Question 19.14: Calculating the Effect of Complex-Ion Formation on Solubilit......

Plan (a) After writing the dissolution equation and the ion-product expression, we use the given K_{sp} to solve for S, the molar solubility of AgBr. (b) In hypo, Ag^+ forms a complex ion with S_2O_3^{2−}, which shifts the equilibrium and dissolves more AgBr. We write the complex-ion equation and add it to the equation for dissolving AgBr to obtain the overall equation for dissolving AgBr in hypo. We multiply K_{sp} by K_f to find K_{\text{overall}}. To find the solubility of AgBr in hypo, we set up a reaction table, with S  =  [Ag(S_2O_3)_2^{3−}], substitute into the expression for K_{\text{overall}}, and solve for S.

Solution (a) Solubility in water. Writing the equation for the saturated solution and the ion-product expression:
            AgBr(s)  \xrightleftharpoons[]{}  Ag^+(aq)  +  Br^−(aq)              K_{sp}  =  [Ag^+][Br^−]
Solving for solubility (S) directly from the equation: We know that
            S  =  [AgBr]_{\text{dissolved}}  =  [Ag^+]  =  [Br^−]
Thus,
            K_{sp}  =  [Ag^+][Br^−] =  S^2  =  5.0×10^{−13}
so
            S  =  7.1×10^{−7}  M
(b) Solubility in 1.0 M hypo. Writing the overall equation:

                      AgBr(s)  \xrightleftharpoons[]{}  \cancel{Ag^+(aq)}  +  Br^−(aq)
            \cancel{Ag^+(aq)}  +  2S_2O_3^{2−}(aq)  \xrightleftharpoons[]{}  Ag(S_2O_3)_2^{3−}(aq)
           \overline{AgBr(s)  +  2S_2O_3^{2−}(aq)  \xrightleftharpoons[]{}  Ag(S_2O_3)_2^{3−}(aq)  +  Br^−(aq)}

Calculating K_{\text{overall}}:
          K_{\text{overall}}  =  \frac{[Ag(S_2O_3)_2^{3−}][Br^−]}{[S_2O_3^{2−}]^2}  =  K_{sp}  ×  K_f  =  (5.0×10^{−13})(4.7×10^{13})  =  24
Setting up a reaction table (Table 1), with S  =  [AgBr]_{\text{dissolved}}  =  [Ag(S_2O_3)_2^{3−}]:

Substituting the values into the expression for K_{\text{overall}} and solving for S:
          K_{\text{overall}}  =  \frac{[Ag(S_2O_3)_2^{3−}][Br^−]}{[S_2O_3^{2−}]^2}  =  \frac{S^2}{(1.0  M  −  2S)^2}  =  24
Taking the square root of both sides gives

          \frac{S}{1.0  M  −  2S}  =  \sqrt{24}  =  4.9\text{      so    } S  =  4.9  M  −  9.8S \text{        and         }10.8S  =  4.9  M
          [Ag(S_2O_3)_2^{3−}]  =  S  =  0.45  M

Check (a) From the number of ions in the formula of AgBr, we know that S  =  \sqrt{K_{sp}}, so the order of magnitude seems right: ∼ \sqrt{10^{−14}}  ≈  10^{−7}. (b) The K_{\text{overall}} seems correct: the exponents cancel, and 5 × 5 = 25. Most importantly, the answer makes sense because the photographic process requires the remaining AgBr to be washed off the film and the large K_{\text{overall}} confirms that. We can check S by rounding and working backward to find K_{\text{overall}}: from the reaction table, we find that
          [(S_2O_3)^{2−}]  =  1.0  M  −  2S  =  1.0  M  −  2(0.45  M)  =  1.0  M  −  0.90  M  =  0.1  M
so K_{\text{overall}}  ≈  (0.45)^2/(0.1)^2  =  20, within rounding of the calculated value.