Answer
Solution $X$: $[OH−]=1×10^{-5}M$
Solution $Y$: $[OH−]=1×10^{-7}M$
Work Step by Step
As we have calculated on 10.79b:
Solution X: $[H_3O^+]=1×10^{−9}M$
Solution Y: $[H_3O^+]=1×10^{−7}M$
For the first solution (X):
1. Write the Kw expression for the water equilibrium. Then substitute the value for $[H_3O^+]$, and solve for $[OH^−]$:
$[H_3O^+][OH^−]=K_w=10^{−14}$
$1×10^{−9}\times[OH−]=10^{−14}$
$[OH−]=\frac{10^{−14}}{1×10^{−9}}$
$[OH−]=1×10^{-5}M$
For the second solution (Y):
2. Write the Kw expression for the water equilibrium. Then substitute the value for $[H_3O^+]$, and solve for $[OH^−]$:
$[H_3O^+][OH^−]=K_w=10^{−14}$
$1×10^{−7}\times[OH−]=10^{−14}$
$[OH−]=\frac{10^{−14}}{1×10^{−7}}$
$[OH−]=1×10^{-7}M$
