Question

Suppose a 0.019 M aqueous solution of oxalic acid (H,C,0,) is prepared. Calculate the equilibrium molarity of C204 find information on the properties of oxalic acid in the ALEKS Data resource You' Round your answer to 2 significant digits.

Answer 1

Answer: The concentration of
`C_2O_4^(2-)` at equilibrium is
`1.52* 10^(-4)M`

Step-by-step explanation:

We are given:

Molarity of oxalic acid solution = 0.019 M

Oxalic acid
`(H_2C_2O_4)` is a weak acid and will dissociate 2 hydrogen ions.

• The chemical equation for the first dissociation of oxalic acid follows:

`H_2C_2O_4(aq.)\rightleftharpoons H^+(aq.)+HC_2O_4^-(aq.)`

Initial: 0.019

At eqllm: 0.019-x x x

The expression of first equilibrium constant equation follows:

`Ka_1=([H^+][HC_2O_4^(-)])/([H_2C_2O_4])`

We know that:

`Ka_1\text{ for }H_2C_2O_4=0.056`

Putting values in above equation, we get:

`0.056=(x* x)/((0.019-x))\\\\x=-0.071,0.015`

Neglecting the negative value of 'x', because concentration cannot be negative.

• The chemical equation for the second dissociation of oxalic acid:

`HC_2O_4^-(aq.)\rightarrow H^+(aq.)+C_2O_4^(2-)(aq.)`

Initial: 0.015

At eqllm: 0.015-y 0.015+y y

The expression of second equilibrium constant equation follows:

`Ka_2=([H^+][C_2O_4^(2-)])/([HC_2O_4^-])`

We know that:

`Ka_2\text{ for }H_2C_2O_4=1.55* 10^(-4)`

Putting values in above equation, we get:

`1.55* 10^(-4)=((0.015+y)* y)/((0.015-y))\\\\y=-0.015,0.000152`

Neglecting the negative value of 'y', because concentration cannot be negative.

So, equilibrium concentration of oxalate ion = y = 0.000152 M

Hence, the concentration of
`C_2O_4^(2-)` at equilibrium is
`1.52* 10^(-4)M`