Answer : The equilibrium concentration of
at
is,
.
Solution : Given,
Equilibrium constant, 
Initial concentration of
= 0.260 m
Let, the 'x' mol/L of
are formed and at same time 'x' mol/L of
are also formed.
The equilibrium reaction is,

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]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BC_2H_3O_2%5E-%5D%7D%7B%5BHC_2H_3O_2%5D%7D)
Now put all the given values in this expression, we get

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

Therefore, the equilibrium concentration of
at
is,
.