AcidBase Equilibria and Solubility Equilibria Chapter 16 Copyright

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. The presence of a common ion suppresses the ionization of a weak acid or a weak base. Consider mixture of CH 3 COONa (strong electrolyte) and CH 3 COOH (weak acid). CH 3 COONa (s) Na+ (aq) + CH 3 COO- (aq) CH 3 COOH (aq) H+ (aq) + CH 3 COO- (aq) common ion 16. 2

Consider mixture of salt Na. A and weak acid HA. Na. A (s) Na+ (aq) + A- (aq) HA (aq) H+ (aq) + A- (aq) [H+] Ka [HA] = [A-] -log [H+] = -log Ka - log [HA] [A-] -] [A -log [H+] = -log Ka + log [HA] [A-] p. H = p. Ka + log [HA] [H+][A-] Ka = [HA] Henderson-Hasselbalch equation p. H = p. Ka + log [conjugate base] [acid] p. Ka = -log Ka 16. 2

What is the p. H of a solution containing 0. 30 M HCOOH and 0. 52 M HCOOK? Mixture of weak acid and conjugate base! HCOOH (aq) Initial (M) Change (M) Equilibrium (M) Common ion effect 0. 30 – x 0. 30 0. 52 + x 0. 52 H+ (aq) + HCOO- (aq) 0. 30 0. 00 0. 52 -x +x +x 0. 30 - x x 0. 52 + x [HCOO-] p. H = p. Ka + log [HCOOH] [0. 52] = 4. 01 p. H = 3. 77 + log [0. 30] HCOOH p. Ka = 3. 77 16. 2

What is the p. H of a buffer made by mixing 1. 00 L of 0. 020 M benzoic acid, HC 7 H 5 O 2, with 3. 00 L of 0. 060 M sodium benzoate, Na. C 7 H 5 O 2? Ka for benzoic acid is 6. 3 × 10 -5. This problem involves dilution first. Once we know the concentrations of benzoic acid and benzoate ion, we can use the acid equilibrium to solve for x. We will use HBz to represent benzoic acid and Bzto represent benzoate ion.

Initial Change Equilibrium HBz(aq) + H 2 O(l) 0. 0050 H 3 O+(aq) + 0 Bz-(aq) 0. 045 –x +x +x 0. 0050 – x x 0. 045 + x

Calculate the p. H of a 0. 10 M HF solution to which sufficient sodium fluoride has been added to make the concentration 0. 20 M Na. F. Ka for HF is 6. 8 × 10 -4. We will use the acid equilibrium for HF. Na. F provides the conjugate base, F-, so [F-] = 0. 20 M.

Initial Change Equilibrium HF(aq) + H 2 O(l) 0. 10 H 3 O+(aq) + 0 F-(aq) 0. 20 –x +x +x 0. 10 – x x 0. 20 + x

A buffer solution is a solution of: 1. A weak acid or a weak base and 2. The salt of the weak acid or weak base Both must be present! A buffer solution has the ability to resist changes in p. H upon the addition of small amounts of either acid or base. Consider an equal molar mixture of CH 3 COOH and CH 3 COONa Add strong acid H+ (aq) + CH 3 COO- (aq) Add strong base OH- (aq) + CH 3 COOH (aq) CH 3 COO- (aq) + H 2 O (l) 16. 3

Henderson–Hasselbalch Equation Buffers at a specific p. H can be prepared using the Henderson-Hasselbalch Equation. Buffers are prepared from a conjugate acid-base pair in which the ionization is approximately equal to the desired H 3 O+ concentration. Consider a buffer made up of a weak acid HA and its conjugate base A- The acid-ionization constant is

The preceding equation can be used to derive an equation for the p. H of the buffer. Take the negative logarithm of both sides of the equation. The p. Ka of a weak acid is defined in a manner similar to p. H and p. OH p. Ka = - log Ka

The equation can then be expressed as: This equation is generally shown as: This equation relates buffer p. H for different concentrations of conjugate acid and base and is known as the Henderson-Hasselbalch equation.

