Answer
a. 7.00
b. 2.38
c. 4.00
d. 5.93
Work Step by Step
a. $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 1.0 \times 10^{-7} } = 1.0 \times 10^{-7} \space M$$
$$pH = -log[H_3O^+] = -log( 1.0 \times 10^{-7} ) = 7.00 $$
b. $$pH = -log[H_3O^+] = -log( 4.2 \times 10^{-3} ) = 2.38 $$
c. $$pH = -log[H_3O^+] = -log( 0.0001 ) = 4.00 $$
d. $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 8.5 \times 10^{-9} } = 1.18 \times 10^{-6} \space M$$
$$pH = -log[H_3O^+] = -log( 1.18 \times 10^{-6} ) = 5.93 $$
