pH of acids, bases and salts
Exercise 12

Using the table of acid-base couples, calculate the $pH$ of a 2 % ($d$ = 1.0216) sodium phosphate solution named $S$
$PO_4^{3-}$: weak base
If $d$ = 1.0216, then $\rho$ = 1.0216 $\frac{g}{mL}$
Let's take $1\;L$ of the solution $S$:
$m_S$ $=$ $\rho \cdot V_S$ $ =$ $ \rho \cdot 1000 $ =
1021.6$\; g$
$m_{Na_3PO_4}$ =
$\frac{\%_{Na_3PO_4}\cdot m_S}{100} $ =
$\frac{2 \cdot1021.6}{100} $ =
20.43$\; g$
$n_{Na_3PO_4}$ $=$ $\frac{m_{Na_3PO_4}}{M_{Na_3PO_4}}$ =
$\frac{20.43}{163.96}$ =
0.125$ \;mol$
$c_{Na_3PO_4}$ = $\frac{n_{Na_3PO_4}}{V_S}$ =
$\frac{0.125}{1}$ =
0.125 $\frac{mol}{L}$
$c_{Na_3PO_4}$ $=$ $c_{PO_4^{3-}}=c$
Given $z=[OH^-]$
The equation
$z^2$ $+$ $K_bz$ $-$ $K_bc$ $=$ $0$
becomes:
$z^2$ $+$ $10^{-1.7}z$ $-$ $10^{-1.7} 0.125$ $= $ $0$
and produces:
$z$ $=$ 4.09 10-2
and so:
$pOH$ $=$ $-log\; z $ = 1.39
$pH$ $=$ $14$ $-$ $pOH$ = 12.61