The rate equation for the acid-catalysed reaction between iodine and propanone is: rate = k [H^+] [C_3H_6O] The rate of reaction was measured for a mixture of iodine, propanone and sulfuric acid at pH = 0.70 In a second mixture the concentration of the sulfuric acid was different but the concentrations of iodine and propanone were unchanged - AQA - A-Level Chemistry - Question 27 - 2017 - Paper 3

To determine the pH of the second mixture, we start by analyzing how the rate changes with pH.

1. The initial rate at pH 0.70 can be calculated. This corresponds to a hydrogen ion concentration of:

   $[H^+] = 10^{-0.70} \approx 0.20 \, M$

2. In the second mixture, the rate of reaction is a quarter of the original rate, meaning:

   $\text{New rate} = \frac{1}{4} \times \text{rate at pH 0.70}$

3. Since the concentrations of iodine and propanone are unchanged, the only variable affecting the rate is the hydrogen ion concentration. We can express the new rate with the new hydrogen concentration as:

   $\text{New rate} \propto [H^+]^a$

   where $a$ is the order of reaction with respect to hydrogen ions, which is assumed to be 1 for simplicity. Thus:

   $\frac{1}{4} = \left( \frac{[H^+]_{new}}{[H^+]_{original}} \right)^a = \left( \frac{[H^+]_{new}}{0.20} \right)^a$

4. Taking the fourth root gives us:

   $[H^+]_{new} = \frac{0.20}{\sqrt[4]{4}} = \frac{0.20}{\sqrt{2}} \approx 0.14 \, M$

5. Finally, we calculate the pH from the new concentration:

   $pH = -\log_{10}(0.14) \approx 0.85$

6. The options provided are: A. 1.00, B. 1.30, C. 1.40, D. 2.80. The closest value to 0.85 is B. 1.30.