Solutions of sodium thiosulfate are used to dissolve unexposed \(\operatorname{AgBr}\left(K_{\mathrm{sp}}=5.0 \times 10^{-13}\right)\) in the developing process for black-and-white film. What mass of \(\mathrm{AgBr}\) can dissolve in \(1.00 \mathrm{L}\) of \(0.500 M\) $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} ? \mathrm{Ag}^{+}\( reacts with \)\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ to form a complex ion: $$\begin{aligned} \mathrm{Ag}^{+}(a q)+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) & \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}(a q) \\ K &=2.9 \times 10^{13} \end{aligned}$$

First, write the expression for the solubility product constant (Ksp) for AgBr: \[K_{sp} = [\mathrm{Ag}^+] [\mathrm{Br}^-]\] Next, write the expression for the equilibrium constant (K) for the formation of the complex ion: \[K = \frac{[\mathrm{Ag}(S_2O_3)_2^{3-}]}{[\mathrm{Ag}^+][\mathrm{S_2O_3}^{2-}]^2}\]

The concentration of Ag+ ions at equilibrium can be found by solving for [Ag+] in the expressions for K and Ksp: Rearrange the expression for K to get the concentration of Ag+ ions: \[[\mathrm{Ag}^+] = \frac{[\mathrm{Ag}(S_2O_3)_2^{3-}]}{K[\mathrm{S_2O_3}]^2}\] We know that the complex ion will form from Ag+ produced from the dissolution of the AgBr, so the concentration of the complex ion is equal to the concentration of Ag+ ions at equilibrium: \[[\mathrm{Ag}(S_2O_3)_2^{3-}] = [\mathrm{Ag}^+]\] Now, we can plug in the values for K and the concentration of S2O3^2- (from the given 0.500 M Na2S2O3 solution) into the expression for [Ag+]: \[[\mathrm{Ag}^+] = \frac{[\mathrm{Ag}^+]}{(2.9 \times 10^{13})(0.500)^2}\] Solve for [Ag+]: \[[\mathrm{Ag}^+] = 3.88 \times 10^{-14} \, \text{M}\]

Now that we have the concentration of Ag+ ions at equilibrium, we can use the Ksp expression to determine the concentration of Br^- ions at equilibrium: \[[\mathrm{Br}^-] = \frac{K_{sp}}{[\mathrm{Ag}^+]}\] Plug in values for Ksp and [Ag+]: \[[\mathrm{Br}^-] = \frac{5.0 \times 10^{-13}}{3.88 \times 10^{-14}}\] Solve for [Br^-]: \[[\mathrm{Br}^-] = 12.89 \, \text{M}\] Since the molar ratio between AgBr and Br^- ions is 1:1, the concentration of dissolved AgBr is the same as the concentration of Br^- ions. To find the mass of dissolved AgBr, multiply the concentration by the volume of the solution and the molar mass of AgBr: \[\text{Mass of AgBr} = [\mathrm{Br}^-] \times \text{Volume} \times \text{Molar mass of AgBr}\] Plug in values: \[\text{Mass of AgBr} = (12.89 \, \text{M}) (1.00 \, \text{L}) (187.77 \, \frac{\text{g}}{\text{mol}})\] Calculate the mass of AgBr: \[\text{Mass of AgBr} = 2420.48 \, \text{g}\] In 1.00 L of 0.500 M Na2S2O3, 2420.48 g of AgBr can dissolve.