Question: What is the [H 3 O+] for a buffer solution that is 0. 250 M in acid and 0. 600 M in the corresponding salt if the weak acid Ka = 5. 80 x 10 -7? Use as example the following equation of a conjugate acidbase pair p. Ka = - log(5. 80 x 10 -7) = -(-6. 24) = 6. 24 [H 3 O+] = p. Ka + log [A-]/[HA] = 6. 24 + log 0. 600/0. 250 [H 3 O+] = 10 -6. 62 = 2. 40 x 10 -7 [H 3 O+] = 2. 40 x 10 -7

HCl + CH 3 COO- H+ + Cl. CH 3 COOH + Cl- 16. 3

Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na 2 CO 3/Na. HCO 3 (a) KF is a weak acid and F- is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO 32 - is a weak base and HCO 3 - is its conjugate acid buffer solution 16. 3

Calculate the p. H of the 0. 30 M NH 3/0. 36 M NH 4 Cl buffer system. What is the p. H after the addition of 20. 0 m. L of 0. 050 M Na. OH to 80. 0 m. L of the buffer solution? NH 4+ (aq) [NH 3] p. H = p. Ka + log [NH 4+] start (moles) end (moles) H+ (aq) + NH 3 (aq) p. Ka = 9. 25 0. 029 0. 001 NH 4+ (aq) + OH- (aq) 0. 028 0. 0 [0. 30] p. H = 9. 25 + log = 9. 17 [0. 36] 0. 024 H 2 O (l) + NH 3 (aq) 0. 025 final volume = 80. 0 m. L + 20. 0 m. L = 100 m. L [NH 4 +] 0. 028 0. 025 = [NH 3] = 0. 10 [0. 25] p. H = 9. 25 + log = 9. 20 [0. 28] 16. 3

Chemistry In Action: Maintaining the p. H of Blood 16. 3

Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL The indicator changes color (pink) 4. 7

An acid–base titration curve is a plot of the p. H of a solution of acid (or base) against the volume of added base (or acid). Such curves are used to gain insight into the titration process. You can use the titration curve to choose an indicator that will show when the titration is complete.

This titration plot shows the titration of a strong acid with a strong base. Note that the p. H at the equivalence point is 7. 0.

The equivalence point is the point in a titration when a stoichiometric amount of reactant has been added. The indicator must change color near the p. H at the equivalence point.

When solving problems in which an acid reacts with a base, we first need to determine the limiting reactant. This requires that we use moles in the calculation. The next step is to review what remains at the end of the reaction. In the case of a strong acid and a strong base, the solution is neutral. When a weak acid or weak base is involved, the product is a salt. After finding the salt concentration, we use a hydrolysis equilibrium to find the p. H.

Calculate the p. OH and the p. H of a solution in which 10. 0 m. L of 0. 100 M HCl is added to 25. 0 m. L of 0. 100 M Na. OH. First we calculate the moles of HCl and Na. OH. Then we determine what remains after the reaction.

The overall reaction is HCl + Na. OH H 2 O + Na. Cl Moles HCl = (0. 100 M)(10. 0 × 10 -3 L) = 1. 00 × 10 -3 mol Moles Na. OH = (0. 100 M)(25. 0 × 10 -3 L) = 2. 50 × 10 -3 mol All of the HCl reacts, leaving 1. 50 × 10 -3 mol Na. OH. The new volume is 10. 0 m. L + 25. 0 m. L = 35. 0 m. L.

Calculate the [OH ] and the p. H at the equivalence point for the titration of 500. m. L of 0. 10 M propionic acid with 0. 050 M calcium hydroxide. (This can be used to prepare a preservative for bread. ) Ka = 1. 3 10 5. First we need to find the moles of propionic acid to find the volume of calcium hydroxide. Then we will examine the salt that is produced and use the hydrolysis equilibrium to find the p. H. We will use HPr to represent propionic acid and Pr- to represent propionate ion.

Moles propionic acid = (0. 10 M)(500. × 10 -3 L) = 0. 050 mol HPr To find the moles of calcium hydroxide, we need the balanced chemical equation: 2 HPr + Ca(OH)2 Ca(Pr)2 + 2 H 2 O Moles calcium hydroxide =

Now we find the volume of Ca(OH)2 required to give 0. 025 mol: The new volume is 500. × 10 -3 L + 0. 50 L = 1. 00 L The concentration of Pr- is 2(0. 025 M) = 0. 050 M

The hydrolysis reaction is Pr- + H 2 O HPr + OH- Initial Change Equilibrium Pr-(aq) + 0. 050 H 2 O(l) HPr(aq) + 0 OH-(aq) 0 –x +x +x 0. 050 – x x x

