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:
[tex]\mathsf{C_3H_6ClCO_2H }[/tex] ⇄ [tex]\mathsf{C_3H_6ClCO_2^-}[/tex] + [tex]\mathsf{H^+}[/tex]
The ICE table can be computed as:
[tex]\mathsf{C_3H_6ClCO_2H }[/tex] ⇄ [tex]\mathsf{C_3H_6ClCO_2^-}[/tex] + [tex]\mathsf{H^+}[/tex]
Initial 0.76 - -
Change -x +x +x
Equilibrium 0.76 - x x x
[tex]K_a = \dfrac{[\mathsf{C_3H_6ClCO_2^-}] [\mathsf{H^+}]}{\mathsf{[C_3H_6ClCO_2H ]}}[/tex]
[tex]K_a = \dfrac{[x] [x]}{ [0.76-x]}[/tex]
where:
[tex]K_a = 3.02*10^{-5}[/tex]
[tex]3.02*10^{-5} = \dfrac{x^2}{ [0.76-x]}[/tex]
however, the value of x is so negligible:
0.76 -x = 0.76
Then:
[tex]3.02*10^{-5}*0.76 = x^2[/tex]
[tex]x=\sqrt{3.02*10^{-5}*0.76 }[/tex]
x = 0.00479 M
∴
[tex]x = \mathsf{[C_3H_6ClCO_2^-] = [H^+]=}[/tex] 0.00479 M
[tex]\mathsf{C_3H_6ClCO_2H }[/tex] = (0.76 - 0.00479) M
= 0.75521 M
Finally, the percentage of the acid dissociated is;
= ( 0.00479 / 0.76) × 100
= 0.630 M
