# How do you determine the pH of a solution that is 3.85% KOH by mass? Assume that the solution has density of 1.01 g/mL.

Dec 16, 2016

$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right] = 13.84$

#### Explanation:

In $1$ $m L$ of solution, $\left[N a O H\right] = \text{moles of NaOH"/"volume of solution}$

=(1.01*gxx3.85%)/(56.11*g*mol^-1)xx1/(1xx10^-3L)

$= 0.693 \cdot m o l \cdot {L}^{-} 1$

Now $p O H = - {\log}_{10} \left[H {O}^{-}\right]$, $p O H = - {\log}_{10} \left(0.693\right)$

$= 0.159$

But in water, we know that $p H + p O H = 14$,

And thus $14 - 0.159 = p H = 13.84$

The high value of $p H$ is reasonable for a solution of potash.

See here for more on the relationship between $p H$ and acidity.