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

Table of contents
- Solution 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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