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 applies to salts with the of a weak base. The calculations are the same as the last two chapters. 4 Logical Steps As we approach these problems: 1. Identify the major species in solution, and consider their acidity or basicity. 2. Identify the important equilibrium that is the source of H+ and therefore determines ph. 3. Tabulate the concentrations at equilibrium (ICE chart) 4. Use the equilibrium-constant expression to calculate [H+] and then ph. Example What is the ph of a solution made by adding 0.30 mole of acetic acid and 0.30 mole of sodium acetate to enough water to make 1.0 Liter of solution? Appendix D gives Ka = 1.8 E -5. Try this one Calculate the ph of a solution containing 0.085 M nitrous acid (Ka = 4.5 E -4) and 0.10 M potassium nitrite. How about these Calculate the fluoride ion concentration and ph of a solution that is 0.20 M in HF and 0.10 M in HCl. Calculate the formate ion concentration and ph of a solution that is 0.50 M in formic acid (HCOOH, Ka = 1.8 E -4) and 0.10 M nitric acid. 1
17.2 Buffered Solutions A solution that Either a and its salt or a and its salt. We can make a buffer of any ph by varying the concentrations of these solutions. Which of the following conjugate acid-base pairs will not function as a buffer? HCOOH and HCOO -1? HCO 3-1 and CO 3-2? HNO 3 and NO 3-1? How buffers work Let s consider a buffer composed of a weak acid (HX) and one of its salts (MX). Write the equation for the dissociation of this weak acid: Write the corresponding acid-dissociation constant expression: Rearrange, solving for [H+]: What do we have? [H+], and thus ph, is determined by two factors: The value of The ratio of the concentrations of the conjugate acid-base pair: As long as [HX]/[X-] in the buffer compared to the amount of H+ or OHadded, the ratio doesn t change much, and thus the change in ph is small. Give it some thought What happens when NaOH is added to a buffer composed of HC 2 H 3 O 2 and C 2 H 3 O 2-1? What happens when HCl is added to this buffer? Calculating the ph of a Buffer Ka = so The [H + ] depends taking the negative log of both sides ph = ph = ph = 2
This is called the equation ph = ph = Let s try one What is the ph of a buffer that is 0.12 M in lactic acid (HC3H5O3) and 0.10 M in sodium lactate? For lactic acid, Ka = 1.4 E -4. We can use the ICE CHART or the Henderson-Hasselbalch equation. Preparing a Buffer How many moles of ammonium chloride must be added to 2.0 Liters of a 0.10 M ammonia solution to form a buffer whose ph is 9.00? Assume that the addition of ammonium chloride does not change the volume of the solution. Another one Calculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid (HC7H5O2) to produce a ph of 4.00. Buffer Capacity The ph of a buffered solution is determined by the ratio As long as this doesn t change much the ph won t change much. The more concentrated these two are the more H + and OH - the solution will be able to absorb. Larger concentrations bigger ph Range The ph range over which the buffer acts effectively. Resist a change in ph in either direction when 3
Optimal ph of a buffer is when Try to select a buffer whose acid form has a pka close the desired ph. This will give us a usable ph range of + or one ph unit of pka. Give this some thought What is the optimal ph buffered by a solution containing acetic acid and sodium acetate? Ka for acetic acid is 1.8 E -5. Adding a strong acid or base Do the first. A strong base will grab protons from the weak acid reducing [HA] 0 A strong acid will add its proton to the anion of the salt reducing [A - ] 0 Then do the equilibrium problem, most easily done with the Henderson-Hasselbalch equation. Calculating ph Changes in Buffers A buffer is made by adding 0.30 mole acetic acid and 0.30 mole sodium acetate to enough water to make 1.0 L solution. The ph of the buffer is 4.74. Calculate the ph of this solution after 0.20 mol of NaOH is added For comparison, calculate the ph that would result if 0.20 mole of NaOH was added to 1.0 Liter of pure water. Practice This Determine The ph of the original buffer described in the problem above after the addition of 0.02 mol HCl o To compare, calculate the ph of the solution what would result from the addition of 0.02 mol HCl to 1.0 L pure water. 4
17.3 Titrations Millimole (mmol) = 1/1000 mol Molarity = mmol/ml = mol/l Makes calculations easier because we will rarely add Liters of solution. Adding a solution of known concentration until the substance being tested is consumed. This is called the Graph of ph vs. ml is a Titration Curves Can be used to determine the equivalence point in the titration Can also be used to determine the Ka of a weak acid or the Kb of the weak base being titrated. 3 Kinds of Titrations: 1. 2. 3. Strong acid with Strong Base Do the There is no. They both dissociate completely. The titration of 50.0 ml of 0.200 M HNO 3 with 0.100 M NaOH Analyze the ph Graph of Strong Acid with Strong Base (Draw it in) Equivalence at ph 7 4 Regions of Curve 1. determined by the initial concentration of the strong acid. Example: 0.1 M HCl has ph =1 2. Between the : as NaOH added, ph rises slowly at first, then rapidly when it nears the equivalence point. In this case, the ph is determined by how much HCl has not yet been neutralized. 3. : An equal number of moles of NaOH and HCl have reacted, leaving only a solution of their salt, NaCl. ph = 7 because the cation of a strong base and the anion of a strong acid do not hydrolyze and therefore do not affect ph. 4. : ph of the solution is determined by the concentration of the excess NaOH in the solution. 5
