 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/5/b/ii/solution

Let

x = [MeX]

and

x_{0} = 1 0^{- 4} M

A: OH $^{-}$ starts in 100-fold excess, so its consumption changes its concentration by at most about 1%. Set $h ≃ h_{0} = 0.01 M$ : $- d x / d t = k h_{0} x$ , a pseudo-first-order reaction law.
B: initially $h_{0} = x_{0}$ . Stoichiometry preserves $h = x$ while hydroxide dominates water autoionization, giving $- d x / d t = k x^{2}$ . It is not pseudo-first-order reaction merely because its initial pH equals C's.
C: adequate buffer capacity keeps $h = 1 0^{- 4} M$ constant despite consumption: $- d x / d t = kh x$ , again pseudo-first-order reaction. Assume buffer components do not react with MeX or change the effective rate constant appreciably.

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