= Solution
\b[A and C:] integrate $dx/x=-k h_{\rm const}\,dt$:
$$
\boxed{x(t)=x_0e^{-k h_{\rm const}t}},
$$
using $h_{\rm const}=10^{-2}\,\mathrm M$ for A and $10^{-4}\,\mathrm M$ for C.
\b[B:] integrate $dx/x^2=-k\,dt$:
$$
\frac1{x(t)}-\frac1{x_0}=kt,\qquad
\boxed{x(t)=\frac{x_0}{1+kx_0t}}.
$$
For completeness, unbuffered unequal initial <concentrations> obey $h=x+\delta$, $\delta=h_0-x_0$, and the exact expression is
$$
\ln\frac{h(t)x_0}{x(t)h_0}=k\delta t.
$$
It approaches the equimolar second-order expression as $\delta\to0$ and the excess-hydroxide exponential when depletion is negligible.
The unequal-concentration expression is the <unequal-concentration second-order integration> formula.
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