General Chemistry: Principles and Modern Applications (10th Edition)

Published by Pearson Prentice Hal
ISBN 10: 0132064529
ISBN 13: 978-0-13206-452-1

Chapter 16 - Acids and Bases - Example 16-3 - Calculating Ion Concentrations in an Aqueous Solution of a Strong Acid - Page 707: Practice Example B

Answer

$$pH =1.466 $$

Work Step by Step

$$Volume = 535 \space mL \times \frac{1 \space L}{1000 \space mL} = 0.535 \space L$$ $$Pressure = 747 \space mmHg \times \frac{1 \space atm}{760 \space mmHg} = 0.983 \space atm$$ $$Temperature/K = 26.5 + 273.15 = 299.7 \space K$$ 1. According to the Ideal Gas Law: $$n = \frac{PV}{RT} = \frac{( 0.983 \space atm )( 0.535 \space L )}{( 0.08206 \space atm \space L \space mol^{-1} \space K^{-1} )( 299.7 \space K)}$$ $$n = 0.0214 \space mol$$ 2. Calculate the molarity of the new solution: $$[HCl] = \frac{0.0214 \space mol}{625 \space mL} \times \frac{1000 \space mL}{1 \space L} = 0.0342 \space M$$ 3. Since HCl is a strong acid: $[H_3O^+] = [HCl]$ $$pH = -log[H_3O^+] = -log( 0.0342 ) = 1.466 $$
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