pH of acids, bases and salts

Exercise 7

    

Using the table of acid-base couples, calculate the $pH$ of a 5 %  ($d$ = 1.0408)  iron(III) chloride solution named $S$.

$Fe(H_2O)_6^{3+}: $ weak acid If $d$ = 1.0408, then $\rho$ = 1.0408 $\frac{g}{mL}$ Let's take $1\;L$ of solution $S$: $m_S$ $=$ $\rho \cdot V_S $ $=$ $ \rho \cdot 1000 $ = 1040.8$\; g$ $m_{FeCl_3}$ = $\frac{\%_{FeCl_3}\cdot m_S}{100} $ = $\frac{5 \cdot1040.8}{100} $ = 52.04$\; g$; $n_{FeCl_3}=\frac{m_{FeCl_3}}{M_{FeCl_3}}$ = $\frac{52.04}{162.22}$ = 0.321$\; mol$ $c_{Fe(H_2O)_6^{3+}}$ = $c_{FeCl_3}$ = $\frac{n_{FeCl_3}}{V_S}$ = $\frac{0.321}{1}$ = 0.321 $\frac{mol}{L}$ Given $y=[H_3O^+]$ The equation $y^2$ $+$ $K_a\;y$ $-$ $K_a\;c$ $=$ $0$ becomes: $y^2$ $+$ $10^{-3}y$ $-$ $10^{-3} 0.321$ $=$ $0$ and produces: $y=$ 1.74 10^-2 and so: $pH$ $=$ $-log\; y$ $ =$ 1.759