Strong Acid-Strong Base Titrations Na. OH (aq) + HCl (aq) OH- (aq) + H+ (aq) H 2 O (l) + Na. Cl (aq) H 2 O (l) 16. 4

Weak Acid-Strong Base Titrations CH 3 COOH (aq) + Na. OH (aq) CH 3 COONa (aq) + H 2 O (l) CH 3 COOH (aq) + OH- (aq) CH 3 COO- (aq) + H 2 O (l) At equivalence point (p. H > 7): CH 3 COO- (aq) + H 2 O (l) OH- (aq) + CH 3 COOH (aq) 16. 4

Strong Acid-Weak Base Titrations HCl (aq) + NH 3 (aq) H+ (aq) + NH 3 (aq) NH 4 Cl (aq) At equivalence point (p. H < 7): NH 4+ (aq) + H 2 O (l) NH 3 (aq) + H+ (aq) 16. 4

Exactly 100 m. L of 0. 10 M HNO 2 are titrated with a 0. 10 M Na. OH solution. What is the p. H at the equivalence point ? start (moles) 0. 01 HNO 2 (aq) + OH- (aq) 0. 0 NO 2 - (aq) + H 2 O (l) end (moles) 0. 01 Final volume = 200 m. L [NO 2 -] = = 0. 05 M 0. 200 NO 2 - (aq) + H 2 O (l) OH- (aq) + HNO 2 (aq) Initial (M) Change (M) Equilibrium (M) 0. 05 0. 00 -x +x +x 0. 05 - x x x [OH-][HNO 2] x 2 -11 = 2. 2 x 10 Kb = = [NO 2 -] 0. 05 -x p. OH = 5. 98 0. 05 – x 0. 05 x 1. 05 x 10 -6 = [OH-] p. H = 14 – p. OH = 8. 02

Acid-Base Indicators HIn (aq) H+ (aq) + In- (aq) [HIn] 10 Color of acid (HIn) predominates [In ] [HIn] -) predominates Color of conjugate base (In 10 [In-] 16. 5

p. H 16. 5

The titration curve of a strong acid with a strong base. 16. 5

Which indicator(s) would you use for a titration of HNO 2 with KOH ? Weak acid titrated with strong base. At equivalence point, will have conjugate base of weak acid. At equivalence point, p. H > 7 Use cresol red or phenolphthalein 16. 5

To deal quantitatively with an equilibrium, you must know the equilibrium constant. We will look at the equilibria of slightly soluble (or nearly insoluble) ionic compounds and show you can determine their equilibrium constants. Once you find these values for various ionic compounds, you can use them to answer questions about solubility or precipitation.

When an ionic compound is insoluble or slightly soluble, an equilibrium is established: MX(s) M+(aq) + X-(aq) The equilibrium constant for this type of reaction is called the solubility-product constant, Ksp. For the above reaction, Ksp = [M+][X-]

Solubility Equilibria Ag. Cl (s) Ksp = [Ag+][Cl-] Mg. F 2 (s) Ag 2 CO 3 (s) Ca 3(PO 4)2 (s) Ag+ (aq) + Cl- (aq) Ksp is the solubility product constant Mg 2+ (aq) + 2 F- (aq) Ksp = [Mg 2+][F-]2 2 Ag+ (aq) + CO 32 - (aq) Ksp = [Ag+]2[CO 32 -] 3 Ca 2+ (aq) + 2 PO 43 - (aq) Ksp = [Ca 2+]3[PO 43 -]2 Dissolution of an ionic solid in aqueous solution: Q < Ksp Unsaturated solution Q = Ksp Saturated solution Q > Ksp Supersaturated solution No precipitate Precipitate will form 16. 6

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Molar solubility (mol/L) is the number of moles of solute dissolved in 1 L of a saturated solution. Solubility (g/L) is the number of grams of solute dissolved in 1 L of a saturated solution. 16. 6

What is the solubility of silver chloride in g/L ? Ag. Cl (s) Initial (M) Change (M) Equilibrium (M) [Ag+] = 1. 3 x 10 -5 M Ag+ (aq) + Cl- (aq) 0. 00 +s +s s s [Cl-] = 1. 3 x 10 -5 M Ksp = 1. 6 x 10 -10 Ksp = [Ag+][Cl-] Ksp = s 2 s = Ksp s = 1. 3 x 10 -5 mol Ag. Cl 143. 35 g Ag. Cl Solubility of Ag. Cl = x = 1. 9 x 10 -3 g/L 1 L soln 1 mol Ag. Cl 16. 6

