Chemistry: Molecular Approach (4th Edition)

Published by Pearson
ISBN 10: 0134112830
ISBN 13: 978-0-13411-283-1

Chapter 16 - Exercises - Page 772: 59

Answer

a. The mass of HI is equal to 1.8 g b. The mass of HI is equal to 0.57 g c. The mass of HI is equal to 0.045 g

Work Step by Step

1. Calculate the molar mass of HI: $ HI $ : ( 1.008 $\times$ 1 )+ ( 126.9 $\times$ 1 )= 127.9 g/mol a. $$[H_3O^+] = 10^{-pH} = 10^{-1.25} = 5.6 \times 10^{-2} \space M$$ Since HI is a strong acid, $[HI] = 5.6 \times 10^{-2} \space M$ $$0.250 \space L \times \frac{5.6 \times 10^{-2} \space mol}{ 1 \space L} \times \frac{127.9 \space g}{1 \space mol}= 1.8 \space g$$ b. $$[H_3O^+] = 10^{-pH} = 10^{-1.75} = 1.8 \times 10^{-2} \space M$$ Since HI is a strong acid, $[HI] = 1.8 \times 10^{-2} \space M$ $$0.250 \space L \times \frac{1.8 \times 10^{-2} \space mol}{ 1 \space L} \times \frac{127.9 \space g}{1 \space mol}= 0.57 \space g$$ c. $$[H_3O^+] = 10^{-pH} = 10^{-2.85} = 1.4 \times 10^{-3} \space M$$ Since HI is a strong acid, $[HI] = 1.4 \times 10^{-3} \space M$ $$0.250 \space L \times \frac{1.4 \times 10^{-3} \space mol}{ 1 \space L} \times \frac{127.9 \space g}{1 \space mol}= 0.045 \space g$$
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