Answers: 1
![What is the pOH of a solution at 25.0∘C with [H3O+]=4.8×10−6 M?](/tpl/images/0717/5611/6660c.jpg)
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of a solution can be found from the hydroxide ion concentration ![\rm \left[OH^{-}\right]](/tpl/images/0717/5611/0cbc6.png)
with the following equation:![\displaystyle \rm pOH = -\log_{10} \rm \left[OH^{-}\right]](/tpl/images/0717/5611/2fb11.png)
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, relates the hydroxide ion concentration ![\rm \left[{H_3O}^{+}\right]](/tpl/images/0717/5611/3118a.png)
:![K_\text{w} = \rm \left[{H_3O}^{+}\right] \cdot \rm \left[OH^{-}\right]](/tpl/images/0717/5611/ac757.png)
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, 
. For this particular ![\rm \left[{H_3O}^{+}\right] = 4.8 \times 10^{-6}\; \rm mol \cdot L^{-1}](/tpl/images/0717/5611/eba66.png)
.![\begin{aligned}\left[\mathrm{OH}^{-}\right] &= \frac{K_\text{w}}{\rm \left[{H_3O}^{+}\right]} \\ &= \frac{1.0 \times 10^{-14}}{4.8 \times 10^{-6}}\; \rm mol\cdot L^{-1} \approx 2.08333 \times 10^{-9}\; \rm mol\cdot L^{-1}\end{aligned}](/tpl/images/0717/5611/cbcf8.png)
.![\begin{aligned}\rm pOH &= -\log_{10} \rm \left[OH^{-}\right] \\ &\approx -\log_{10} \left(4.8 \times 10^{-6}\right) \approx 8.68\end{aligned}](/tpl/images/0717/5611/01d36.png)
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