solubility Equilibria 1) Lead (II) chloride has a gram solubility of 4.4 g/l. Calculate K sp for this salt. 2) If [Br - ] eq = M, what is K sp for PbBr 2? 3) K sp for iron (III) hydroxide is found to be 4.0 x Calculate the molar and gram solubility. 4) The solubility of calcium carbonate is found experimentally to be 9.3x10-3 g/l. Calculate K sp for calcium carbonate. Examples Solubility/K sp Practice 1) PbCl 2 : 4.4 g/l x (1 mol / g) = M K sp = 4x 3 = 4 ( ) 3 = 1.6x10-5 2) If [Br - ] eq = M, then [Pb 2+ ] = M, K sp = [Pb 2+ ][Br - ] 2 = (0.0118)(0.0236) 2 = 6.57 x ) Fe(OH) 3 : K sp = [Fe 3+ ][OH - ] 3 = 27x 4 = 4.0x10-38 x = Solubility/K sp Practice Answers. x = 2.0x10-10 M, 2.1x10-8 g/l = 1.962x10-10 M x ( g/mol) 4) 9.3x10-3 g/l x (1 mol / g) = 9.293x10-5 M = x K sp = [Ca 2+ ][CO 3 2- ] = x 2 = (9.293x10-5 ) 2 = 8.6x10-9 Factors Affecting Solubility Common Ion Effect What will happen if we add solid NaF to a solution of saturated CaF 2? CaF 2 (s) Ca 2+ (aq) + 2F - (aq) Common ion effect: add more F - to precipitate more CaF 2 from solution. How does adding a common ion affect the solubility of the solid? A common ion will decrease the solubility of a solid (increase the amount of solid/precipitate that forms). 9
10 Factors Affecting Solubility Common Ion Effect Calculate the molar solubility of calcium fluoride (in pure water) (K sp = 3.5 x ). x = 2.1 x 10-4 M Calculate the molar solubility of calcium fluoride in M sodium fluoride. x = 3.5 x 10-7 M Example Predicting Precipitation of Ionic Compounds To determine if a precipitate will form as a product of a double replacement reaction or if all of a salt will dissolve in a given amount of water, we calculate Q sp and compare to K sp. Predict products when HCl (aq) is added to Pb(NO 3 ) 2 (aq) 1) Q < K sp ; no precipitate forms (unsaturated) concentration of ions is not high enough to precipitate 2) Q = K sp ; no precipitate forms (saturated) solution has maximum conc. of ions to dissolve. 3) Q > K sp ; precipitate forms (supersaturated; ppt forms) excess ions in solution will cause precipitation Precipitation of Ionic Compounds A solution contains 1.5 x 10-2 M hydrochloric acid and 1.2 x 10-2 M lead(ii) nitrate. Will a precipitate form? K sp (PbCl 2 ) = 1.6 x HCl (aq) + Pb(NO 3 ) 2 (aq) PbCl 2 (s) + 2 HNO 3 (aq) PbCl 2 (s) Pb 2+ (aq) + 2Cl - (aq) Q sp = [Pb 2+ ][Cl - ] 2 = (1.2 x 10-2 )(1.5 x 10-2 ) 2 Q sp = 2.7 x 10-6 Q sp < K sp No precipitate will form. Examples Dissolution of Ionic Compounds 245 mg of magnesium carbonate is placed in 1.00 L of water. Will it all dissolve? K sp = 4.0 x 10-5 Two methods to solve: 1) Calculate gram solubility from K sp. Compare to given mass of solid (same as what we ve already worked). 2) Find molarity of solid. Calculate Q sp. Compare to K sp to determine if precipitate will occur. If Q sp > K sp, supersaturated and salt won t dissolve If Q sp < K sp, unsaturated and salt will dissolve Dissolution of Ionic Compounds Dissolution of Ionic Compounds 245 mg of magnesium carbonate is placed in 1.00 L of water. Will it all dissolve? K sp = 4.0 x 10-5 Method 1: Calculate gram solubility from K sp. Compare to given mass of solid (same as what we ve already worked). K sp = [Mg 2+ ][CO 3 2- ] = x 2 = 4.0 x 10-5 x = (4.0 x 10-5 ) = 6.325x10-3 M x = 6.325x10-3 M x (84.32 g/1 mol) = 0.53 g/l g/l < 0.53 g/l unsat., salt will dissolve 245 mg of magnesium carbonate is placed in 1.00 L of water. Will it all dissolve? K sp = 4.0 x 10-5 Method 2: Find molarity of solid. Calculate Q sp. Compare to K sp to determine if precipitate will occur. 245 mg = g/l x (1 mol/84.32 g) = x 10-3 M K sp = [Mg 2+ ][CO 3 2- ] = x 2 = 4.0 x 10-5 Q sp = (2.906 x 10-3 ) 2 = x 10-6 Q sp < K sp, unsaturated and salt will dissolve 10
