## Chemistry: Molecular Approach (4th Edition)

Published by Pearson

# Chapter 16 - Exercises - Page 772: 61

#### Answer

The pH of this HCl solution is equal to 2.21.

#### Work Step by Step

1. Convert the volume to L: $$Volume = 224 \space mL \times \frac{1 \space L}{1000 \space mL} = 0.224 \space L$$ 2 . Convert the temperature to K: $$T/K = 27.2 + 273.15 = 300.3$$ 3 . Find the amount of moles of this gas: $$n = \frac{PV}{RT} = \frac{( 1.02 )( 0.224 )}{(0.0821)( 300.3 )}$$ $n = 0.00927 \space mol$ 4. Calculate the concentration in 1.5 L: $$M = \frac{9.27 \times 10^{-3}}{1.5} = 6.18 \times 10^{-3} \space M$$ 5. Since $HCl$ is a strong acid, all its concentration is going to produce $H_3O^+$: $$[H_3O^+] = [ HCl ] = 6.18 \times 10^{-3} \space M$$ 6. Calculate the pH: $$pH = -log[H_3O^+] = -log( 6.18 \times 10^{-3} ) = 2.21$$

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