Question

What is the pH of a 0.75M acetic acid solution if Kc = 1.8 x 10-6 ?

Answer 1

2.9352

Step-by-step explanation:

Given that:

`K_(a)=1.8* 10^(-6)`

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:

`K_(a)=\frac {\left [ H^(+) \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}`

`1.8* 10^(-6)=\frac {x^2}{0.75-x}`

For quadratic equation as:

`110^(6)x^2+1.8x-1.35=0`

Solving for x, we get:

x = 0.001161 M

pH = -log[H⁺] = -log(0.001161) = 2.9352