11 Factors Affecting Solubility - ph Mg(OH) 2 (s) Mg 2+ (aq) + 2OH (aq) K sp = [Mg 2+ ][OH ] 2 = 1.2 x ; weak base because insoluble x = 1.4 x 10 4 M At equilibrium: [OH ] = 2(1.4 x 10 4 M ) = 2.8 x 10 4 M ph = = Qualitative Analysis Lab: involves the principle of selective precipitation and can be used to identify the types of ions present in a solution. This is the qualitative analysis lab you ll be doing. In a solution with a ph of less than 10.45, the solubility of Mg(OH) 2 increases. Why? Acid added will react with OH - and remove it from solution. This will shift the equilibrium to the right and increase solubility. Example Selective Precipitation Some compounds can be separated based on selective precipitation. This is the separation of precipitates based on different K sp values. For example, AgCl (K sp = 1.6x10-10 ), AgBr (K sp = 7.7x10-13 ), and AgI (K sp = 8.3x10-17 ) all precipitate. Because silver iodide s K sp value is so low (it is less soluble than the other salts), it would precipitate before the other two. Almost all of the AgI would precipitate before AgCl and AgBr would. Complex Ions Complex ion formation: A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. Because AgCl is an acidic salt, adding base will dissolve it. AgCl (s) + 2 NH 3 (aq) Ag(NH 3 ) 2 + (aq) + Cl - (aq) Ammonia is called a ligand in this compound. Ligands are small neutral molecules or ions that often act as Lewis bases. Adding acid to the product will break the coordinate covalent bond between silver and ammonia to reprecipitate AgCl. Ag + (aq) + Cl - (aq) AgCl (s) Formation Constant (K f ) values Table 15.2: Common Complex Ions and Their Formation Constants Substance K f at 25 o C [Cd(CN) 4 ] 2-1.3x10 7 Ag(NH 3 ) x10 7 [AlF 6 ] 3-7.0x Lewis Acids and Bases A Lewis base is a substance that can donate a pair of electrons. A Lewis acid is a substance that can accept a pair of electrons to form a coordinate covalent bond. The larger the formation constant, the more stable the complex ion is. More values can be found in Appendix K (Formation Constants for Complex Ions)
Chapter 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 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 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 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 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 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 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 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 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 informationAqueous Equilibria Pearson Education, Inc. Mr. Matthew Totaro Legacy High School AP Chemistry
2012 Pearson Education, Inc. Mr. Matthew Totaro Legacy High School AP Chemistry 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) + C 2 H 3 O 2 (aq) If
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 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 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 informationSOLUBILITY PRODUCT (K sp ) Slightly Soluble Salts & ph AND BUFFERS (Part Two)
SOLUBILITY PRODUCT (K sp ) Slightly Soluble Salts & ph AND BUFFERS (Part Two) ADEng. PRGORAMME Chemistry for Engineers Prepared by M. J. McNeil, MPhil. Department of Pure and Applied Sciences Portmore
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 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 informationMore About Chemical Equilibria
1 More About Chemical Equilibria Acid-Base & Precipitation Reactions Chapter 15 & 16 1 Objectives Chapter 15 Define the Common Ion Effect (15.1) Define buffer and show how a buffer controls ph of a solution
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 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 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 informationLecture Presentation. Chapter 16. Aqueous Ionic Equilibrium. Sherril Soman Grand Valley State University Pearson Education, Inc.
Lecture Presentation Chapter 16 Aqueous Ionic Equilibrium Sherril Soman Grand Valley State University The Danger of Antifreeze Each year, thousands of pets and wildlife species die from consuming antifreeze.
