DP Chemistry (first assessment 2025)

Question 19M.2.HL.TZ2.5c

Date	May 2019	Marks available	[Maximum mark: 3]	Reference code	19M.2.HL.TZ2.5c
Level	HL	Paper	2	Time zone	TZ2
Command term	Calculate	Question number	c	Adapted from	N/A

[Maximum mark: 3]

19M.2.HL.TZ2.5c

(c)

At 298 K the concentration of aqueous carbon dioxide in carbonated water is 0.200 mol dm⁻³ and the pK_a for Equilibrium (2) is 6.36.

Calculate the pH of carbonated water.

[3]

Markscheme

«K_a =» 10^–6.36/4.37 × 10^–7 = $\frac{{{{[{{\text{H}}^ + }]}^2}}}{{[{\text{C}}{{\text{O}}_2}]}}$
OR
«K_a =» 10^–6.36/4.37 × 10^–7 = $\frac{{{{[{{\text{H}}^ + }]}^2}}}{{0.200}}$ [✔]

[H⁺] « $\sqrt {0.200 \times 4.37 \times {{10}^{ - 7}}}$ » = 2.95 × 10^–4 «mol dm^–3» [✔]
«pH =» 3.53 [✔]

Note: Award [3] for correct final answer.

Examiners report

Most candidates calculated the pH of the aqueous CO₂. Some candidates attempted to use the Henderson-Hasselback equation and others used the quadratic expression to calculate [H⁺] (these two options were very common in the Spanish scripts) getting incorrect solutions. These answers usually ended in pH of approx. 1 which candidates should realize cannot be correct for soda water.

Syllabus sections

Reactivity 3. What are the mechanisms of chemical change? » Reactivity 3.1—Proton transfer reactions » Reactivity 3.1.11—For a conjugate acid–base pair, the relationship Ka Å~ Kb = Kw can be derived from the expressions for Ka and Kb. Solve problems involving these values.

Question 19M.2.HL.TZ2.5c

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