Exactly 0. 133 mg of Ag. Br will dissolve in 1. 00 L of water. What is the value of Ksp for Ag. Br? Solubility equilibrium: Ag. Br(s) Ag+(aq) + Br-(aq) Solubility-product constant expression: Ksp = [Ag+][Br-]

The solubility is given as 0. 133 mg/1. 00 L, but Ksp uses molarity: Ag. Br(s) Ag+(aq) + Br-(aq) Initial 0 0 Change +x +x Equilibrium x x

[Ag+] = [Cl-] = x = 7. 083 × 10 -7 M Ksp = (7. 083 × 10 -7)2 Ksp = 5. 02 × 10 -13

An experimenter finds that the solubility of barium fluoride is 1. 1 g in 1. 00 L of water at 25°C. What is the value of Ksp for barium fluoride, Ba. F 2, at this temperature? Solubility equilibrium: Ba. F 2(s) Ba 2+(aq) + 2 F-(aq) Solubility-product constant expression: Ksp = [Ba 2+][F-]2

The solubility is given as 1. 1 g/1. 00 L, but Ksp uses molarity: Ba. F 2(s) Initial Change Equilibrium Ba 2+(aq) + 2 F-(aq) 0 0 +x +2 x x 2 x

[Ba 2+] = x = 6. 27 × 10 -3 M [F-] = 2 x = 2(6. 27 × 10 -3) = 1. 25 × 10 -2 M Ksp = (6. 27 × 10 -3)(1. 25 × 10 -2)2 Ksp = 9. 8 × 10 -7

When Ksp is known, we can find the molar solubility.

Calomel, whose chemical name is mercury(I) chloride, Hg 2 Cl 2, was once used in medicine (as a laxative and diuretic). It has a Ksp equal to 1. 3 10 18. What is the solubility of Hg 2 Cl 2 in grams per liter? Solubility equilibrium: Hg 2 Cl 2(s) Hg 22+(aq) + 2 Cl-(aq) Solubility-product constant expression: Ksp = [Hg 22+][Cl-]2

Hg 2 Cl 2(s) Hg 22+(aq) + 2 Cl-(aq) Initial 0 0 Change +x +2 x Equilibrium x 2 x Ksp = x(2 x)2 Ksp = x(4 x 2) Ksp = 4 x 3 1. 3 × 10 -18 = 4 x 3 = 3. 25 × 10 -19 x = 6. 88 × 10 -7 M

The molar solubility is 6. 9 × 10 -7 M, but we also need the solubility in g/L:

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If 2. 00 m. L of 0. 200 M Na. OH are added to 1. 00 L of 0. 100 M Ca. Cl 2, will a precipitate form? The ions present in solution are Na+, OH-, Ca 2+, Cl-. Only possible precipitate is Ca(OH)2 (solubility rules). Is Q > Ksp for Ca(OH)2? [Ca 2+]0 = 0. 100 M [OH-]0 = 4. 0 x 10 -4 M Q = [Ca 2+]0[OH-]02 = 0. 10 x (4. 0 x 10 -4)2 = 1. 6 x 10 -8 Ksp = [Ca 2+][OH-]2 = 8. 0 x 10 -6 Q < Ksp No precipitate will form 16. 6

We can use the reaction quotient, Q, to determine whether precipitation will occur.

One form of kidney stones is calcium phosphate, Ca 3(PO 4)2, which has a Ksp of 1. 0 10 26. A sample of urine contains 1. 0 10 3 M Ca 2+ and 1. 0 10 8 M PO 43 ion. Calculate Qc and predict whether Ca 3(PO 4)2 will precipitate.

Ca 3(PO 4)2(s) 3 Ca 2+(aq) + 2 PO 43 -(aq) Ksp = [Ca 2+]3 [PO 43 -]2 Ksp = 1. 0 × 10 -26 Qc = (1. 0 × 10 -3)3 (1. 0 × 10 -8)2 Qc = 1. 0 × 10 -25 Ksp < Qc A precipitate will form.

When a problem gives the amounts and concentrations of two samples that are then mixed, the first step in solving the problem is to calculate the new initial concentrations.

