# What is pOH of a solution prepared by adding a 15.8*mg mass of potassium hydroxide to 11*mL of solution?

May 9, 2017

$p O H = 1.60$

#### Explanation:

Now..........$p O H = - {\log}_{10} \left[H {O}^{-}\right]$...........

And we have $15.8 \cdot m g$ $K O H$ dissolved in $11 \cdot m L$ of solution.

$\left[H {O}^{-}\right] = \frac{\frac{15.8 \times {10}^{-} 3 \cdot g}{56.11 \cdot g \cdot m o {l}^{-} 1}}{11 \cdot m L \times {10}^{-} 3 \cdot m L \cdot {L}^{-} 1}$

$= 2.56 \times {10}^{-} 2 \cdot m o l \cdot {L}^{-} 1$.

But by definition, $p O H = - {\log}_{10} \left(2.56 \times {10}^{-} 2\right) = 1.60$

What is $p H$ of this solution? You should be able to answer immediately. Why? Because in aqueous solution, $p H + p O H = 14$.

Also see this old problem here for more treatment.