Answer
a. 7.70
b. 1.30
c. 10.54
d. 11.73
Work Step by Step
a. $$pH = -log[H_3O^+] = -log( 2.0 \times 10^{-8} ) = 7.70 $$
b. $$pH = -log[H_3O^+] = -log( 5.0 \times 10^{-2} ) = 1.30 $$
c. $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 3.5 \times 10^{-4} } = 2.9 \times 10^{-11} \space M$$
$$pH = -log[H_3O^+] = -log( 2.9 \times 10^{-11} ) = 10.54 $$
d.$$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 5.4 \times 10^{-3} } = 1.85 \times 10^{-12} \space M$$
$$pH = -log[H_3O^+] = -log( 1.85 \times 10^{-12} ) = 11.73 $$
