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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.

Short Answer

Expert verified
The equilibrium constant for the dissociation of bismuth subsalicylate in water is approximately \(3.3 \times 10^{-57}\).

Step by step solution

01

Write down the dissociation equation and the respective equilibrium equation

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.
02

Use the given information and stoichiometry to find the concentrations at equilibrium

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: \[[\mathrm{C}_{7} \mathrm{H}_{4} \mathrm{O}_{3}{ }^{2-}] = [\mathrm{Bi}^{3+}] = [\mathrm{OH}^{-}] = 3.2 \times 10^{-19}\;mol/L\]
03

Calculate K' using the equilibrium concentrations

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 \(3.3 \times 10^{-57}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dissociation Reaction
A dissociation reaction occurs when a compound breaks apart into its component ions. This process is common in solutions and is usually reversible, meaning the ions can recombine to form the original compound. When bismuth subsalicylate, the active ingredient in Pepto-Bismol, is added to water, it dissociates into three types of ions:
  • Bi3+
  • C7H4O32-
  • OH-
The dissociation of this compound in water is represented by the equation: \[\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{BiO}_{4}(s) + \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{7} \mathrm{H}_{4} \mathrm{O}_{3}{ }^{2-}(aq) + \mathrm{Bi}^{3+}(aq) + \mathrm{OH}^{-}(aq)\] The double arrow in the equation indicates that the reaction can proceed in both directions, meaning it can both dissociate and recombine. The extent to which the compound dissociates is a measure of the equilibrium. Understanding this balance is vital in chemistry, especially when calculating the equilibrium constant.
Stoichiometry
Stoichiometry involves the calculation of reactants and products in chemical reactions. It is based on the conservation of mass where the mass of the reactants equals the mass of the products. In the dissociation of bismuth subsalicylate, stoichiometry tells us how many moles of each product ion are produced per mole of reactant. For this reaction, the stoichiometric relationship can be understood as each mole of bismuth subsalicylate dissociating to form:
  • One mole of C7H4O32-
  • One mole of Bi3+
  • One mole of OH-
Applying these ratios helps us determine that when 3.2 x 10-19 mol/L of bismuth subsalicylate reacts, it produces an equivalent amount of each dissociated ion in the solution.
Bismuth Subsalicylate
Bismuth subsalicylate is widely known as the active ingredient in over-the-counter medications like Pepto-Bismol, which is used to treat temporary discomforts of the stomach and gastrointestinal tract. Chemically, it is a compound that, when dissolved in water, undergoes a dissociation reaction. This reaction is particularly interesting because it creates ions that contribute to the medication's function. The release of
  • C7H4O32-
  • Bi3+
  • OH-
ions suggests how the medication may neutralize stomach acid and settle the stomach. Understanding this dissociation at a molecular level provides insight into how such medications perform their role and underscores the importance of equilibrium constants in pharmaceutical applications.
Equilibrium Concentrations
Equilibrium concentrations refer to the concentrations of the reactants and products at a point where the forward and reverse reactions occur at the same rate. In our given reaction, once equilibrium is reached, the concentrations of the product ions remain constant, meaning the process of dissociation and recombination continues without changing the concentrations further. For the dissociation of bismuth subsalicylate, the equilibrium concentrations of:
  • C7H4O32-
  • Bi3+
  • OH-
are measured to be 3.2 x 10-19 mol/L. This calculation, achieved using stoichiometry and the law of mass action, allows us to find the equilibrium constant \(K'\). It represents the balance point of the reaction according to the equation: \[K' = [\mathrm{C}_{7} \mathrm{H}_{4} \mathrm{O}_{3}{ }^{2-}][\mathrm{Bi}^{3+}][\mathrm{OH}^{-}]\]This value helps chemists predict how the reaction will behave under different conditions, offering insights into chemical processes and their efficiencies.

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Most popular questions from this chapter

Nanotechnology has become an important field, with applications ranging from high-density data storage to the design of "nano machines." One common building block of nanostructured architectures is manganese oxide nanoparticles. The particles can be formed from manganese oxalate nanorods, the formation of which can be described as follows: \(\mathrm{Mn}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MnC}_{2} \mathrm{O}_{4}(a q) \quad K_{1}=7.9 \times 10^{3}\) \(\mathrm{MnC}_{2} \mathrm{O}_{4}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}(a q)\) $$ K_{2}=7.9 \times 10^{1} $$ Calculate the value for the overall formation constant for \(\mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}\) $$ K=\frac{\left[\mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}\right]}{\left[\mathrm{Mn}^{2+}\right]\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right]^{2}} $$

As sodium chloride solution is added to a solution of silver nitrate, a white precipitate forms. Ammonia is added to the mixture and the precipitate dissolves. When potassium bromide solution is then added, a pale yellow precipitate appears. When a solution of sodium thiosulfate is added, the yellow precipitate dissolves. Finally, potassium iodide is added to the solution and a yellow precipitate forms. Write equations for all the changes mentioned above. What conclusions can you draw concerning the sizes of the \(K_{\text {sp }}\) values for \(\mathrm{AgCl}, \mathrm{AgBr}\), and \(\mathrm{AgI}\) ?

A \(50.0-\mathrm{mL}\) sample of \(0.0413 \mathrm{MAgNO}_{3}(a q)\) is added to \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaIO}_{3}(a q)\). Calculate the \(\left[\mathrm{Ag}^{+}\right]\) at equilibrium in the resulting solution. \(\left[K_{\text {sp }}\right.\) for \(\mathrm{AgIO}_{3}(s)=3.17 \times 10^{-8}\).]

Sulfide precipitates are generally grouped as sulfides insoluble in acidic solution and sulfides insoluble in basic solution. Explain why there is a difference between the two groups of sulfide precipitates.

Devise as many ways as you can to experimentally determine the \(K_{\mathrm{sp}}\) value of a solid. Explain why each of these would work.

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