Question

In a aqueous solution of 4-chlorobutanoic acid , what is the percentage of -chlorobutanoic acid that is dissociated

Answer 1

Step-by-step explanation:

Let assume that the missing aqueous solution of 4-chlorobutanoic acid = 0.76 M

Then, the dissociation of 4-chlorobutanoic acid can be expressed as:

`\mathsf{C_3H_6ClCO_2H }` ⇄
`\mathsf{C_3H_6ClCO_2^-}` +
`\mathsf{H^+}`

The ICE table can be computed as:

`\mathsf{C_3H_6ClCO_2H }` ⇄
`\mathsf{C_3H_6ClCO_2^-}` +
`\mathsf{H^+}`

Initial 0.76 - -

Change -x +x +x

Equilibrium 0.76 - x x x

`K_a = \frac{[\mathsf{C_3H_6ClCO_2^-}] [\mathsf{H^+}]}{\mathsf{[C_3H_6ClCO_2H ]}}`

`K_a = ([x] [x])/( [0.76-x])`

where:

`K_a = 3.02*10^(-5)`

`3.02*10^(-5) = (x^2)/( [0.76-x])`

however, the value of x is so negligible:

0.76 -x = 0.76

Then:

`3.02*10^(-5)*0.76 = x^2`

`x=\sqrt{3.02*10^(-5)*0.76 }`

x = 0.00479 M

∴

`x = \mathsf{[C_3H_6ClCO_2^-] = [H^+]=}` 0.00479 M

`\mathsf{C_3H_6ClCO_2H }` = (0.76 - 0.00479) M

= 0.75521 M

Finally, the percentage of the acid dissociated is;

= ( 0.00479 / 0.76) × 100

= 0.630 M