Question

The equilibrium constant for the dissociation of acetic acid at 25oC is 1.85x10-5. What will be the equilibrium concentration of H+ at 25oC, if the initial concentration of acetic acid was 0.100 M? CH3COOH (aq) ↔ CH3COO- (aq) + H+ (aq) Give your answer to two significant figures and in decimal form.

Answer 1

Answer:

Equilibrium concentration of $H^{+}$ is $1.4\times 10^{-3}M$

Explanation:

Construct an ICE table to calculate changes in concentration at equilibrium.

   $CH_{3}COOH\rightleftharpoons CH_{3}COO^{-}+H^{+}$

I:     0.1                                                          0                     0

C:     -x                                                          +x                   +x

E:    0.1-x                                                       x                      x

$\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}$= $K_{a}(CH_{3}COOH)$

(species inside third bracket represent equilibrium concentrations)

$\frac{x^{2}}{0.1-x}=1.85\times 10^{-5}$

or, $x^{2}+(1.85\times 10^{-5})x-(1.85\times 10^{-6})=0$

So, $x=\frac{-(1.85\times 10^{-5})+\sqrt{(1.85\times 10^{-5})^{2}+(4\times 1.85\times 10^{-6})}}{2}$M

or, $x=1.4\times 10^{-3}M$

So, equilibrium concentration of $H^{+}$=$[H^{+}]$=x=$1.4\times 10^{-3}M$