Respuesta :
Answer: The equilibrium concentration of [tex]H_3O^+[/tex] ion is [tex]8.3064\times 10^{-2}M[/tex]
Explanation:
We are given:
Molarity of oxalic acid solution = 0.20 M
Oxalic acid [tex](H_2C_2O_4)[/tex] is a weak acid and will dissociate 2 hydrogen ions.
- The chemical equation for the first dissociation of oxalic acid follows:
[tex]H_2C_2O_4(aq.)+H_2O\rightleftharpoons H_3O^+(aq.)+HC_2O_4^-(aq.)[/tex]
Initial: 0.20
At eqllm: 0.20-x x x
The expression of first equilibrium constant equation follows:
[tex]Ka_1=\frac{[H_3O^+][HC_2O_4^{-}]}{[H_2C_2O_4]}[/tex]
We know that:
[tex]Ka_1\text{ for }H_2C_2O_4=0.059[/tex]
Putting values in above equation, we get:
[tex]0.059=\frac{x\times x}{(0.20-x)}\\\\x=-0.142,0.083[/tex]
Neglecting the negative value of 'x', because concentration cannot be negative.
So, equilibrium concentration of hydronium ion = x = 0.083 M
- The chemical equation for the second dissociation of oxalic acid:
[tex]HC_2O_4^-(aq.)+H_2O\rightarrow H_3O^+(aq.)+C_2O_4^{2-}(aq.)[/tex]
Initial: 0.083
At eqllm: 0.083-y 0.083+y y
The expression of second equilibrium constant equation follows:
[tex]Ka_2=\frac{[H_3O^+][C_2O_4^{2-}]}{[HC_2O_4^-]}[/tex]
We know that:
[tex]Ka_2\text{ for }H_2C_2O_4=6.4\times 10^{-5}[/tex]
Putting values in above equation, we get:
[tex]6.4\times 10^{-5}=\frac{(0.083+y)\times y}{(0.083-y)}\\\\y=-0.083,0.0000639[/tex]
Neglecting the negative value of 'x', because concentration cannot be negative.
So, equilibrium concentration of hydronium ion = y = 0.0000639 M
Total concentration of hydronium ion = [x + y] = [0.083 + 0.0000639] = 0.0830639 M
Hence, the equilibrium concentration of [tex]H_3O^+[/tex] ion is [tex]8.3064\times 10^{-2}M[/tex]
