15.3 Multiple equilibria  (Page 3/8)

Prevention of precipitation of mg(oh) ₂

Solution

Reaction (1): $\text{Mg}{(\text{OH})}_2(s)\rightleftharpoons {\text{Mg}}^{\text{2+}}(aq)+2{\text{OH}}^{\text{−}}(aq);K_{\text{sp}}=8.9\times {10}^{\text{−12}}$

Reaction (2): ${\text{NH}}_3(aq)+{\text{H}}_2\text{O}(l)\rightleftharpoons {\text{NH}}_4{}^{\text{+}}(aq)+{\text{OH}}^{\text{−}}(aq)K_{\text{b}}=1.8\times {10}^{\text{−5}}$

To prevent the formation of solid Mg(OH) ₂ , we must adjust the concentration of OH ^– so that the reaction quotient for Equation (1), Q = [Mg ²⁺ ][OH ^– ] ² , is less than K _sp for Mg(OH) ₂ . (To simplify the calculation, we determine the concentration of OH ^– when Q = K _sp .) [OH ^– ] can be reduced by the addition of ${\text{NH}}_4{}^{\mathrm{+}},$ which shifts Reaction (2) to the left and reduces [OH ^– ].

1. We determine the [OH ^– ] at which Q = K _sp when [Mg ²⁺ ] = 0.10 M:

   $Q=[{\text{Mg}}^{\text{2+}}]{[{\text{OH}}^{\text{−}}]}^2=(0.10){[{\text{OH}}^{\text{−}}]}^2=8.9\times {10}^{\text{−12}}$
   $[{\text{OH}}^{\text{−}}]=9.4\times {10}^{\text{−6}}M$

   Solid Mg(OH) ₂ will not form in this solution when [OH ^– ] is less than 1.2 $\times$ 10 ^–5 M .

2. We calculate the $\mathit{[}N{H}_4{}^{+}\mathit{]}$ needed to decrease [OH ^– ] to 1.2 $\times$ 10 ^–5 M when [NH ₃ ] = 0.10.

   $K_{\text{b}}=\frac{[{\text{NH}}_4{}^{\mathrm{+}}][{\text{OH}}^{\text{−}}]}{[{\text{NH}}_3]}=\frac{[{\text{NH}}_4{}^{\mathrm{+}}](9.4\times {10}^{\text{−6}})}{0.10}=1.8\times {10}^{\text{−5}}$
   $[{\text{NH}}_4{}^{\mathrm{+}}]=0.19M$

When $[{\text{NH}}_4{}^{\mathrm{+}}]$ equals 0.19 M , [OH ^– ] will be 9.4 $\times$ 10 ^–6 M . Any $[{\text{NH}}_4{}^{\mathrm{+}}]$ greater than 0.19 M will reduce [OH ^– ] below 9.4 $\times$ 10 ^–6 M and prevent the formation of Mg(OH) ₂ .

Check your learning

and calculate the concentration of hydronium ion required to prevent the precipitation of ZnS in a solution that is 0.050 M in Zn ²⁺ and saturated with H ₂ S (0.10 M H ₂ S).