# At equilibrium, it was found that CH_3COOH = 0.8537 M, CH_3COO^- = 0.0040 M, and H_3O^+ = 0.0040 M . How do you calculate the acidity constant of CH_3COOH and the pH value of the solution?

Oct 30, 2016

pH=2.40;pK_a=4.73.

#### Explanation:

$p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$ $=$ $- {\log}_{10} \left(0.004\right) = 2.40$

Acetic acid undergoes the following reaction:

${H}_{3} C - C {O}_{2} H + {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} C - C {O}_{2}^{-} + {H}_{3} {O}^{+}$

And we write the equilibrium expression in the usual way:

([H_3O^+][""^(-)OAc])/([HOAc]) $=$ ${K}_{a}$

The concentration of water does not appear in this expression, because $\left[{H}_{2} O\right]$ is effectively constant.

$=$ $\frac{0.004 \cdot m o l \cdot {L}^{-} 1 \times 0.004 \cdot m o l \cdot {L}^{-} 1}{0.8537 \cdot m o l \cdot {L}^{-} 1}$ $=$ $1.87 \times {10}^{-} 5$

$p {K}_{a} = - {\log}_{10} {K}_{a} = - {\log}_{10} \left(1.87 \times {10}^{-} 5\right)$ $=$ $4.73$.