Sample Problem Calculate the ph when the following quantities of 0.10 M NaOH solution have been added to 50 ml of 0.10 M HCl solution: 49.0 ml 51.0 ml Weak acid with Strong base Four Areas of the Graph: 1. found by calculations involving Ka (Section 16 2. Between : 1.Consider the neutralization of the acid by OH- to determine [HX] and [X-]. 2. Calculate the ph of this buffer pair (Section 17.2) 3. : The X- ion is a weak base and therefore affects the ph at the equivalence point. >7 4. : ph is determined by the concentration of the excess NaOH. The X- ion is negligible compared to OH- in this titration. Draw the titration curve here: Summary Strong acid and base just stoichiometry. Determine Ka, use for 0 ml base Weak acid before equivalence point Stoichiometry first Then Henderson-Hasselbach Weak acid at equivalence point Kb Weak base after equivalence - leftover strong base. Summary Determine Ka, use for 0 ml acid. Weak base before equivalence point. Stoichiometry first Then Henderson-Hasselbach Weak base at equivalence point Ka. Weak base after equivalence - leftover strong acid. 6
Sample Problems Calculate the ph of the solution formed when 45.0 ml of 0.10 M NaOH is added to 50 ml of 0.10 M HC 2 H 3 O 2. Ka = 1.8 E -5. Calulate the ph in the solution formed by adding 10.0 ml of 0.05 M NaOH to 40 ml of 0.025 M benzoic acid. Ka = 6.3 E -5. Calculate the ph in the solution formed by adding 10 ml of 0.10 M HCl to 20 ml of 0.1 M NH 3. Try these: Calculate the ph at the equivalence point in the titration of 50 ml of 0.1 M acetic acid with 0.1 M sodium hydroxide. Calculate the ph at the equivalence point when 40 ml of 0.025 M benzoic acid is titrated with 0.05 M NaOH 40 ml of 0.10 M ammonia is titrated with 0.10 M HCl 7
Comparison of Curves The ph titration curves for weak acid-strong base titrations differ from those for strong acid-strong base titrations in 3 ways: 1. The solution of the weak acid has a than a solution of a strong acid of the same concentration. 2. The ph change at the rapid-rise portion of the curve near the equivalence point is for the weak acid than it is for the strong acid. 3. The ph at the equivalence point is for the weak acidstrong base titration. Effect of Ka on Curve Use of Indicators Optimally, an indicator would change color at the equivalence point. Not really necessary in practice. ph changes by several units quickly near the equivalence point by a single drop of titrant the indicator needs to change color where the rise is steep. Called its Indicators Weak acids that change color when they become bases. weak acid written Weak base Equilibrium is controlled - when the indicator changes color. Indicators Since it is equilibrium the color change is gradual. It is noticeable when the ratio of [In - ]/[HI] or [HI]/[In - ] is 1/10 Since the Indicator is a weak acid, it has a Ka. ph the indicator changes at is. ph= ph=pka - 1 on the way up Indicators ph= ph=pka+1 on the way down Choose the indicator with a pka at equivalence point if you are titrating with base. Choose the indicator with a pka at equivalence point if you are titrating with acid. 8
Titrations of Polyprotic Acids (Draw the general curve here) Solubility Equilibria Will it all dissolve, and if so, how much? All dissolving is an. If there is not much solid it will all dissolve. As more solid is added the solution will become. The solid will precipitate as fast as it dissolves. Equilibrium General equation M + stands for the cation (usually metal). Nm - stands for the anion (a nonmetal). K = But the concentration of a solid doesn t change. K sp = Called the for each compound. Watch out Solubility is not the same as solubility product. Solubility product is an. it doesn t change except with temperature. Solubility is an equilibrium position for how. A common ion can change this. Calculating K sp from Solubility Solid silver chromate is added to pure water at 25 C. Some of the solid remains undissolved at the bottom of the flask. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved silver chromate and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3 E -4M. Assuming that silver chromate completely dissociates in water and that there are no other important equilibria involving the two ions in the solution, calculate the Ksp for the compound. 9
Another Practice A saturated solution of magnesium hydroxide in contact with undissolved solid is prepared at 25 C. The ph of the solution is found to be 10.17. Assuming that magnesium hydroxide dissociates completely in water and that there are no other simultaneous equilibria involving the two ions in the solution, calculate the Ksp for this compound. Calculating Solubility from Ksp The Ksp for calcium fluoride is 3.9 E -11 at 25 C. Assuming that calcium fluoride dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of calcium fluoride in grams per liter. Another Try The Ksp for LaF3 is 2 E -19. What is the solubility of LaF3 in water in moles per liter? 17.5 Factors that Affect Solubility Solubility is not only affected by temperature, but also by the presence of other solutes. There are 3 factors that affect the solubility of ionic compounds: 1. 2. 3. affects both ph and complexing agents. 10
