AP Chemistry 7.8 Representations of Equilibrium - MCQs - Exam Style Questions
Question
orange yellow
(B) The mixture will become more yellow because \(OH^{-}(aq)\) will reduce the Cr in \(Cr_{2}O_{7}^{2-}(aq)\).
(C) The mixture will become more yellow because \(OH^{-}(aq)\) will shift the equilibrium toward products.
(D) The color of the mixture will not change because \(OH^{-}(aq)\) does not appear in the equilibrium expression.
▶️ Answer/Explanation
1. Analyze the Stress:
\(NaOH(aq)\) is a strong base. Adding it to the solution introduces a high concentration of \(OH^{-}(aq)\) ions.
2. Identify the Interaction:
The \(OH^{-}(aq)\) ions are a base and will react with the acid present in the equilibrium. In this case, the acid is the product \(H_{3}O^{+}(aq)\).
The neutralization reaction is: \(OH^{-}(aq) + H_{3}O^{+}(aq) \rightarrow 2~H_{2}O(l)\)
3. Apply Le Chatelier’s Principle:
This neutralization reaction consumes a product (\(H_{3}O^{+}\)). According to Le Chatelier’s principle, the system will shift to counteract this change. To replace the lost \(H_{3}O^{+}\), the equilibrium will shift to the right (toward the product side).
4. Observe the Result:
A shift to the right consumes the orange \(Cr_{2}O_{7}^{2-}\) reactant and produces more of the yellow \(CrO_{4}^{2-}\) product. Therefore, the mixture will become more yellow.
5. Evaluate Options:
Option (C) correctly states the mixture will become more yellow and provides the correct explanation (the \(OH^{-}\) ions cause the shift by reacting with \(H_{3}O^{+}\), which shifts the equilibrium toward products).
✅ Answer: (C)
Question
\(H_2O(l)+H_2O(l)\rightleftharpoons H_3O^+(aq)+OH^-(aq)\) \( K_w=7.0\times 10^{−14}\) at 55°C
Pure water autoionizes as shown in the equation above. Based on this information, which of the following is correct?
A The autoionization equilibrium for pure water favors the formation of reactants more at 55°C than at 25°C.
B The autoionization equilibrium for pure water produces the same amount of OH− ions at 55°C and 25°C.
C At 55°C , \(pH =-log(\sqrt{K_w})\) for pure water.
D At 55°C , \(pH=−log(K_w)\) for pure water.
▶️Answer/Explanation
Ans:C
Regardless of the temperature, for pure water \(K_w=[H_3O^+][OH^-]\).
At 55°C, \([H_3O^+]=[OH^-]=\sqrt{7.0\times 10^{-14}}\) and \(pH =-log(\sqrt{K_w})\)
Question
\(K_w=[H_3O^+][OH^−]=1.0\times 10^{−14}\) at 25°C
Based on the information above, which of the following is true for a sample of pure water at 25°C ?
A \([H_3O^+]=7.0M\)
B \([OH^−]=1.0\times 10^{−14}M\)
C \(pH=10^{−7}\)
D \(pOH=7.00\)
▶️Answer/Explanation
Ans:D
\(pOH=−log[OH^−]\) , which is equal to \(−log (1.0\times 10^{−7})=7.00\) at 25°C.