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 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 informationChapter 8: Applications of Aqueous Equilibria
Chapter 8: Applications of Aqueous Equilibria 8.1 Solutions of Acids or Bases Containing a Common Ion 8.2 Buffered Solutions 8.3 Exact Treatment of Buffered Solutions 8.4 Buffer Capacity 8.5 Titrations
More informationCHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 16.5 (a) This is a weak acid problem. Setting up the standard equilibrium table: CH 3 COOH(aq) H (aq) CH 3 COO (aq) Initial (): 0.40 0.00 0.00
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 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 informationAdvanced Placement Chemistry Chapters Syllabus
As you work through the chapter, you should be able to: Advanced Placement Chemistry Chapters 14 16 Syllabus Chapter 14 Acids and Bases 1. Describe acid and bases using the Bronsted-Lowry, Arrhenius, and
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 informationAcid-Base Equilibria and Solubility Equilibria
ACIDS-BASES COMMON ION EFFECT SOLUBILITY OF SALTS Acid-Base Equilibria and Solubility Equilibria Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 2 The common
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 7.0: IONIC EQUILIBRIA
Acids and Bases 1 CHAPTER 7.0: IONIC EQUILIBRIA 7.1: Acids and bases Learning outcomes: At the end of this lesson, students should be able to: Define acid and base according to Arrhenius, Bronsted- Lowry
More informationAP CHEMISTRY NOTES 10-1 AQUEOUS EQUILIBRIA: BUFFER SYSTEMS
AP CHEMISTRY NOTES 10-1 AQUEOUS EQUILIBRIA: BUFFER SYSTEMS THE COMMON ION EFFECT The common ion effect occurs when the addition of an ion already present in the system causes the equilibrium to shift away
More informationChemical Equilibrium
Chemical Equilibrium Many reactions are reversible, i.e. they can occur in either direction. A + B AB or AB A + B The point reached in a reversible reaction where the rate of the forward reaction (product
More informationAP Chapter 15 & 16: Acid-Base Equilibria Name
AP Chapter 15 & 16: Acid-Base Equilibria Name Warm-Ups (Show your work for credit) Date 1. Date 2. Date 3. Date 4. Date 5. Date 6. Date 7. Date 8. AP Chapter 15 & 16: Acid-Base Equilibria 2 Warm-Ups (Show
More informationCHM 112 Dr. Kevin Moore
CHM 112 Dr. Kevin Moore Reaction of an acid with a known concentration of base to determine the exact amount of the acid Requires that the equilibrium of the reaction be significantly to the right Determination
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 informationTYPES OF CHEMICAL REACTIONS
TYPES OF CHEMICAL REACTIONS Precipitation Reactions Compounds Soluble Ionic Compounds 1. Group 1A cations and NH 4 + 2. Nitrates (NO 3 ) Acetates (CH 3 COO ) Chlorates (ClO 3 ) Perchlorates (ClO 4 ) Solubility
More informationChemical Equilibrium. Many reactions are, i.e. they can occur in either direction. A + B AB or AB A + B
Chemical Equilibrium Many reactions are, i.e. they can occur in either direction. A + B AB or AB A + B The point reached in a reversible reaction where the rate of the forward reaction (product formation,
More informationChapter 17 Additional Aspects of Aqueous Equilibria
Chapter 17 Additional Aspects of Aqueous Equilibria Water is a common solvent. Dissolved materials can be involved in different types of chemical equilibria. 17.1 The Common Ion Effect Metal ions or salts
More informationHomework: 14, 16, 21, 23, 27, 29, 39, 43, 48, 49, 51, 53, 55, 57, 59, 67, 69, 71, 77, 81, 85, 91, 93, 97, 99, 104b, 105, 107
Homework: 14, 16, 21, 23, 27, 29, 39, 43, 48, 49, 51, 53, 55, 57, 59, 67, 69, 71, 77, 81, 85, 91, 93, 97, 99, 104b, 105, 107 Chapter 15 Applications of Aqueous Equilibria (mainly acid/base & solubility)
More informationCHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA
CHAPTER 16 ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 16.3 (a) This is a weak acid problem. Setting up the standard equilibrium table: CH 3 COOH(aq) H + (aq) + CH 3 COO (aq) Initial (M): 0.40 0.00
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 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 informationChapter 16 Aqueous Ionic Equilibrium
Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro Chapter 16 Aqueous Ionic Equilibrium Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall The Danger of Antifreeze
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 informationChapter 8: Phenomena. Chapter 8: Applications of Aqueous Equilibrium
Chapter 8: Phenomena ph ph ph ph 14 12 10 8 6 4 2 0 14 12 10 8 6 4 2 0 Phenomena: Buffers are sometimes defined as: a solution that resists changes in ph when an acid base is added to it. This definition
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 informationChemical Equilibrium
Chemical Equilibrium THE NATURE OF CHEMICAL EQUILIBRIUM Reversible Reactions In theory, every reaction can continue in two directions, forward and reverse Reversible reaction! chemical reaction in which
More informationEquation Writing for a Neutralization Reaction
Equation Writing for a Neutralization Reaction An Acid-Base reaction is also called a Neutralization reaction because the acid (generates H + or H 3 O + ) and base (generates OH ) properties of the reactants
More informationChem 112, Fall 05 Exam 3A
Before you begin, make sure that your exam has all 10 pages. There are 32 required problems (3 points each, unless noted otherwise) and two extra credit problems (3 points each). Stay focused on your exam.
