Calculate the $pH$ of a 1 % ($d$ = 1.0068) potassium hydroxide solution named $S$
$OH^-$ treated as a strong base
If $d$ = 1.0068, then $\rho$ = 1.0068 $\frac{g}{mL}$
Let's take $1\;L$ of solution $S$:
$m_S=\rho \cdot V_S = \rho \cdot 1000 $ =
1006.8$\;g$
$m_{KOH}$ =
$\frac{\%_{KOH}\cdot m_S}{100} $ =
$\frac{1 \cdot1006.8}{100} $ =
10.07$\; g$
$n_{KOH}=\frac{m_{KOH}}{M_{KOH}}$ =
$\frac{10.07}{68.05}$ =
0.148$ \;mol$
The number of moles of $KOH$ is equal to the number of moles of $OH^-$ !
$[OH^-]$ $=$ $\frac{n_{OH^-}}{V_S}$ =
$\frac{0.148}{1}$ =
0.148 $\frac{mol}{L}$
$pH$ $=$ $14$ $+$ $log\;[OH^-]$ =
$14$ $+$ $log \;$0.148 =
13.17
