Answer: 2.23
Explanation:
The dissociation of acetic acid is as follows:
HCOOH(aq) ⇋ HCOO− (aq) + H+(aq)
The ICE table for the concentrations of ions is given below. From the table, the concentration of HCOO- and H+ can be found out.
HCOOH → HCOO− + H+
Initial Concentration 0.019 M 0 0
Equilibrium Concentration (0.019−x) M x x
Where,
- x is the concentration of the ions at equilibrium.
At equilibrium, dissociation constant can be calculated as follows.
[tex]K_{\mathrm{a}}=\frac{x^{2}}{(0.19-x) \mathrm{M}}[/tex]
At equilibrium, the concentration of x is negligible as compared to that of HCOOH.
Substitute the value of Ka in the above equation.
[tex]\begin{aligned}K_{\mathrm{a}} &=\frac{x^{2}}{0.19-x} \\x &=\sqrt{1.8 \times 10^{-4} \times 0.19} \\&=0.00584 \mathrm{M}\end{aligned}[/tex]
Here, the concentration of hydrogen ion is obtained. From the hydrogen ion concentration, the pH of the solution is found out as follows:
[tex]\begin{aligned}\mathrm{pH} &=-\log \left[\mathrm{H}^{+}\right] \\&=-\log (0.00584 \mathrm{M}) \\&=2.23\end{aligned}[/tex]
Therefore, the pH of 0.19M HCCOH is 2.23
