= Solution
Let $x=[\mathrm{MeX}]$ and $x_0=10^{-4}\,\mathrm M$.
* \b[A:] OH$^-$ starts in 100-fold excess, so its consumption changes its <concentration> by at most about 1%. Set $h\simeq h_0=0.01\,\mathrm M$: $-dx/dt=k h_0x$, a <pseudo-first-order reaction> law.
* \b[B:] initially $h_0=x_0$. <Stoichiometry> preserves $h=x$ while <hydroxide> dominates water autoionization, giving $-dx/dt=kx^2$. It is \b[not] <pseudo-first-order reaction> merely because its initial <pH> equals C's.
* \b[C:] adequate buffer capacity keeps $h=10^{-4}\,\mathrm M$ constant despite consumption: $-dx/dt=khx$, again <pseudo-first-order reaction>. Assume buffer components do not react with MeX or change the effective <rate constant> appreciably.
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