Question

2. The following question is mostly a review of General Chemistry. You may need to consult your old chemistry texts or notes from other classes to refresh your memory on some of these things. If you aren't having much success with questions 1-3, at least do question 4, which doesn't really depend on your answer to 1-3. Consider the following balanced equation for the hydrolysis of hydrated metal ions of a. Write the equilibrium constant expression for this dissociation. (The constant should be called K) b. The K, for the first deprotonation of hydrated Fe is 6.7x10 Calculate the pH of a 0.100 M solution of Fe" in pure water. Assume that the first deprotonation (the one in the above equation) is the only one observed. c. Some metal ions, such as Na', have almost no effect on pH. How small does K, have to be before the added H,O: trom hydrolysis of hydrated metal ions provides only a 10% increase in the H,O concentration over that found in pure water? (This would cause the phi to decrease from 7,00 to 6.96.) d. If a general chemistry student came and asked you why the pH of their NaCl solution was 7.00 but their FeCl, solution was more acidic and their ScCI solution was even more acidic;, how would you explain to them the trend in solution acidities?

Answer 1

For the equation

[M(H₂O)₆]ⁿ⁺(aq) + H₂O(l) $\rightleftharpoons$ [M(H₂O)₅)H]^(n-1)+(aq) + H₃O⁺(aq)

a) The equilibrium expression for this reaction is

Ka = [[M(H₂O)₅)H]^(n-1)+][H₃O⁺]/[[M(H₂O)₆]ⁿ⁺]

b) [M(H₂O)₆]ⁿ⁺(aq) + H₂O(l) $\rightleftharpoons$ [M(H₂O)₅)H]^(n-1)+(aq) + H₃O+ (aq) let us set up an ICE table

Initial 0.1 0 0

change -x +x +x

equilibrium 0.1-x x x

Ka = x²/0.1-x

6.7 x 10^-3 = x²/0.1-x

0.00067 - 0.0067x = x²

x²+0.0067x -0.00067 = 0 solving this quadratic equation we get

one positive root which is 0.0227M this is the concentration of H⁺

pH = -log H⁺

pH = -log 0.0227

pH = 1.64

c) you want the pH to go from 7 to 6.96

H⁺ at pH 6.96 is 10^-6.96 = 1.09 x 10^-7

Let us start with the 0.1 M metal solution

[M(H₂O)₆]ⁿ⁺(aq) + H₂O(l) $\rightleftharpoons$ [M(H₂O)₅)H]^(n-1)+(aq) + H₃O+ (aq)

since H+ = 1.09 x 10^-7 as per the above equation you will expect [M(H₂O)₅)H]^(n-1)+ to also be 1.09 x 10^-7

Ka = (1.09 x 10^-7)²/(0.1-1.09 x 10^-7)

Ka = 1.18 x 10^-13 This should be the Ka