MC test: Calculations involving acids & bases

Multiple choice test on 18.2 Calculations involving acids & bases

Use the following 'quiz' to test your knowledge and understanding of this sub-topic. You will need access to a periodic table (Section 6 of the IB data booklet).

If you get an answer wrong, read through the explanation carefully to learn from your mistakes.

The graph shows how the ionic product of water, Kw varies with temperature.

What will be the concentration of hydrogen ions in pure water at 75 oC?

Kw = [H+(aq)] x [OH−(aq)] = [H+(aq)]2 = 2.0 x 10−13
[H+(aq)] = (2.0 x 10−13)1/2 = 4.5 x 10−7 mol dm−3

 

The graph shows how the ionic product of water, Kw varies with temperature.

What will be the pH of pure water at 60 oC?

Kw = [H+(aq)] x [OH−(aq)] = [H+(aq)]2 = 1.0 x 10−13
[H+(aq)] = (1.0 x 10−13)1/2 = 3.16 x 10−7 mol dm−3
pH = −log10[H+(aq)] = −log10(3.16 x 10−7) = 6.5

or (more simply) pH + pOH = 13 and for pure water pH = pOH so pH = 6.5

 

What is the pOH of a solution of 5.0 x 10−2 mol dm−3 barium hydroxide solution, Ba(OH)2(aq)?

[OH−(aq)] = 2 x 5.0 x 10−1 = 1.0 x 10−1 mol dm−3
pOH = −log10[OH−(aq)] = −log10(1.0 x 10−1) = 1.0

 

Which is the correct expression for the acid dissociation constant, Ka for prussic acid, HCN in water?

HCN(aq) + H2O(l) ⇄ H3O+(aq) + CN−(aq)

Since the reaction is carried out in aqueous solution, the concentration of water is constant and does not appear in the equilibrium expression.

 

Consider the dissociation of 0.1 mol dm−3 ethanoic acid in water.

CH3COOH(aq) + H2O(l) ⇄ H3O+(aq) + CH3COO−(aq) Ka = 1.8 x 10−5 at 298 K

The concentration of ethanoate ions in this solution at 298 K is x mol dm−3

Which is the correct expression in terms of x?

The equilibrium concentration of H3O+(aq) = [CH3COO−(aq)] = x;  and [CH3COOH(aq)] = 0.1 − x
Ka = ([H3O+(aq)] x [CH3COO−(aq)]) ÷ [CH3COOH(aq)] = [H3O+(aq)]2 ÷ [CH3COOH(aq)] = x2 ÷ (0.1 − x)

 

Which row contains the correct combination for a weak acid and its conjugate base?

Row pKa of acid Kb of conjugate base
1 low low
2 low high
3 high low
4 high high

Weak acids have strong conjugate bases. The weaker the acid the higher the pKa value and the stronger the base the greater the base dissociation constant, Kb.

 

A 0.025 mol dm−3 solution of a weak acid has a pH of 4.0

What is the value of the acid dissociation constant, Ka for this acid ?

Ka = [H+(aq)]2 ÷ (0.025 − [H+(aq)]) ≈ [H+(aq)]2 ÷ 0.025 ≈ (10−4)2 ÷ 0.025 ≈ 4.0 x 10−7

 

The Ka value of a weak monoprotic acid at 298 K is 5.0 x 10−5.

What is the concentration of an aqueous solution of the acid that has a pH of 3?

 

Ka = [H+(aq)]2 ÷ ([HA(aq)] − [H+(aq)]) ≈ [H+(aq)]2 ÷ [HA(aq)] ≈ (10−3)2 ÷ [HA(aq)]
[HA(aq)] ≈  10−6 ÷ (5.0 x 10−5) ≈ 2.0 x 10−2 mol dm−3

 

The pH of a 2.0 x 10−3 mol dm−3 aqueous solution of a weak base is 11.

What is the value of the base dissociation constant, Kb of this base.

pH = 11 so pOH = 3 and [OH−(aq)] = 10−3.
Kb ≈ [OH−(aq)]2 ÷ 2.0 x 10−3 ≈ 10−6 ÷ 2.0 x 10−3  ≈ 5.0 x 10−4

 

What is the pH of a 0.010 mol dm−3 solution of a weak base with a base dissociation constant of 1.0 x 10−4

Kb = 1.0 x 10−4 ≈ [OH−(aq)]2 ÷ 1.0 x 10−2
[OH−(aq)] ≈  (1.0 x 10−4 x 1.0 x 10−2)1/2  ≈ 1.0 x 10−3 mol dm−3
or Since [H+(aq)] x [OH−(aq)] = 1.0 x 10−14, [H+(aq)] ≈ 1.0 x 10−11 mol dm−3
pH = −log10[H+(aq)] ≈ 11

 

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