Chemistry, Third edition

(Wang) #1
ACIDS AND BASES IN AQUEOUS SOLUTION 293

Calculations involving weak acids


(i)Calculate the percentage of HCN molecules ionized in a HCN solution of concentration
0.020 mol dm^3. What is the pH of the solution? (Use the data in Table 16.2.)


(ii)Benzoic acid is a natural bacteriostatic preservative, i.e. it helps prevent the growth of
bacteria. Some fruit contains 0.050% by mass of benzoic acid. Estimate the pH of 1.0 dm^3
of the solution produced from 4.0 kg of the fruit. (HintStart by calculating the concentration
of benzoic acid.)


Exercise 16F


Weak acid equilibria


(i)25.0 cm^3 of 0.10 mol dm^3 HCl requires 25.0 cm^3 of 0.10 mol dm^3 NaOH for
neutralization. Even though only 1.3% of ethanoic acid molecules are ionized at room
temperature, 25.0 cm^3 of 0.10 mol dm^3 CH 3 COOH also requires 25 cm^3 of 0.10 mol dm^3
NaOH for neutralization. Explain this.


(ii)The enthalpy change of neutralization (H—N^ ) is the standard enthalpy change whenone mol of
water is made in the neutralization of an acid by a base. Values are:


Acid–base pair H^ —N/kJ mol^1
HNO 3 –KOH 57.3
HCl–NaOH 57.1
CH 3 COOH–NaOH 55.2

Why are the values for the first two pairs approximately the same? Why is H^ —Nfor the
neutralization of ethanoic acid less negative than the others?

Exercise 16G


BOX 16.1


An expression to calculate the pH of a weak acid


We need the two following relationships to derive equation (16.5):


1.The equilibrium concentration of unionized ethanoic acid will be the difference between CA
and the equilibrium hydronium ion concentration:
[CH 3 COOH(aq)]CA[H 3 O(aq)] (16.6)


2.For every molecule of ethanoic acid that ionizes, one hydronium and one ethanoate ion is
produced. Thus,
[CH 3 COO(aq)][H 3 O(aq)] (16.7)


Using the equilibrium expression


Ka(T)

[CH 3 COO(aq)][H 3 O(aq)]
[CH 3 COOH(aq)]

and substituting the right-hand sides of equations (16.6) and (16.7) gives


Ka(T)

[H 3 O(aq)]^2
CA[H 3 O(aq)]

For acids that are only slightly ionized (5%) we are justified in assuming that CA[H 3 O(aq)]
is approximately equal to CA. We then obtain equation (16.5):


Ka(T)

[H 3 O(aq)]^2
CA
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