pH of acids, bases and salts
Exercise 12

Using the table of acid-base couples, calculate the $pH$ of a 1.5 % ($d$ = 1.0138) sodium carbonate solution named $S$
$CO_3^{2-}$: weak base
If $d$ = 1.0138, then $\rho$ = 1.0138 $\frac{g}{mL}$
Let's take $1\;L$ of the solution $S$:
$m_S$ $=$ $\rho \cdot V_S$ $ =$ $ \rho \cdot 1000 $ =
1013.8$\; g$
$m_{Na_2CO_3}$ =
$\frac{\%_{Na_2CO_3}\cdot m_S}{100} $ =
$\frac{1.5 \cdot1013.8}{100} $ =
15.21$\; g$
$n_{Na_2CO_3}$ $=$ $\frac{m_{Na_2CO_3}}{M_{Na_2CO_3}}$ =
$\frac{15.21}{106}$ =
0.143$ \;mol$
$c_{Na_2CO_3}$ = $\frac{n_{Na_2CO_3}}{V_S}$ =
$\frac{0.143}{1}$ =
0.143 $\frac{mol}{L}$
$c_{Na_2CO_3}$ $=$ $c_{CO_3^{2-}}=c$
Given $z=[OH^-]$
The equation
$z^2$ $+$ $K_bz$ $-$ $K_bc$ $=$ $0$
becomes:
$z^2$ $+$ $10^{-3.68}z$ $-$ $10^{-3.68} 0.143$ $= $ $0$
and produces:
$z$ $=$ 5.37 10-3
and so:
$pOH$ $=$ $-log\; z $ = 2.27
$pH$ $=$ $14$ $-$ $pOH$ = 11.73