Common Ion Effect If we try to dissolve the solid in a solution with either the cation or anion already present less will dissolve. Sample Exercise: Calculate the molar solubility of calcium fluoride at 25 C in a solution that is: (Appendix D shows a Ksp for CaF2 = 3.9 E -11) 0.010 M in calcium nitrate 0.010 M in sodium fluoride Practice Some More The value for Ksp for manganese II hydroxide is 1.6 E -13. Calculate the molar solubility of manganese II hydroxide in a solution that contains 0.020 M NaOH. ph and solubility OH - can be a common ion. More soluble in acid. For other anions if they come from a weak acid they are more soluble in an acidic solution than in water. CaC 2 O 4 H + + C 2 O 4-2 Reduces C 2 O 4-2 in acidic solution. General Rule The solubility of slightly soluble salts containing basic anions increases Carbonates, phosphates, cyanides, sulfides, etc. 11
The, the more the solubility is influenced by ph. Salts with anions of negligible basicity (the anions of strong acids) are by ph changes. Predicting the Effect of Acid on Solubility Which of the following substances will be more soluble in acidic solution than in basic solution? Nickel II hydroxide Calcium carbonate Barium fluoride Silver chloride Practice this Write the net ionic equation for the reaction of the following copper (II) compounds with acid: CuS Cu(N3)2 Formation of Complex Ions Metal ions have the ability to act as, (electron pair acceptors), towards water molecules, which act as, (electron pair donors). ions can interact with other molecules, such as NH3. An assembly of a metal ion and the Lewis bases bonded to it, such as Ag(NH3)2 +, is called a. Complex ion Equilibria A charged ion surrounded by. Ligands are Lewis bases using their lone pair to stabilize the. Common ligands are Coordination number is the number of attached ligands. Cu(NH 3 ) 4 +2 has a coordination # of 12
Kf the equilibrium constant for the formation of a complex ion. The magnitude of this number indicates the ion s stability. Larger the number, The addition of each ligand has its own equilibrium Usually the ligand is in large excess. And the individual K s will be large so we can treat them as if they go to equilibrium. The complex ion will be the biggest ion in solution. Practice Exercise Calculate the concentration of Ag+ present in solution at equilibrium when concentrated ammonia is added to a 0.010 M solution of silver nitrate to give an equilibrium concentration of [NH3] = 0.20 M. Neglect the small volume change that occurs when ammonia is added. Try this one Calculate the [Cr +3 ] in equilibrium with Cr(OH)4-1 when 0.010 mol of chromium III nitrate is dissolved in a liter of solution buffered at ph 10.0. General Rule: The solubility of metal salts increases in the presence of suitable Lewis bases, such as 13
Amphoteric Oxides and Bases Some metal oxides and hydroxides that are relatively insoluble in neutral water dissolve in strongly acidic and strongly basic solutions. Because they themselves act as either an. Include those of 17.6 Precipitation and Separation of Ions Use the, Q, to determine the direction in which a reaction must proceed to reach equilibrium. Precipitation Ion Product, Q = If Q>Ksp a If Q<Ksp. If Q =. Predicting Whether a Precipitate forms Will a precipitate form when 0.10 L of 8 E -3 M lead II nitrate is added to 0.40 L 5 E -3 M sodium sulfate? Will a precipitate form when 0.050 L of 2.0 E -2 M NaF is mixed with 0.010 L of 1.0 E -2 M calcium nitrate? Selective Precipitation of Ions Used to separate of metal ions in solutions. Add anions that will only precipitate certain metals at a time. Used to purify mixtures. Sample Problem A solution contains 1.0 E -2 M Ag+ ion and 2.0 E -2 Pb +2 ion. When Cl -1 is added to the solution, both AgCl (Ksp = 1.8 E -10) and PbCl 2 (Ksp = 1.7 E -5) precipitate from the solution. What concentration of Cl -1 is necessary to begin the precipitation of each salt? Which salt precipitates first? 14
Try This One A solution consists of 0.050 M Mg +2 and 0.020 M Cu +2. Which ion will precipitate first as OH - is added to the solution? What concentration of OH - is necessary to begin the precipitation of each cation? [Ksp = 1.8 E -11 for magnesium hydroxide, and Ksp = 2.2 E -20 for copper II hydroxide] Selective Precipitation Often use because in acidic solution Hg +2, Cd +2, Bi +3, Cu +2, Sn +4 will precipitate. In Basic adding OH - solution S -2 will increase so more soluble sulfides will precipitate. Co+2, Zn +2, Mn +2, Ni +2, Fe +2, Cr(OH) 3, Al(OH) 3 Selective precipitation Alkali metals and NH 4 + remain in solution. Putting it all together A sample of 1.25 L of HCl gas at 21 C and 0.950 atm is bubbled through 0.500 L of 0.15 M ammonia solution. Assuming that all the HCl dissolves and that the volume of the solution remains 0.500 L, calculate the ph of the resulting solution. 15