Question

Hc2h3o2(aq)+h2o(l)⇌h3o+(aq)+c2h3o−2(aq) kc=1.8×10−5 at 25∘c part a if a solution initially contains 0.260 m hc2h3o2, what is the equilibrium concentration of h3o+ at 25∘c? express your answer using two significant figures.

Answer 1

Answer : The equilibrium concentration of $H_3O^+$ at $25^oC$ is, $2.154\times 10^{-3}m$.

Solution : Given,

Equilibrium constant, $K_c=1.8\times 10^{-5}$

Initial concentration of $HC_2H_3O_2$ = 0.260 m

Let, the 'x' mol/L of $H_3O^+$ are formed and at same time 'x' mol/L of $HC_2H_3O_2$ are also formed.

The equilibrium reaction is,

                  $HC_2H_3O_2(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+C_2H_3O_2^-(aq)$

Initially                0.260 m                       0                 0

At equilibrium    (0.260 - x)                     x                 x

The expression for equilibrium constant for a given reaction is,

$K_c=\frac{[H_3O^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}$

Now put all the given values in this expression, we get

$1.8\times 10^{-5}=\frac{(x)(x)}{(0.260-x)}$

By rearranging the terms, we get the value of 'x'.

$x=2.154\times 10^{-3}m$

Therefore, the equilibrium concentration of $H_3O^+$ at $25^oC$ is, $2.154\times 10^{-3}m$.