Answer:
2.9352
Explanation:
Given that:
[tex]K_{a}=1.8\times 10^{-6}[/tex]
Concentration = 0.75 M
Consider the ICE take for the dissociation of acetic acid as:
CH₃COOH ⇄ H⁺ + CH₃COO⁻
At t=0 0.75 - -
At t =equilibrium (0.75-x) x x
The expression for dissociation constant of acetic acid is:
[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}[/tex]
[tex]1.8\times 10^{-6}=\frac {x^2}{0.75-x}[/tex]
For quadratic equation as:
[tex]110^{6}x^2+1.8x-1.35=0[/tex]
Solving for x, we get:
x = 0.001161 M
pH = -log[H⁺] = -log(0.001161) = 2.9352
