What is the ["H"_3"O"^(+)] concentration of a "0.318-M" acetic acid solution?

The ${K}_{a}$ of acetic acid is 1.8xx10^(−5).

Apr 8, 2018

Explanation:

The equation for the equilibrium state of acetic acid in water is-

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

${K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \left[C {H}_{3} C O {O}^{-}\right]}{\left[C {H}_{3} C O O H\right]}$

Now, because of the mol ratio of acetate with hydronium, their concentrations must be equal.

$\implies {K}_{a} = \frac{\left[{H}_{3} {O}^{+}\right] \left[{H}_{3} {O}^{+}\right]}{\left[C {H}_{3} C O O H\right]}$

$\implies {K}_{a} = \frac{{\left[{H}_{3} {O}^{+}\right]}^{2}}{\left[C {H}_{3} C O O H\right]}$

=>[H_3O^+]=sqrt(K_a×[CH_3COOH])

Value of K_a= 1.8×10^-5
And,
Concentration of acetic acid is $0.318$ moles per litre of water. Plugging in the values,

=>[H_3O^+]=sqrt(1.8×10^-5×0.318)

$\implies \left[{H}_{3} {O}^{+}\right] = 0.0023924 \ldots$

Thus,
Concentration of hydronium ions is ≈0.0023

Respected @Michael has reminded this little addition:

Assumption: The initial concentration of Acetic acid before dissociation will obviously be less. But after dissociation, that is formation of ions, the concentration change is almost negligible due to small value of ${K}_{a}$.