pH of acids, bases and salts

Exercise 12

    

Using the table of acid-base couples, calculate the $pH$ of a 5 %  ($d$ = 1.0437)  potassium carbonate solution named $S$

$CO_3^{2-}$: weak base If $d$ = 1.0437, then $\rho$ = 1.0437 $\frac{g}{mL}$ Let's take $1\;L$ of the solution $S$: $m_S$ $=$ $\rho \cdot V_S$ $ =$ $ \rho \cdot 1000 $ = 1043.7$\; g$ $m_{K_2CO_3}$ = $\frac{\%_{K_2CO_3}\cdot m_S}{100} $ = $\frac{5 \cdot1043.7}{100} $ = 52.19$\; g$ $n_{K_2CO_3}$ $=$ $\frac{m_{K_2CO_3}}{M_{K_2CO_3}}$ = $\frac{52.19}{138.2}$ = 0.378$ \;mol$ $c_{K_2CO_3}$ = $\frac{n_{K_2CO_3}}{V_S}$ = $\frac{0.378}{1}$ = 0.378 $\frac{mol}{L}$ $c_{K_2CO_3}$ $=$ $c_{CO_3^{2-}}=c$ Given $z=[OH^-]$ The equation $z^2$ $+$ $K_bz$ $-$ $K_bc$ $=$ $0$ becomes: $z^2$ $+$ $10^{-3.68}z$ $-$ $10^{-3.68} 0.378$ $= $ $0$ and produces: $z$ $=$ 8.78 10^-3 and so: $pOH$ $=$ $-log\; z $ = 2.06 $pH$ $=$ $14$ $-$ $pOH$ = 11.94