pH of acids, bases and salts
Exercise 12

Using the table of acid-base couples, calculate the $pH$ of a 3 % ($d$ = 1.0153) sodium acetate (ethanoate) solution named $S$
$CH_3COO^-$: weak base
If $d$ = 1.0153, then $\rho$ = 1.0153 $\frac{g}{mL}$
Let's take $1\;L$ of the solution $S$:
$m_S$ $=$ $\rho \cdot V_S$ $ =$ $ \rho \cdot 1000 $ =
1015.3$\; g$
$m_{CH_3COONa}$ =
$\frac{\%_{CH_3COONa}\cdot m_S}{100} $ =
$\frac{3 \cdot1015.3}{100} $ =
30.46$\; g$
$n_{CH_3COONa}$ $=$ $\frac{m_{CH_3COONa}}{M_{CH_3COONa}}$ =
$\frac{30.46}{82.04}$ =
0.371$ \;mol$
$c_{CH_3COONa}$ = $\frac{n_{CH_3COONa}}{V_S}$ =
$\frac{0.371}{1}$ =
0.371 $\frac{mol}{L}$
$c_{CH_3COONa}$ $=$ $c_{CH_3COO^-}=c$
Given $z=[OH^-]$
The equation
$z^2$ $+$ $K_bz$ $-$ $K_bc$ $=$ $0$
becomes:
$z^2$ $+$ $10^{-9.25}z$ $-$ $10^{-9.25} 0.371$ $= $ $0$
and produces:
$z$ $=$ 1.44 10-5
and so:
$pOH$ $=$ $-log\; z $ = 4.84
$pH$ $=$ $14$ $-$ $pOH$ = 9.16