Calculate the $pH$ of a 3.5 % ($d$ = 1.0373) sodium hydroxide solution named $S$
$OH^-$ treated as a strong base
If $d$ = 1.0373, then $\rho$ = 1.0373 $\frac{g}{mL}$
Let's take $1\;L$ of solution $S$:
$m_S=\rho \cdot V_S = \rho \cdot 1000 $ =
1037.3$\;g$
$m_{NaOH}$ =
$\frac{\%_{NaOH}\cdot m_S}{100} $ =
$\frac{3.5 \cdot1037.3}{100} $ =
36.31$\; g$
$n_{NaOH}=\frac{m_{NaOH}}{M_{NaOH}}$ =
$\frac{36.31}{40}$ =
0.908$ \;mol$
The number of moles of $NaOH$ is equal to the number of moles of $OH^-$ !
$[OH^-]$ $=$ $\frac{n_{OH^-}}{V_S}$ =
$\frac{0.908}{1}$ =
0.908 $\frac{mol}{L}$
$pH$ $=$ $14$ $+$ $log\;[OH^-]$ =
$14$ $+$ $log \;$0.908 =
13.958
