pH of acids, bases and salts

Exercise 12

    

Using the table of acid-base couples, calculate the $pH$ of a 2 %  ($d$ = 1.0103)  sodium acetate (ethanoate) solution named $S$

$CH_3COO^-$: weak base If $d$ = 1.0103, then $\rho$ = 1.0103 $\frac{g}{mL}$ Let's take $1\;L$ of the solution $S$: $m_S$ $=$ $\rho \cdot V_S$ $ =$ $ \rho \cdot 1000 $ = 1010.3$\; g$ $m_{CH_3COONa}$ = $\frac{\%_{CH_3COONa}\cdot m_S}{100} $ = $\frac{2 \cdot1010.3}{100} $ = 20.21$\; g$ $n_{CH_3COONa}$ $=$ $\frac{m_{CH_3COONa}}{M_{CH_3COONa}}$ = $\frac{20.21}{82.04}$ = 0.246$ \;mol$ $c_{CH_3COONa}$ = $\frac{n_{CH_3COONa}}{V_S}$ = $\frac{0.246}{1}$ = 0.246 $\frac{mol}{L}$ $c_{CH_3COONa}$ $=$ $c_{CH_3COO^-}=c$ Given $z=[OH^-]$ The equation $z^2$ $+$ $K_bz$ $-$ $K_bc$ $=$ $0$ becomes: $z^2$ $+$ $10^{-9.25}z$ $-$ $10^{-9.25} 0.246$ $= $ $0$ and produces: $z$ $=$ 1.18 10^-5 and so: $pOH$ $=$ $-log\; z $ = 4.93 $pH$ $=$ $14$ $-$ $pOH$ = 9.07