Acid-base equilibria

WCH14Topic 4.43 questions

Brønsted-Lowry acids and bases, pH calculations, $K_a$ , titration curves, indicators and buffers.

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Original practice questions written in the style of the exam.

Question 14 marks
Calculate the pH of $0.150\,\text{mol dm}^{-3}$ propanoic acid. State one assumption you make.

[ $K_a$ of propanoic acid $= 1.35 \times 10^{-5}\,\text{mol dm}^{-3}$ ]
Mark scheme
- $K_a = \dfrac{[\text{H}^+]^2}{[\text{CH}_3\text{CH}_2\text{COOH}]}$ (1)
Question 25 marks
$20.0\,\text{cm}^3$ of $0.200\,\text{mol dm}^{-3}$ sodium hydroxide is added to $50.0\,\text{cm}^3$ of ethanoic acid to make a buffer solution.

CH2

COOH

]

[H+]2

Assumption:

[\text{H}^+] = [\text{CH}_3\text{CH}_2\text{COO}^-]

(ionisation of water negligible), or the equilibrium concentration of acid equals the initial concentration (ionisation is very small) (1)

[\text{H}^+] = \sqrt{1.35 \times 10^{-5} \times 0.150} = 1.42 \times 10^{-3}\,\text{mol dm}^{-3}

(1)

= 2.85

(1)

50.0cm3

0.200\,\text{mol dm}^{-3}

Calculate the pH of the buffer. [ $K_a$ of ethanoic acid $= 1.74 \times 10^{-5}\,\text{mol dm}^{-3}$ ]

Mark scheme

$n(\text{CH}_3\text{COOH})$ initially $= 0.0100\,\text{mol}$ and $n(\text{OH}^-) = 4.00 \times 10^{-3}\,\text{mol}$ (1)
After reaction: $n(\text{CH}_3\text{COOH}) = 6.00 \times 10^{-3}\,\text{mol}$ and $n$ (1)
$[\text{H}^+] = K_a \times \dfrac{[\text{CH}_3\text{COOH}]}{[\text{CH}_3\text{COO}^-]}$ (1)
$[\text{H}^+] = 1.74 \times 10^{-5} \times \dfrac{6.00}{4.00} = 2.61 \times 10^{-5}\,\text{mol dm}^{-3}$ (1)
pH $= 4.58$ (1)

(

{CH}_{3}

{COO}^{-}

)

=

4.00

\times

10^{- 3}

mol

n(\text{CH}_3\text{COO}^-) = 4.00 \times 10^{-3}\,\text{mol}

n (CH_{3} COO^{-}) = 4.00 \times 1 0^{- 3} mol

[CH3COO−]

[CH3COOH]

4.006.00

2.61×

10−5mol dm−3