The active ingredient of Pepto-Bismol is the compound bismuth subsalicylate, which undergoes the following dissociation when added to water: $$\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{BiO}_{4}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{7} \mathrm{H}_{4} \mathrm{O}_{3}^{2-}(a q) +\mathrm{Bi}^{3+}(a q)+\mathrm{OH}^{-}(a q) \qquad K=?$$ 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.

We are given the dissociation reaction of bismuth subsalicylate in water: \[C_7H_5BiO_4(s) + H_2O(l) \rightleftharpoons C_7H_4O_3^{2-}(aq) + Bi^{3+}(aq) + OH^-(aq)\] Our goal is to calculate the equilibrium constant, K, for this reaction, given the maximum amount of bismuth subsalicylate that can react \(3.2 \times 10^{-19} \ \text{mol/L}\).

Let's set up a table to help us keep track of the initial concentrations, the change in the concentrations during the reaction, and the equilibrium concentrations: | Compound | Initial (M) | Change (M) | Equilibrium (M) | |-------------------|-------------|------------|-----------------| | C_7H_5BiO_4 | - | - | - | | C_7H_4O_3^{2-} | 0 | +x | x | | Bi^{3+} | 0 | +x | x | | OH^- | 0 | +x | x | The initial concentration of \(C_7H_4O_3^{2-}\), \(Bi^{3+}\), and \(OH^-\) are all zero. The reaction proceeds, and we assume that x moles per liter of the compounds have formed at equilibrium.

We are given that the maximum amount of bismuth subsalicylate that reacts by this reaction is \(3.2 \times 10^{-19}\ \text{mol/L}\). This implies that at equilibrium, the concentrations of \(C_7H_4O_3^{2-}\), \(Bi^{3+}\), and \(OH^-\) are: \[x = 3.2 \times 10^{-19} \ \text{mol/L}\] Now, we'll define the equilibrium concentrations of the three ions.

The equilibrium constant K for the given reaction can be written as: \[K = \frac{[C_7H_4O_3^{2-}][Bi^{3+}][OH^-]}{1}\] We don't have the concentration for bismuth subsalicylate and water in the expression, because the former is a solid and the latter a liquid.

Now, we will substitute the equilibrium concentrations that we calculated earlier into the equilibrium constant expression and solve for K: \[K = \frac{(3.2 \times 10^{-19})(3.2 \times 10^{-19})(3.2 \times 10^{-19})}{1} = (3.2 \times 10^{-19})^3\] \[K = 3.2768 \times 10^{-57}\] Hence, the equilibrium constant for the reaction is approximately \(K = 3.28 \times 10^{-57}\).