Answer
(a) $[H_3O^+] = 4.8 \times 10^{- 4}M$: Acidic solution.
(b) $[H_3O^+] = 1.3 \times 10^{- 6}M$: Acidic solution.
(c) $[H_3O^+] = 4.8 \times 10^{- 11}M$: Basic solution.
(d) $[H_3O^+] = 1.0 \times 10^{- 12}M$: Basic solution.
Work Step by Step
If $[H_3O^+] > [OH^-]$, the solution is acidic.
If $[OH^-] > [H_3O^+]$, the solution is basic.
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(a)
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 2.1 \times 10^{- 11} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 2.1 \times 10^{- 11}}$
$[H_3O^+] = 4.8 \times 10^{- 4}$
(b)
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 7.5 \times 10^{- 9} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 7.5 \times 10^{- 9}}$
$[H_3O^+] = 1.3 \times 10^{- 6}$
(c)
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 2.1 \times 10^{- 4} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 2.1 \times 10^{- 4}}$
$[H_3O^+] = 4.8 \times 10^{- 11}$
(d)
$[OH^-] * [H_3O^+] = Kw = 10^{-14}$
$ 1.0 \times 10^{- 2} * [H_3O^+] = 10^{-14}$
$[H_3O^+] = \frac{10^{-14}}{ 1 \times 10^{- 2}}$
$[H_3O^+] = 1.0 \times 10^{- 12}$