Exactly 0. 400 L of 0. 50 M Pb 2+ and 1. 60 L of 2. 50 10 2 M Cl are mixed together to form 2. 00 L of solution. Calculate Qc and predict whether Pb. Cl 2 will precipitate. Ksp for Pb. Cl 2 is 1. 6 10 5.

Pb. Cl 2(s) Pb 2+(aq) + 2 Cl-(aq) Ksp = [Pb 2+] [Cl-]2 = 1. 6 × 10 -8 Qc = (0. 100)(0. 0200)2 = 4. 00 × 10 -5 Ksp < Qc A precipitate will form.

What concentration of Ag is required to precipitate ONLY Ag. Br in a solution that contains both Br- and Cl- at a concentration of 0. 02 M? Ag. Br (s) Ag+ (aq) + Br- (aq) Ksp = 7. 7 x 10 -13 Ksp = [Ag+][Br-] -13 K 7. 7 x 10 sp -11 M = = 3. 9 x 10 [Ag+] = 0. 020 [Br-] Ag. Cl (s) [Ag+] Ag+ (aq) + Cl- (aq) Ksp = 1. 6 x 10 -10 Ksp = [Ag+][Cl-] Ksp 1. 6 x 10 -10 -9 M = = 8. 0 x 10 = 0. 020 [Cl-] 3. 9 x 10 -11 M < [Ag+] < 8. 0 x 10 -9 M 16. 7

The Common Ion Effect and Solubility The presence of a common ion decreases the solubility of the salt. What is the molar solubility of Ag. Br in (a) pure water and (b) 0. 0010 M Na. Br? Na. Br (s) Na+ (aq) + Br- (aq) Ag. Br (s) Ag+ (aq) + Br- (aq) [Br-] = 0. 0010 M Ksp = 7. 7 x 10 -13 Ag. Br (s) Ag+ (aq) + Br- (aq) s 2 = Ksp [Ag+] = s s = 8. 8 x 10 -7 [Br-] = 0. 0010 + s 0. 0010 Ksp = 0. 0010 x s s = 7. 7 x 10 -10 16. 8

p. H and Solubility • • • The presence of a common ion decreases the solubility. Insoluble bases dissolve in acidic solutions Insoluble acids dissolve in basic solutions add remove Mg(OH)2 (s) Mg 2+ (aq) + 2 OH- (aq) Ksp = [Mg 2+][OH-]2 = 1. 2 x 10 -11 Ksp = (s)(2 s)2 = 4 s 3 = 1. 2 x 10 -11 s = 1. 4 x 10 -4 M [OH-] = 2 s = 2. 8 x 10 -4 M p. OH = 3. 55 p. H = 10. 45 16. 9 At p. H less than 10. 45 Lower [OH-] OH- (aq) + H+ (aq) H 2 O (l) Increase solubility of Mg(OH)2 At p. H greater than 10. 45 Raise [OH-] Decrease solubility of Mg(OH)2

Complex Ion Equilibria and Solubility A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. Co. Cl 24 (aq) Co 2+ (aq) + 4 Cl- (aq) The formation constant or stability constant (Kf) is the equilibrium constant for the complex ion formation. Co(H 2 O)2+ 6 Co. Cl 24 Kf = [Co. Cl 42 - ] [Co 2+][Cl-]4 Kf stability of complex 16. 10

Metal ions that form complex ions include Ag+, Cd 2+, Cu 2+, Fe 3+, Ni 2+, and Zn 2+. Complexing agents, called ligands, are Lewis bases. They include CN-, NH 3, S 2 O 32 -, and OH-. In each case, an equilibrium is established, called the complex-ion formation equilibrium.

Ag+(aq) + 2 NH 3(aq) Ag(NH 3)2+(aq) Zn 2+(aq) + 4 OH-(aq) Zn(OH)42 -(aq)

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Qualitative Analysis of Cations 16. 11

Flame Test for Cations lithium sodium potassium copper 16. 11

Chemistry In Action: How an Eggshell is Formed Ca 2+ (aq) + CO 32 - (aq) CO 2 (g) + H 2 O (l) carbonic Ca. CO 3 (s) H 2 CO 3 (aq) anhydrase H 2 CO 3 (aq) H+ (aq) + HCO 3 - (aq) H+ (aq) + CO 32 - (aq)