Answer
a. pH = 4.0
b. pH = 8.5
c. pH = 9.0
d. pH = 3.40
e. pH = 7.17
f. pH = 10.92
Work Step by Step
a.$$pH = -log[H_3O^+] = -log( 1 \times 10^{-4} ) = 4.0 $$
b. $$pH = -log[H_3O^+] = -log( 3 \times 10^{-9} ) = 8.5 $$
c. $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 1 \times 10^{-5} } = 1. \times 10^{-9} \space M$$
$$pH = -log[H_3O^+] = -log( 1. \times 10^{-9} ) = 9.0 $$
d. $$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 2.5 \times 10^{-11} } = 4.0 \times 10^{-4} \space M$$
$$pH = -log[H_3O^+] = -log( 4.0 \times 10^{-4} ) = 3.40 $$
e.$$pH = -log[H_3O^+] = -log( 6.7 \times 10^{-8} ) = 7.17 $$
$$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 8.2 \times 10^{-4} } = 1.2 \times 10^{-11} \space M$$
f.$$pH = -log[H_3O^+] = -log( 1.2 \times 10^{-11} ) = 10.92 $$
