Question

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

Answer 1

Answer:

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  

$K_a = \dfrac{[\mathsf{C_3H_6ClCO_2^-}] [\mathsf{H^+}]}{\mathsf{[C_3H_6ClCO_2H ]}}$

$K_a = \dfrac{[x] [x]}{ [0.76-x]}$

where:

$K_a = 3.02*10^{-5}$

$3.02*10^{-5} = \dfrac{x^2}{ [0.76-x]}$

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