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

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii
- iii b
  - Solution iii
- iv b
  - Solution iv

i

Solution

At

2 5^{\circ} C

,

[O H^{-}] = 1 0^{pH - 14} mol d m^{- 3}

in the usual dilute-solution thermodynamic activity approximation. Therefore A: $1 0^{- 2} M$ ; B: $1 0^{- 4} M$ ; C: $1 0^{- 4} M$ . These are initial concentrations; only C is actively buffered at its stated value.

ii

Solution

Let

x = [MeX]

and

x_{0} = 1 0^{- 4} M

.

A: OH $^{-}$ starts in 100-fold excess, so its consumption changes its concentration by at most about 1%. Set $h ≃ h_{0} = 0.01 M$ : $- d x / d t = k h_{0} x$ , a pseudo-first-order reaction law.
B: initially $h_{0} = x_{0}$ . Stoichiometry preserves $h = x$ while hydroxide dominates water autoionization, giving $- d x / d t = k x^{2}$ . It is not pseudo-first-order reaction merely because its initial pH equals C's.
C: adequate buffer capacity keeps $h = 1 0^{- 4} M$ constant despite consumption: $- d x / d t = kh x$ , again pseudo-first-order reaction. Assume buffer components do not react with MeX or change the effective rate constant appreciably.

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.

The unequal-concentration expression is the unequal-concentration second-order integration formula.

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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