$pH$ of acids, bases and salts
Exercise 6

Using the table of acid-base couples, calculate the pH of a 2 % (d = 1.0011) acetic acid solution.
Weak acid
If $d$ = 1.0011, then $\rho$ = 1.0011 $\frac{g}{mL}$
Let's take $1\;L$ of solution $S$:
$m_S$ $=$ $\rho \cdot V_S$ $ =$ $ \rho \cdot 1000 $ =
1001.1$\; g$
$m_{}$ =
$\frac{\%_{}\cdot m_S}{100} $ =
$\frac{2 \cdot1001.1}{100} $ =
20.02$\; g$
$n_{}=\frac{m_{}}{M_{}}$ =
$\frac{20.02}{60.05}$ =
0.333$\; mol$
$c_{}$ = $\frac{n_{}}{V_S}$ =
$\frac{0.333}{1}$ =
0.333 $\frac{mol}{L}$
Given $y=[H_3O^+]$
The equation
$y^2$ $+$ $K_a\;y$ $-$ $K_a\;c$ $=$ $0$
becomes:
$y^2$ $+$ $10^{-4.75}y$ $-$ $10^{-4.75} 0.333$ $=$ $0$
and produces:
$y$ $=$ 2.43 10-3
and so:
$pH$ $=$ $ -log\; y$ $ =$ 2.614