A brief note on Buffers, Salt Hydrolysis and Solubility . Read it before a test. Try each question at the end before you open its answer.
Salt hydrolysis An ion from a weak acid or a weak base reacts with water. Strong acid and strong base: neutral. Weak acid and strong base: basic. Strong acid and weak base: acidic. K h = K w / K a K_h = K_w / K_a K h = K w / K a (or K w / K b K_w / K_b K w / K b ) and h = K h / C h = \sqrt{K_h / C} h = K h / C . Weak acid and strong base: pH = 7 + 1 2 ( p K a + log C ) \text{pH} = 7 + \tfrac{1}{2}(\text{p}K_a + \log C) pH = 7 + 2 1 ( p K a + log C ) . Strong acid and weak base: pH = 7 − 1 2 ( p K b + log C ) \text{pH} = 7 - \tfrac{1}{2}(\text{p}K_b + \log C) pH = 7 − 2 1 ( p K b + log C ) . Weak acid and weak base: pH = 7 + 1 2 ( p K a − p K b ) \text{pH} = 7 + \tfrac{1}{2}(\text{p}K_a - \text{p}K_b) pH = 7 + 2 1 ( p K a − p K b ) . Buffers Common ion effect: a shared ion pushes back the ionisation of a weak electrolyte. Acidic buffer: weak acid and its salt. Basic buffer: weak base and its salt. pH = p K a + log [ salt ] [ acid ] \text{pH} = \text{p}K_a + \log\dfrac{[\text{salt}]}{[\text{acid}]} pH = p K a + log [ acid ] [ salt ] ; pOH = p K b + log [ salt ] [ base ] \text{pOH} = \text{p}K_b + \log\dfrac{[\text{salt}]}{[\text{base}]} pOH = p K b + log [ base ] [ salt ] . Best within one pH unit of p K a \text{p}K_a p K a . Blood: H 2 CO 3 / HCO 3 − \text{H}_2\text{CO}_3/\text{HCO}_3^- H 2 CO 3 / HCO 3 − . Solubility product A x B y \text{A}_x\text{B}_y A x B y : K s p = [ A ] x [ B ] y = x x y y s x + y K_{sp} = [\text{A}]^x[\text{B}]^y = x^x y^y s^{x+y} K s p = [ A ] x [ B ] y = x x y y s x + y . AB: s 2 s^2 s 2 . AB 2 \text{AB}_2 AB 2 : 4 s 3 4s^3 4 s 3 . A 2 B 3 \text{A}_2\text{B}_3 A 2 B 3 : 108 s 5 108s^5 108 s 5 . A common ion lowers s s s : for AB in c c c mol/L of a common ion, s ≈ K s p / c s \approx K_{sp} / c s ≈ K s p / c . Q s p > K s p Q_{sp} > K_{sp} Q s p > K s p : a precipitate forms. A worked example Silver carbonate, Ag 2 CO 3 \text{Ag}_2\text{CO}_3 Ag 2 CO 3 , has K s p = 8.1 × 10 − 12 K_{sp} = 8.1 \times 10^{-12} K s p = 8.1 × 1 0 − 12 . How much dissolves in a litre of water?
Type. It is A 2 B \text{A}_2\text{B} A 2 B : [ Ag + ] = 2 s [\text{Ag}^+] = 2s [ Ag + ] = 2 s and [ CO 3 2 − ] = s [\text{CO}_3^{2-}] = s [ CO 3 2 − ] = s . Ksp. K s p = ( 2 s ) 2 s = 4 s 3 K_{sp} = (2s)^2 s = 4s^3 K s p = ( 2 s ) 2 s = 4 s 3 . Solve. s 3 = 8.1 × 10 − 12 4 = 2.03 × 10 − 12 s^3 = \dfrac{8.1 \times 10^{-12}}{4} = 2.03 \times 10^{-12} s 3 = 4 8.1 × 1 0 − 12 = 2.03 × 1 0 − 12 , so s = 1.27 × 10 − 4 mol/L s = 1.27 \times 10^{-4}\ \text{mol/L} s = 1.27 × 1 0 − 4 mol/L . Practice with new numbers
A program to try The program finds the molar solubility of three salts of different types from their K s p K_{sp} K s p .
Run it. Which salt is the most soluble?
1. Add aluminium hydroxide, Al(OH) 3 \text{Al(OH)}_3 Al(OH) 3 , with K s p = 1.3 × 10 − 33 K_{sp} = 1.3 \times 10^{-33} K s p = 1.3 × 1 0 − 33 . What are its x and y?
Mistakes to avoid Every salt is neutral. Only a salt of a strong acid and a strong base is. The Henderson fraction upside down. The salt goes on top for an acidic buffer. K s p = s 2 K_{sp} = s^2 K s p = s 2 for every salt. Count the ions in the formula. Comparing K s p K_{sp} K s p of different salt types. Compare their solubilities instead. Forgetting dilution on mixing. Equal volumes halve each concentration. Check yourself Is a solution of ammonium chloride acidic or basic? Stuck? Show a hint Show the answerA buffer has twice as much salt as acid. Is its pH above or below p K a \text{p}K_a p K a ? Stuck? Show a hint Show the answerWhy does AgCl dissolve less in potassium chloride solution than in water? Stuck? Show a hint Show the answerWrite K s p K_{sp} K s p of a salt of type AB 3 \text{AB}_3 AB 3 in terms of s s s . Stuck? Show a hint Show the answer