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Date	May 2022	Marks available	3	Reference code	22M.1.SL.TZ2.4
Level	Standard Level	Paper	Paper 1	Time zone	Time zone 2
Command term	Justify and Determine	Question number	4	Adapted from	N/A

Question

The pH of a solution measures its acidity and can be determined using the formula pH $= - \log_{10} C$ , where $C$ is the concentration of hydronium ions in the solution, measured in moles per litre. A lower pH indicates a more acidic solution.

The concentration of hydronium ions in a particular type of coffee is $1.3 \times 10^{- 5}$ moles per litre.

A different, unknown, liquid has $10$ times the concentration of hydronium ions of the coffee in part (a).

Calculate the pH of the coffee.

[2]

Determine whether the unknown liquid is more or less acidic than the coffee. Justify your answer mathematically.

[3]

Markscheme

(pH =) $- \log_{10} (1.3 \times 10^{- 5})$ (M1)

$4.89 (4.88605 \dots)$ A1

[2 marks]

EITHER

calculating pH

(pH =) $- \log_{10} (10 \times 1.3 \times 10^{- 5})$ (M1)

$3.89 (3.88605 \dots)$ A1

( $3.89 < 4.89$ , therefore) the unknown liquid is more acidic (than coffee). A1

Note: Follow through within the part for the final A1. A correct conclusion must be supported by a mathematical justification linking the $C$ -value to the pH level to earn the final A1; a comparison of $C$ -values only earns M0A0A0.

referencing the graph

The graph of $y = - \log_{10} (x)$ shows that as the value of $x$ increases, the value of $y$ decreases. M1

Since the $C$ -value ( $x$ -value) of the unknown liquid is larger than that of the coffee, the pH level ( $y$ -value) is lower. R1

The unknown liquid is more acidic (than coffee). A1

[3 marks]

Examiners report

Evaluation of logarithms was well done, although the notation when substituting into the logarithmic formula was not always correct, with several candidates including a multiplication sign between the base and the argument. Even when the substitution was done correctly, some candidates still used multiplication, so not fully understanding logarithmic notation.

Several candidates multiplied their answer to part (a) by 10 rather than multiplying the C-value by 10, and several attempted to compare the C-values rather than calculating the pH of the unknown liquid. Most were able to make a correct contextual interpretation of their result.

Syllabus sections

Topic 1—Number and algebra » SL 1.5—Intro to logs

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