Answer
(a) $[H_3O^+] = 3.7 \times 10^{- 3}M$, Acidic solution;
(b) $[H_3O^+] = 4.0 \times 10^{- 13}M$, Basic solution;
(c) $[H_3O^+] = 9.1 \times 10^{- 5}M$, Acidic solution;
(d) $[H_3O^+] = 3.0 \times 10^{- 11}M$, Basic solution
Work Step by Step
If $[H_3O^+] > [OH^-]$, the solution is acidic.
If $[OH^-] > [H_3O^+]$, the solution is basic.
--------
(a)
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 2.7 \times 10^{- 12} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 2.7 \times 10^{- 12}}$
$[H_3O^+] = 3.7 \times 10^{- 3}$
(b)
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 2.5 \times 10^{- 2} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 2.5 \times 10^{- 2}}$
$[H_3O^+] = 4.0 \times 10^{- 13}$
(c)
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 1.1 \times 10^{- 10} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 1.1 \times 10^{- 10}}$
$[H_3O^+] = 9.1 \times 10^{- 5}$
(d)
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 3.3 \times 10^{- 4} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 3.3 \times 10^{- 4}}$
$[H_3O^+] = 3.0 \times 10^{- 11}$
