Questions
\(Cr_{2 }O_{7}^{2-}+3H_{2}O_{(l)}\leftrightharpoons 2CrO^{2-}_{4}(aq)+2H_{3}O^{+}(aq)\)
(Orange) (Yellow)
The equilibrium system represented by the equation above initially contains equal concentrations of \(Cr_{2 }O_{7}^{2-}\)(aq) and \(CrO_{4}^{2-}\)(aq). Which of the following statements correctly predicts the result of adding a sample of 6.0 M NaOH(aq) to the system, and provides an explanation?
(A) The mixture will become more orange because OH-(aq) will oxidize the Cr in \(CrO_{4}^{2-}\)(aq).
(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
Ans: C
The given equilibrium system involves the following reaction:
\[ \mathrm{Cr}_2\mathrm{O}_7^{2-}(aq) + 3\mathrm{H}_2\mathrm{O}(l) \rightleftharpoons 2\mathrm{CrO}_4^{2-}(aq) + 2\mathrm{H}_3\mathrm{O}^+(aq) \]
Initially, the concentrations of \( \mathrm{Cr}_2\mathrm{O}_7^{2-} \) and \( \mathrm{CrO}_4^{2-} \) are equal. Let’s consider the effect of adding a sample of \( 6.0 \, \mathrm{M} \, \mathrm{NaOH(aq)} \) to the system.
Reaction with \( \mathrm{NaOH} \):
\( \mathrm{NaOH(aq)} \) dissociates into \( \mathrm{Na}^+ \) and \( \mathrm{OH}^- \) ions.
The \( \mathrm{OH}^- \) ions react with \( \mathrm{H}_3\mathrm{O}^+ \) ions in the system, leading to the formation of water (\( \mathrm{H}_2\mathrm{O} \)).
This reaction shifts the equilibrium to the right, favoring the formation of more \( \mathrm{CrO}_4^{2-} \) ions.
Effect on Color:
Initially, the system contains both orange \( \mathrm{Cr}_2\mathrm{O}_7^{2-} \) and yellow \( \mathrm{CrO}_4^{2-} \).
As more \( \mathrm{CrO}_4^{2-} \) ions are formed due to the addition of \( \mathrm{NaOH} \), the mixture will become more yellow.
Therefore, the correct statement is C The mixture will become more yellow because \( \mathrm{OH}^- \) will shift the equilibrium toward products. The color change from orange to yellow indicates the increased concentration of \( \mathrm{CrO}_4^{2-} \) ions.
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.
Question
\(2H_2O(l)\rightleftharpoons H^3O^+(aq)+OH^-(aq) \) ΔH°\(=+56kJ/mol_{rxn}\)
The endothermic autoionization of pure water is represented by the chemical equation shown above. The pH of pure water is measured to be 7.00 at 25.0°C and 6.02 at 100.0°C. Which of the following statements best explains these observations?
A At the higher temperature water dissociates less, \([H_3O^+]<[OH^−]\), and the water becomes basic.
B At the higher temperature water dissociates less, \([H_3O^+]=[OH^−]\), and the water remains neutral.
C At the higher temperature water dissociates more, \([H_3O^+]>[OH^−]\), and the water becomes acidic.
D At the higher temperature water dissociates more, \([H_3O^+]=[OH^−]\), and the water remains neutral.
▶️Answer/Explanation
Ans:D
Because the reaction is endothermic, increasing temperature will increase \(K_w\). So the water dissociates more, and \([H_3O^+]\) and \([OH^−]\) will always be equal in pure water.