15.3 Multiple equilibria  (Page 4/8)

Solution

Reaction (1): $\text{AgBr}(s)\rightleftharpoons {\text{Ag}}^{\text{+}}(aq)+{\text{Br}}^{\text{−}}(aq)K_{\text{sp}}=5.0\times {10}^{\text{−13}}$

Reaction (2): ${\text{Ag}}^{\text{+}}(aq)+{\text{S}}_2{\text{O}}_3{}^{\text{2−}}(aq)\rightleftharpoons \text{Ag}{({\text{S}}_2{\text{O}}_3)}_2{}^{\text{3−}}(aq)K_{\text{f}}=4.7\times {10}^{13}$

In order for 1.00 g of AgBr to dissolve, the [Ag ⁺ ] in the solution that results must be low enough for Q for Reaction (1) to be smaller than K _sp for this reaction. We reduce [Ag ⁺ ] by adding ${\text{S}}_2{\text{O}}_3{}^{\text{2−}}$ and thus cause Reaction (2) to shift to the right. We need the following steps to determine what mass of Na ₂ S ₂ O ₃ is needed to provide the necessary ${\text{S}}_2{\text{O}}_3{}^{\text{2−}}.$

1. We calculate the [Br ^– ] produced by the complete dissolution of 1.00 g of AgBr (5.33 $\times$ 10 ^–3 mol AgBr) in 1.00 L of solution:

   $[{\text{Br}}^{\text{−}}]=5.33\times {10}^{\text{−3}}M$

2. We use [Br ^– ] and K _sp to determine the maximum possible concentration of Ag ⁺ that can be present without causing reprecipitation of AgBr:

   $[{\text{Ag}}^{\mathrm{+}}]=9.4\times {10}^{\text{−11}}M$

3. We determine the $\mathit{[}{S}_2{O}_3{}^{\mathit{2-}}\mathit{]}$ required to make [Ag ⁺ ] = 9.4 $\times$ 10 ^–11 M after the remaining Ag ⁺ ion has reacted with $S_2{O}_3{}^{\mathit{2-}}$ according to the equation:

   ${\text{Ag}}^{\mathrm{+}}+2{\text{S}}_2{\text{O}}_3{}^{\text{2−}}\rightleftharpoons \text{Ag}{({\text{S}}_2{\text{O}}_3)}_2{}^{\text{3−}}{K}_{\text{f}}=4.7\times {10}^{13}$

   Because 5.33 $\times$ 10 ^–3 mol of AgBr dissolves:

   $(5.33\times {10}^{\text{−3}})-(9.4\times {10}^{\text{−11}})=5.33\times {10}^{\text{−3}}\text{mol}\text{Ag}{({\text{S}}_2{\text{O}}_3)}_2{}^{\text{3−}}$

   Thus, at equilibrium: $[\text{Ag}{({\text{S}}_2{\text{O}}_3)}_2{}^{\text{3−}}]$ = 5.33 $\times$ 10 ^–3 M , [Ag ⁺ ] = 9.4 $\times$ 10 ^–11 M , and Q = K _f = 4.7 $\times$ 10 ¹³ :

   $K_{\text{f}}=\frac{[\text{Ag}{({\text{S}}_2{\text{O}}_3)}_2{}^{\text{3−}}]}{[{\text{Ag}}^{\mathrm{+}}][{\text{S}}_2{\text{O}}_3{}^{\text{2−}}{]}^2}=4.7\times {10}^{13}$
   $[{\text{S}}_2{\text{O}}_3{}^{\text{2−}}]=1.1\times {10}^{\text{−3}}M$

   When $[{\text{S}}_2{\text{O}}_3{}^{\text{2−}}]$ is 1.1 $\times$ 10 ^–3 M , [Ag ⁺ ] is 9.4 $\times$ 10 ^–11 M and all AgBr remains dissolved.

4. We determine the total number of moles of $S_2{O}_3{}^{\mathit{2-}}$ that must be added to the solution. This equals the amount that reacts with Ag ⁺ to form $\text{Ag}{({\text{S}}_2{\text{O}}_3)}_2{}^{\text{3−}}$ plus the amount of free ${\text{S}}_2{\text{O}}_3{}^{\text{2−}}$ in solution at equilibrium. To form 5.33 $\times$ 10 ^–3 mol of $\text{Ag}{({\text{S}}_2{\text{O}}_3)}_2{}^{\text{3−}}$ requires 2 $\times$ (5.33 $\times$ 10 ^–3 ) mol of ${\text{S}}_2{\text{O}}_3{}^{\text{2−}}.$ In addition, 1.1 $\times$ 10 ^–3 mol of unreacted ${\text{S}}_2{\text{O}}_3{}^{\text{2−}}$ is present (Step 3). Thus, the total amount of ${\text{S}}_2{\text{O}}_3{}^{\text{2−}}$ that must be added is:

   
   
   $2\times (5.33\times {10}^{\text{−3}}\text{mol}{\text{S}}_2{\text{O}}_3{}^{\text{2−}})+1.1\times {10}^{\text{−3}}\text{mol}{\text{S}}_2{\text{O}}_3{}^{\text{2−}}=1.18\times {10}^{\text{−2}}\text{mol}{\text{S}}_2{\text{O}}_3{}^{\text{2−}}$

5. We determine the mass of Na ₂ S ₂ O ₃ required to give 1.18 $\times$ 10 ^–2 mol $S_2{O}_3{}^{\mathit{2-}}$ using the molar mass of Na ₂ S ₂ O ₃ :

   $1.18\times {10}^{\text{−2}}\text{mol}{\text{S}}_2{\text{O}}_3{}^{\text{2−}}\times \frac{158.1\text{g}{\text{Na}}_2{\text{S}}_2{\text{O}}_3}{1\text{mol}{\text{Na}}_2{\text{S}}_2{\text{O}}_3}=1.9\text{g}{\text{Na}}_2{\text{S}}_2{\text{O}}_3$

   Thus, 1.00 L of a solution prepared from 1.9 g Na ₂ S ₂ O ₃ dissolves 1.0 g of AgBr.

Check your learning

Dissolution versus weak electrolyte formation

We can determine how to shift the concentration of ions in the equilibrium between a slightly soluble solid and a solution of its ions by applying Le Châtelier’s principle. For example, one way to control the concentration of manganese(II) ion, Mn ²⁺ , in a solution is to adjust the pH of the solution and, consequently, to manipulate the equilibrium between the slightly soluble solid manganese(II) hydroxide, manganese(II) ion, and hydroxide ion:

This could be important to a laundry because clothing washed in water that has a manganese concentration exceeding 0.1 mg per liter may be stained by the manganese. We can reduce the concentration of manganese by increasing the concentration of hydroxide ion. We could add, for example, a small amount of NaOH or some other base such as the silicates found in many laundry detergents. As the concentration of OH ^– ion increases, the equilibrium responds by shifting to the left and reducing the concentration of Mn ²⁺ ion while increasing the amount of solid Mn(OH) ₂ in the equilibrium mixture, as predicted by Le Châtelier’s principle.