Calculate the $pH$ of a 2 % ($d$ = 1.0207) sodium hydroxide solution named $S$
$OH^-$ treated as a strong base
If $d$ = 1.0207, then $\rho$ = 1.0207 $\frac{g}{mL}$
Let's take $1\;L$ of solution $S$:
$m_S=\rho \cdot V_S = \rho \cdot 1000 $ =
1020.7$\;g$
$m_{NaOH}$ =
$\frac{\%_{NaOH}\cdot m_S}{100} $ =
$\frac{2 \cdot1020.7}{100} $ =
20.41$\; g$
$n_{NaOH}=\frac{m_{NaOH}}{M_{NaOH}}$ =
$\frac{20.41}{40}$ =
0.51$ \;mol$
The number of moles of $NaOH$ is equal to the number of moles of $OH^-$ !
$[OH^-]$ $=$ $\frac{n_{OH^-}}{V_S}$ =
$\frac{0.51}{1}$ =
0.51 $\frac{mol}{L}$
$pH$ $=$ $14$ $+$ $log\;[OH^-]$ =
$14$ $+$ $log \;$0.51 =
13.708
