pH of acids, bases and salts

Exercise 12

    

Using the table of acid-base couples, calculate the $pH$ of a 2.5 %  ($d$ = 1.0275)  sodium phosphate solution named $S$

$PO_4^{3-}$: weak base If $d$ = 1.0275, then $\rho$ = 1.0275 $\frac{g}{mL}$ Let's take $1\;L$ of the solution $S$: $m_S$ $=$ $\rho \cdot V_S$ $ =$ $ \rho \cdot 1000 $ = 1027.5$\; g$ $m_{Na_3PO_4}$ = $\frac{\%_{Na_3PO_4}\cdot m_S}{100} $ = $\frac{2.5 \cdot1027.5}{100} $ = 25.69$\; g$ $n_{Na_3PO_4}$ $=$ $\frac{m_{Na_3PO_4}}{M_{Na_3PO_4}}$ = $\frac{25.69}{163.96}$ = 0.157$ \;mol$ $c_{Na_3PO_4}$ = $\frac{n_{Na_3PO_4}}{V_S}$ = $\frac{0.157}{1}$ = 0.157 $\frac{mol}{L}$ $c_{Na_3PO_4}$ $=$ $c_{PO_4^{3-}}=c$ Given $z=[OH^-]$ The equation $z^2$ $+$ $K_bz$ $-$ $K_bc$ $=$ $0$ becomes: $z^2$ $+$ $10^{-1.7}z$ $-$ $10^{-1.7} 0.157$ $= $ $0$ and produces: $z$ $=$ 4.68 10^-2 and so: $pOH$ $=$ $-log\; z $ = 1.33 $pH$ $=$ $14$ $-$ $pOH$ = 12.67