Answer
Solution A: $[OH^-] = 1\times 10^{-10}M$
Solution B: $[OH^-] = 1 \times 10^{-8}M$
Work Step by Step
As we have calculated on 10.79b:
Solution A: $[H_3O^+] = 1 \times 10^{- 4}M$
Solution B: $[H_3O^+] = 1 \times 10^{-6}M$
For the first solution (A):
1. Write the $K_w$ expression for the water equilibrium. Then substitute the value for $[H_3O^+]$, and solve for $[OH^-]$:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$1 \times 10^{-4} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{1 \times 10^{-4}}$
$[OH^-] = 1\times 10^{-10}M$
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For the second solution (B):
2. Write the $K_w$ expression for the water equilibrium. Then substitute the value for $[H_3O^+]$, and solve for $[OH^-]$:
$[H_3O^+] * [OH^-] = Kw = 10^{-14}$
$1 \times 10^{-8} * [OH^-] = 10^{-14}$
$[OH^-] = \frac{10^{-14}}{1 \times 10^{-8}}$
$[OH^-] = 1 \times 10^{-8}M$
