Calculate the $pH$ of a 3 % ($d$ = 1.0318) sodium hydroxide solution named $S$
$OH^-$ treated as a strong base
If $d$ = 1.0318, then $\rho$ = 1.0318 $\frac{g}{mL}$
Let's take $1\;L$ of solution $S$:
$m_S=\rho \cdot V_S = \rho \cdot 1000 $ =
1031.8$\;g$
$m_{NaOH}$ =
$\frac{\%_{NaOH}\cdot m_S}{100} $ =
$\frac{3 \cdot1031.8}{100} $ =
30.95$\; g$
$n_{NaOH}=\frac{m_{NaOH}}{M_{NaOH}}$ =
$\frac{30.95}{40}$ =
0.774$ \;mol$
The number of moles of $NaOH$ is equal to the number of moles of $OH^-$ !
$[OH^-]$ $=$ $\frac{n_{OH^-}}{V_S}$ =
$\frac{0.774}{1}$ =
0.774 $\frac{mol}{L}$
$pH$ $=$ $14$ $+$ $log\;[OH^-]$ =
$14$ $+$ $log \;$0.774 =
13.889
