Question 18.14: Predicting the Relative Acidity of a Salt Solution from Ka a......

Plan (a) The formula consists of the small, highly charged, and therefore weakly acidic, Zn^{2+} cation and the weakly basic HCOO^− anion of the weak acid HCOOH. To determine the relative acidity of the solution, we write equations that show the reactions of the ions with water, and then find K_a of Zn^{2+} (in Appendix C) and calculate K_b of HCOO^− (from K_a of HCOOH in Appendix C) to see which ion reacts with water to a greater extent.

(b) K^+ is from the strong base KOH and does not react with water; the anion, HSO_3^{-}, is the first anion of the weak polyprotic acid H_2SO_3. We write the reactions of this anion acting as a weak acid and as a weak base in water; we find its K_a (in Appendix C) and calculate its K_b (from the K_a of its conjugate acid, H_2SO_3, in Appendix C) to determine if the ion is a stronger acid or a stronger base.

Solution (a) Writing the reactions with water:

             Zn(H_2O)_6^{2+}(aq)  +  H_2O(l) \xrightleftharpoons[]{}  Zn(H_2O)_5OH^+(aq)  +  H_3O^+(aq)

             HCOO^−(aq)  +  H_2O(l) \xrightleftharpoons[]{}  HCOOH(aq) +  OH^−(aq)

Obtaining K_a and K_b of the ions: From Appendix C, the K_a of Zn(H_2O)_6^{2+}(aq) is 1×10^{−9}. We obtain K_a of HCOOH and solve for K_b of HCOO^−:

             K_b \text{ of }HCOO^−  =  \frac{K_w}{K_a \text{ of } HCOOH}  =  \frac{1.0×10^{−14}}{1.8×10^{−4}}  =  5.6×10^{−11}

K_a of Zn(H_2O)_6^{2+}  >  K_b of HCOO^{−}, so the solution is acidic.

(b) Writing the reactions with water:
             HSO_3^−(aq)  +  H_2O(l) \xrightleftharpoons[]{}  SO_3^{2−}(aq)  +  H_3O^+(aq)        (acid dissociation)
             HSO_3^−(aq)  +  H_2O(l) \xrightleftharpoons[]{} H_2SO_3(aq)  +  OH^−(aq)        (base reaction)
Obtaining K_a and K_b of HSO_3^−: From Appendix C, K_a is 6.5×10^{−8}. We obtain K_a of H_2SO_3 and solve for K_b of HSO_3^−:

K_a  >  K_b, so the solution is acidic.