Answer
a. 8.0
b. 5.3
c. 12.0
d. 11.92
e. 1.33
f. 8.59
Work Step by Step
a. $$pH = -log[H_3O^+] = -log( 1 \times 10^{-8} ) = 8.0 $$
b. $$pH = -log[H_3O^+] = -log( 5 \times 10^{-6} ) = 5.3 $$
c. $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 0.01 } = 1. \times 10^{-12} \space M$$
$$pH = -log[H_3O^+] = -log( 1. \times 10^{-12} ) = 12.0 $$
d. $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 8.0 \times 10^{-3} } = 1.2 \times 10^{-12} \space M$$
$$pH = -log[H_3O^+] = -log( 1.2 \times 10^{-12} ) = 11.92 $$
e. $$pH = -log[H_3O^+] = -log( 4.7 \times 10^{-2} ) = 1.33 $$
f. $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 3.9 \times 10^{-6} } = 2.6 \times 10^{-9} \space M$$
$$pH = -log[H_3O^+] = -log( 2.6 \times 10^{-9} ) = 8.59 $$
