$pH$ of acids, bases and salts

Exercise 6

    

Using the table of acid-base couples, calculate the pH of a 2 %  (d = 1.0023) lactic acid solution.

Weak acid If $d$ = 1.0023, then $\rho$ = 1.0023 $\frac{g}{mL}$ Let's take $1\;L$ of solution $S$: $m_S$ $=$ $\rho \cdot V_S$ $ =$ $ \rho \cdot 1000 $ = 1002.3$\; g$ $m_{}$ = $\frac{\%_{}\cdot m_S}{100} $ = $\frac{2 \cdot1002.3}{100} $ = 20.05$\; g$ $n_{}=\frac{m_{}}{M_{}}$ = $\frac{20.05}{90.08}$ = 0.223$\; mol$ $c_{}$ = $\frac{n_{}}{V_S}$ = $\frac{0.223}{1}$ = 0.223 $\frac{mol}{L}$ Given $y=[H_3O^+]$ The equation $y^2$ $+$ $K_a\;y$ $-$ $K_a\;c$ $=$ $0$ becomes: $y^2$ $+$ $10^{-3.3}y$ $-$ $10^{-3.3} 0.223$ $=$ $0$ and produces: $y$ $=$ 1.03 10^-2 and so: $pH$ $=$ $ -log\; y$ $ =$ 1.987