The rate equation for the acid-catalysed iodination of propanone is:
rate=k[H+][CH3COCH3] \text{rate} = k [\text{H}^+] [\text{CH}_3\text{COCH}_3] rate=k[H+][CH3COCH3]In an experiment, the initial rate of this reaction was measured at 298 K298\text{ K}298 K using a mixture of propanone and hydrochloric acid. The pH\text{pH}pH of this reaction mixture was 2.702.702.70.
In a second experiment at the same temperature, the concentration of propanone was kept constant, but the concentration of hydrochloric acid was different. The rate of this second reaction was found to be exactly one-fifth (15\frac{1}{5}51) of the rate in the first experiment.
Determine the pH\text{pH}pH of this second reaction mixture. Give your answer to 2 decimal places.
