Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 18 - Ionic Equilibria I: Acids and Bases - Exercises - The Autoionization of Water - Page 743: 14

Answer

6(b) $$[H_3O^+] = 2.2 \times 10^{-13} \space M$$ 7(a) $$[H_3O^+] = 4.8 \times 10^{-13} \space M$$ 8(b) $$[H_3O^+] = 8.5 \times 10^{-13} \space M$$

Work Step by Step

6(b) $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 0.045 } = 2.2 \times 10^{-13} \space M$$ 7(a) $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 0.0210 } = 4.8 \times 10^{-13} \space M$$ 8(b) $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 0.0117 } = 8.5 \times 10^{-13} \space M$$ --- All these concentrations are lower than $[H_3O^+] = 1.0 \times 10^{-7}$, which makes sense, since they are all solutions made of basic compounds, and therefore, basic solutions.
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