The active ingredient of Pepto-Bismol is the compound bismuth subsalicylate, which undergoes the following dissociation when added to water: \(\begin{aligned} \mathrm{C}_{7} \mathrm{H}_{5} \mathrm{BiO}_{4}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{7} \mathrm{H}_{4} \mathrm{O}_{3}{ }^{2-}(a q) \\\ &+\mathrm{Bi}^{3+}(a q)+\mathrm{OH}^{-}(a q) \quad K=? \end{aligned}\) If the maximum amount of bismuth subsalicylate that reacts by this reaction is \(3.2 \times 10^{-19} \mathrm{~mol} / \mathrm{L}\), calculate the equilibrium constant for the preceding reaction.

The dissociation equation is given as: \[\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{BiO}_{4}(s) + \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{7} \mathrm{H}_{4} \mathrm{O}_{3}{ }^{2-}(a q) + \mathrm{Bi}^{3+}(a q) + \mathrm{OH}^{-}(a q)\] We can represent the equilibrium equation using the equilibrium constant K: \[K = \frac{[\mathrm{C}_{7} \mathrm{H}_{4} \mathrm{O}_{3}{ }^{2-}][\mathrm{Bi}^{3+}][\mathrm{OH}^{-}]}{[\mathrm{H}_{2}\mathrm{O}]}\] However, since the concentration of water ([H2O]) remains relatively constant, we can express the equilibrium constant as: \[K' = [\mathrm{C}_{7} \mathrm{H}_{4} \mathrm{O}_{3}{ }^{2-}][\mathrm{Bi}^{3+}][\mathrm{OH}^{-}]\] In this form, K’ will be the equilibrium constant that we need to find.

The maximum amount of bismuth subsalicylate reacted is \(3.2 \times 10^{-19}\; mol/L\). According to the stoichiometry of the reaction, for every 1 molecule of bismuth subsalicylate that dissociates, 1 molecule of \(\mathrm{C}_7\mathrm{H}_4\mathrm{O}_3^{2-}\), 1 molecule of \(\mathrm{Bi}^{3+}\), and 1 molecule of \(\mathrm{OH}^-\) are produced. So, at equilibrium, the concentration of these ions is the same as the mols of bismuth subsalicylate reacted: \[[\mathrm{C}_{7} \mathrm{H}_{4} \mathrm{O}_{3}{ }^{2-}] = [\mathrm{Bi}^{3+}] = [\mathrm{OH}^{-}] = 3.2 \times 10^{-19}\;mol/L\]

Now substitute the equilibrium concentrations found in step 2 into the equation for K’: \[K' = (3.2 \times 10^{-19})(3.2 \times 10^{-19})(3.2 \times 10^{-19})\] Solve for K': \[K' \approx 3.3 \times 10^{-57}\] Thus, the equilibrium constant for the dissociation of bismuth subsalicylate in water is approximately \(3.3 \times 10^{-57}\).