IB Chemistry HL 3.1.17 Hl Buffer Ph and Composition Questions

Use real HL exam questions to calculate buffer pH, set the required conjugate ratio and explain what dilution changes or leaves unchanged.

Exam points

Calculate buffer pH or a conjugate concentration using pKa or pKb, equilibrium constants and the Henderson–Hasselbalch relationship.
Determine the acid-to-conjugate-base or base-to-conjugate-acid composition/ratio required for a target buffer pH, including partial neutralization and mixture quantities.
Explain the effect of dilution on buffer pH by distinguishing an unchanged conjugate-pair ratio from reduced total concentrations and buffer capacity.

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Study 3.1.17 (HL)—Buffer pH and composition conceptsOpen the corresponding Concept lesson for explanations, examples and exam connections.

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Nitrous acid, $\mathrm{HNO}_{2}$ , is a weak acid which can be used to make acidic buffers.

A $1.00 \mathrm{~mol} \mathrm{dm}^{-3}$ solution of nitrous acid was prepared. This solution contains two conjugate acid-base pairs.

This $1.00 \mathrm{~mol} \mathrm{dm}^{-3}$ solution of nitrous acid was used to prepare a buffer with pH 3.00 .

Calculate the concentration of the conjugate base of nitrous acid required to make this buffer. The $\mathrm{pK}_{\mathrm{a}}$ of nitrous acid is 3.25.

Concentration of conjugate base:

（a）

（ii）

$K_{a}<10^{-3.25}>=5.62 \times 10^{-4}$

« $\left[\mathrm{H}^{+}\right]=10^{-3.00}$ » =0.001 «mol dm ${ }^{-3}$ »
$\text { « }\left[\mathrm{A}^{-}\right]=\left(5.62 \times 10^{-4} \times 1.00\right) / 0.001 »=0.562 \text { «mol } \mathrm{dm}^{-3} \text { » }$

Alternative solution：
$\mathrm{pH}=\mathrm{p} K_{a}+\log _{10}[\text { salt }] /[\text { acid }]$ $\text { « } \log _{10}[\text { salt }] /[\text { acid }] \text { 》 }=-0.25$

OR
«[salt]/[acid]» = 0.562、

《 $\left[\mathrm{A}^{-}\right]=0.562 / 1.00 »=0.562$ «mol $\mathrm{dm}^{-3}$ » ↓

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IB Chemistry HL 3.1.17 Hl Buffer Ph and Composition Questions question 1