 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/iv/solution

x = 0.1 x_{0}

t_{90, A} ≃ \frac{ln 10}{k ( 1 0 ^{- 2} M )}, t_{90, B} = \frac{9}{k ( 1 0 ^{- 4} M )},

t_{90, C} = \frac{ln 10}{k ( 1 0 ^{- 4} M )} .

The corresponding numerical concentration factors are approximately

230.3

90000

and

23026 M^{- 1}

, each divided by

k

. Thus A is about 100 times faster than C at this conversion, while B takes about

9/ ln 10 = 3.91

times as long as C because its hydroxide is depleted.

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