Which of the following could be added to 100 mL of 0.01 mol L⁻¹ hydrochloric acid solution to change its pH to 4? (A) 900 mL of water (B) 900 mL of 0.01 mol L⁻¹ hydrochloric acid (C) 9900 mL of water (D) 9900 mL of 0.01 mol L⁻¹ hydrochloric acid - HSC - SSCE Chemistry - Question 12 - 2016 - Paper 1

To achieve a pH of 4, the concentration of H⁺ ions should be:

Using the dilution equation: $C_1V_1 = C_2V_2$ where:

• $C_1 = 0.01 \text{ mol L⁻¹}$ (initial concentration)
• $C_2 = 10^{-4} \text{ mol L⁻¹}$ (final concentration)
• $V_1 = 100 \text{ mL} = 0.1 \text{ L}$ (initial volume) We need to find $V_2$, the final volume needed to achieve this concentration. Plugging in the values: $0.01 \times 0.1 = 10^{-4} \times V_2$ Solving for $V_2$ gives: $V_2 = \frac{0.01 \times 0.1}{10^{-4}} = 100 \text{ L}$

Since the final volume needed is 100 L and the initial volume is 0.1 L, the volume of water to add is: $100 \text{ L} - 0.1 \text{ L} = 99.9 \text{ L} = 9900 \text{ mL}$

From the provided options, adding 9900 mL of water (option C) would effectively change the pH to 4.