More informationSolubility Equilibria
Chapter 17 SOLUBILITY EQUILIBRIA (Part II) Dr. Al Saadi 1 Solubility Equilibria The concept of chemical equilibrium helps to predict how much of a specific ionic compound (salt) will dissolve in water.
More informationReaction Classes. Precipitation Reactions
Reaction Classes Precipitation: synthesis of an ionic solid a solid precipitate forms when aqueous solutions of certain ions are mixed AcidBase: proton transfer reactions acid donates a proton to a base,
More informationChapter 17. Additional Aspects of Aqueous Equilibria. Lecture Presentation. John D. Bookstaver St. Charles Community College Cottleville, MO
Lecture Presentation Chapter 17 Additional Aspects of John D. Bookstaver St. Charles Community College Cottleville, MO The Common-Ion Effect Consider a solution of acetic acid: CH 3 COOH(aq) + H 2 O(l)
More informationThe ph of aqueous salt solutions
The ph of aqueous salt solutions Sometimes (most times), the salt of an acid-base neutralization reaction can influence the acid/base properties of water. NaCl dissolved in water: ph = 7 NaC 2 H 3 O 2
More informationChapter 17 Additional Aspects of Aqueous Equilibria (Part A)
Chapter 17 Additional Aspects of Aqueous Equilibria (Part A) Often, there are many equilibria going on in an aqueous solution. So, we must determine the dominant equilibrium (i.e. the equilibrium reaction
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 informationLecture #11-Buffers and Titrations The Common Ion Effect
Lecture #11-Buffers and Titrations The Common Ion Effect The Common Ion Effect Shift in position of an equilibrium caused by the addition of an ion taking part in the reaction HA(aq) + H2O(l) A - (aq)
More informationWhat is the ph of a 0.25 M solution of acetic acid (K a = 1.8 x 10-5 )?
1 of 17 After completing this chapter, you should, at a minimum, be able to do the following. This information can be found in my lecture notes for this and other chapters and also in your text. Correctly
More informationThe Common Ion Effect
Chapter 17 ACID BASE EQUILIBRIA (Part I) Dr. Al Saadi 1 17.1 The Common Ion Effect A phenomenon known as the common ion effect states that: When a compound containing an ion in common with an already dissolved
More informationChapter 4 Reactions in Aqueous Solution
Chapter 4 Reactions in Aqueous Solution Homework Chapter 4 11, 15, 21, 23, 27, 29, 35, 41, 45, 47, 51, 55, 57, 61, 63, 73, 75, 81, 85 1 2 Chapter Objectives Solution To understand the nature of ionic substances
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 informationChapter 10. Acids, Bases, and Salts
Chapter 10 Acids, Bases, and Salts Topics we ll be looking at in this chapter Arrhenius theory of acids and bases Bronsted-Lowry acid-base theory Mono-, di- and tri-protic acids Strengths of acids and
More informationSection 4: Aqueous Reactions
Section 4: Aqueous Reactions 1. Solution composition 2. Electrolytes and nonelectrolytes 3. Acids, bases, and salts 4. Neutralization ti reactions 5. Precipitation reactions 6. Oxidation/reduction reactions
More informationAP Chemistry. Reactions in Solution
AP Chemistry Reactions in Solution S o l u t i o n s solution: a homogeneous mixture of two or more substances -- The solvent is present in greatest quantity. -- Any other substance present is called a.
