pH of acids, bases and salts

Exercise 7

    

Using the table of acid-base couples, calculate the $pH$ of a 1.5 %  ($d$ = 1.0137)  zinc sulfate solution named $S$.

$Zn(H_2O)_6^{2+}: $ weak acid If $d$ = 1.0137, then $\rho$ = 1.0137 $\frac{g}{mL}$ Let's take $1\;L$ of solution $S$: $m_S$ $=$ $\rho \cdot V_S $ $=$ $ \rho \cdot 1000 $ = 1013.7$\; g$ $m_{ZnSO_4}$ = $\frac{\%_{ZnSO_4}\cdot m_S}{100} $ = $\frac{1.5 \cdot1013.7}{100} $ = 15.21$\; g$; $n_{ZnSO_4}=\frac{m_{ZnSO_4}}{M_{ZnSO_4}}$ = $\frac{15.21}{161.44}$ = 0.094$\; mol$ $c_{Zn(H_2O)_6^{2+}}$ = $c_{ZnSO_4}$ = $\frac{n_{ZnSO_4}}{V_S}$ = $\frac{0.094}{1}$ = 0.094 $\frac{mol}{L}$ Given $y=[H_3O^+]$ The equation $y^2$ $+$ $K_a\;y$ $-$ $K_a\;c$ $=$ $0$ becomes: $y^2$ $+$ $10^{-8.96}y$ $-$ $10^{-8.96} 0.094$ $=$ $0$ and produces: $y=$ 1.02 10^-5 and so: $pH$ $=$ $-log\; y$ $ =$ 4.991