 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/iii/solution

A and C: integrate

d x / x = - k h_{const} d t

x (t) = x_{0} e^{- k h_{const} t},

using

h_{const} = 1 0^{- 2} M

for A and

1 0^{- 4} M

for C.

B: integrate

d x / x^{2} = - k d t

\frac{1}{x ( t )} - \frac{1}{x _{0}} = k t, x (t) = \frac{x _{0}}{1 + k x _{0} t} .

For completeness, unbuffered unequal initial concentrations obey

h = x + δ

δ = h_{0} - x_{0}

, and the exact expression is

ln \frac{h ( t ) x _{0}}{x ( t ) h _{0}} = k δ t .

It approaches the equimolar second-order expression as

δ \to 0

and the excess-hydroxide exponential when depletion is negligible.

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