More informationADVANCED PLACEMENT CHEMISTRY ACIDS, BASES, AND AQUEOUS EQUILIBRIA
ADVANCED PLACEMENT CHEMISTRY ACIDS, BASES, AND AQUEOUS EQUILIBRIA Acids- taste sour Bases(alkali)- taste bitter and feel slippery Arrhenius concept- acids produce hydrogen ions in aqueous solution while
More informationCHEMISTRY Matter and Change
CHEMISTRY Matter and Change UNIT 18 Table Of Contents Section 18.1 Introduction to Acids and Bases Unit 18: Acids and Bases Section 18.2 Section 18.3 Section 18.4 Strengths of Acids and Bases Hydrogen
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 informationACIDS AND BASES. HCl(g) = hydrogen chloride HCl(aq) = hydrochloric acid HCl(g) H + (aq) + Cl (aq) ARRHENIUS THEORY
ACIDS AND BASES A. CHARACTERISTICS OF ACIDS AND BASES 1. Acids and bases are both ionic compounds that are dissolved in water. Since acids and bases both form ionic solutions, their solutions conduct electricity
More informationGrade A buffer: is a solution that resists changes in its ph upon small additions of acid or base.sq1
Chapter 15 Lesson Plan Grade 12 402. The presence of a common ion decreases the dissociation. BQ1 Calculate the ph of 0.10M CH 3 COOH. Ka = 1.8 10-5. [H + ] = = ( )( ) = 1.34 10-3 M ph = 2.87 Calculate
More informationCh. 17 Applications of Aqueous Equilibria: Buffers and Titrations
Ch. 17 Applications of Aqueous Equilibria: Buffers and Titrations Sec 1 The Common-Ion Effect: The dissociation of a weak electrolyte decreases when a strong electrolyte that has an ion in common with
More informationName Date Class ACID-BASE THEORIES
19.1 ACID-BASE THEORIES Section Review Objectives Define the properties of acids and bases Compare and contrast acids and bases as defined by the theories of Arrhenius, Brønsted-Lowry, and Lewis Vocabulary
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 informationSCHOOL YEAR CH- 13 IONS IN AQUEOUS SOLUTIONS AND COLLIGATIVE PROPERTIES SUBJECT: CHEMISTRY GRADE : 11 TEST A
SCHOOL YEAR 2017-18 NAME: CH- 13 IONS IN AQUEOUS SOLUTIONS AND COLLIGATIVE PROPERTIES SUBJECT: CHEMISTRY GRADE : 11 TEST A Choose the best answer from the options that follow each question. 1. A solute
More informationChem 103 Exam #1. Identify the letter of the choice that best completes the statement or answers the question. Multiple Choice
Chem 103 Exam #1 Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. 1. Which of the following can act as a Bronsted-Lowry base, but not as a Bronsted-Lowry
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 informationSolubility & Net Ionic review
Solubility & Net Ionic review Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Which of the following statements is/are correct? 1. All ionic compounds
More informationAcid-Base Equilibria
Acid-Base Equilibria 1. Classify each of the following species as an acid, a base, or amphoteric in aqueous solution: (a) H 2 O; (b) CH 3 CH 2 ; (c) PO 4 3 ; (d) C 6 H 5 NH 3 2. Write the proton transfer
More informationCHEMISTRY 1102 NOTES ZUMDAHL CHAPTER 15 - APPLICATIONS OF AQUEOUS EQUILIBRIA
CHEMISTRY 1102 NOTES ZUMDAHL CHAPTER 15 - APPLICATIONS OF AQUEOUS EQUILIBRIA The introduction refers to the formation of stalactites and stalagmites as an example of the material in this chapter: CO 2
More informationChapter 4. The Major Classes of Chemical Reactions 4-1
Chapter 4 The Major Classes of Chemical Reactions 4-1 The Major Classes of Chemical Reactions 4.1 The Role of Water as a Solvent 4.2 Writing Equations for Aqueous Ionic Reactions 4.3 Precipitation Reactions
More information5. What is the percent ionization of a 1.4 M HC 2 H 3 O 2 solution (K a = ) at 25 C? A) 0.50% B) 0.36% C) 0.30% D) 0.18% E) 2.
