Answer
(a) $[OH^-] = 7.7 \times 10^{-12}M$: Acidic solution;
(b) $[OH^-] = 1.1 \times 10^{-3}M$: Basic solution;
(c) $[OH^-] = 1.9 \times 10^{-11}M$: Acidic solution;
(d) $[OH^-] = 1.6 \times 10^{-6}M$: Basic solution;
Work Step by Step
If $[H_3O^+] > [OH^-]$, the solution is acidic.
If $[OH^-] > [H_3O^+]$, the solution is basic.
1. Calculate the hydroxide ion concentration:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$1.3 \times 10^{-3} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{1.3 \times 10^{-3}}$
$[OH^-] = 7.7 \times 10^{-12}$
2. Calculate the hydroxide ion concentration:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$9.1 \times 10^{-12} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{9.1 \times 10^{-12}}$
$[OH^-] = 1.1 \times 10^{-3}$
3. Calculate the hydroxide ion concentration:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$5.2 \times 10^{-4} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{5.2 \times 10^{-4}}$
$[OH^-] = 1.9 \times 10^{-11}$
4. Calculate the hydroxide ion concentration:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$6.1 \times 10^{-9} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{6.1 \times 10^{-9}}$
$[OH^-] = 1.6 \times 10^{-6}$