Name: Date: 1. For which of the following equilibria does K c correspond to an acid-ionization constant, K a? A) NH 3 (aq) + H 3 O + (aq) NH 4 + (aq) + H 2 O(l) B) NH 4 + (aq) + H 2 O(l) NH 3 (aq) + H
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 informationAdvanced Chemistry Practice Problems
Aqueous Equilibria: olar Solubility and the Common Ion Effect 1. Question: Which of the following compounds will decrease the solubility of lead(ii) bromide in water? a. Lead(II) nitrate b. Sodium chloride
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 AP Chemistry 2014-15 North Nova Education Centre Mr. Gauthier
More informationChapter 4. Types of Chemical Reactions and Solution Stoichiometry
Chapter 4 Types of Chemical Reactions and Solution Stoichiometry Chapter 4 Table of Contents 4.1 Water, the Common Solvent 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes 4.3 The Composition
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 informationChem Chapter 18: Sect 1-3 Common Ion Effect; Buffers ; Acid-Base Titrations Sect 4-5 Ionic solubility Sect 6-7 Complex Formation
Chem 106 3--011 Chapter 18: Sect 1-3 Common Ion Effect; Buffers ; Acid-Base Titrations Sect 4-5 Ionic solubility Sect 6-7 Complex Formation 3//011 1 The net ionic equation for the reaction of KOH(aq) and
More informationExam Practice. Chapters
Exam Practice Chapters 16.6 17 1 Chapter 16 Chemical Equilibrium Concepts of: Weak bases Percent ionization Relationship between K a and K b Using structure to approximate strength of acids Strength of
More informationChapter 17: Additional Aspects of Aqueous Equilibria
Chapter 17: Additional Aspects of Aqueous Equilibria -Buffer Problems -Titrations -Precipitation Rx Common Ion Effect- The effect that a common ion (from two different sources) has on an equilibrium. (LeChatelier's
More informationChapter 16. Equilibria in Aqueous Systems
Chapter 16 Equilibria in Aqueous Systems Buffers! buffers are solutions that resist changes in ph when an acid or base is added! they act by neutralizing the added acid or base! but just like everything
More informationChapter 4 Types of Chemical Reaction and Solution Stoichiometry
Chapter 4 Types of Chemical Reaction and Solution Stoichiometry Water, the Common Solvent One of the most important substances on Earth. Can dissolve many different substances. A polar molecule because
More informationChemistry 102 Discussion #8, Chapter 14_key Student name TA name Section
Chemistry 102 Discussion #8, Chapter 14_key Student name TA name Section 1. If 1.0 liter solution has 5.6mol HCl, 5.mol NaOH and 0.0mol NaA is added together what is the ph when the resulting solution
More informationph + poh = 14 G = G (products) G (reactants) G = H T S (T in Kelvin) 1. Which of the following combinations would provide buffer solutions?
JASPERSE CHEM 210 PRACTICE TEST 3 VERSION 3 Ch. 17: Additional Aqueous Equilibria Ch. 18: Thermodynamics: Directionality of Chemical Reactions Key Equations: For weak acids alone in water: [H + ] = K a
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 FIFTEEN APPLICATIONS OF AQUEOUS EQUILIBRIA. For Review
CHAPTER FIFTEEN APPLICATIONS OF AQUEOUS EQUILIBRIA For Review 1. A common ion is an ion that appears in an equilibrium reaction but came from a source other than that reaction. Addition of a common ion
More information2018 Version. Chemistry AS C3.6 Aqueous Systems
2018 Version Chemistry AS 91392 C3.6 Aqueous Systems Achievement Criteria - Solubility Demonstrate understanding of equilibrium principles in aqueous systems Aqueous systems are limited to those involving
More informationChapter 9: Acids, Bases, and Salts
Chapter 9: Acids, Bases, and Salts 1 ARRHENIUS ACID An Arrhenius acid is any substance that provides hydrogen ions, H +, when dissolved in water. ARRHENIUS BASE An Arrhenius base is any substance that
More informationChemistry Lab Equilibrium Practice Test
Chemistry Lab Equilibrium Practice Test Basic Concepts of Equilibrium and Le Chatelier s Principle 1. Which statement is correct about a system at equilibrium? (A) The forward and reverse reactions occur
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 informationMake a mixture of a weak acid and its conjugate base (as the SALT) Make a mixture of a weak base and its conjugate acid (as the SALT)
175 BUFFERS - resist ph change caused by either the addition of strong acid/base OR by dilution Made in one of two ways: Make a mixture of a weak acid and its conjugate base (as the SALT) Make a mixture
More information