Question

A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic acid that is in the dissociated form in his solution. Express your answer as a percentage. You will probably find some useful data in the ALEKS Data resource. Round your answer to 2 significant digits. 1% x 5 ?

Answer 1

0.00091%

Step-by-step explanation:

The fraction of 4-chlorobutanoic acid that would dissociatr in aqueos solution is a function of Ionization percentage. It is obtained by

Ka = [dissociated acid] / [original acid] x 100%

The equation of the reaction is

C₂HClCOOH + H20 (aq) = C₂HClCOO⁻ + H₃O⁺ pka =4.69

But pKa = - log Ka

4.69 = - Log Ka

10⁻⁴⁶⁹ = Ka

Taking the antilog of the equation we get the ionization constant

Ka = 0.0000204 M

=2.04 x 10⁻⁵M

At the beginning of the reaction we have the following concentrations

1.8M C₂HClCOOH : 0M (zero molar) C₂HClCOO⁻ : 0M (zero molar) H₃O⁺

At equilibrium, we have,

(1.8M -x) xM C₂HClCOO⁻ and xM H₃0⁺

Therefore,

Ka = [C2HClCOO-] [H30+] / [C2HClCOOH],

Inputing the value of Ka

Ka .[C2HClCOOH] = [CHCLCOO-] [H3O+]

0.0000204 (1.8 - X) = (x).(x)

x² = (0.00003672 - 0.0000204 X)

= (3.672 x10-5 - 2.04 x10⁻⁵X)

x² +2.04x10⁻⁵ x = 3.672 x 10⁻⁵

x² + 2.04 x 10⁻⁵x - 3.672 x 10⁻⁵ = 0

x = 0.00001632

= 1.632 x 10⁻⁵

Inputing back into equation 1

1.8 - x = [H3O+]

1.8 - 0.00001632 = 1.7999837

It therefore implies that only 0.00001632M of 4-chloroutanoic acid dissciated at equilibrium, we can now calculate the percentage dissociation by

Percentage dissociation = 0.00001632 / 1.8M x 100%

= (1.632 x 10⁻⁵/1.8 ) x 100%

= 0.00090667%

= 0.00091%

Answer 2

Percentage dissociated = 0.41%

Step-by-step explanation:

The chemical equation for the reaction is:

`C_3H_6ClCO_2H_((aq)) \to C_3H_6ClCO_2^-_((aq))+ H^+_((aq))`

The ICE table is then shown as:

`C_3H_6ClCO_2H_((aq)) \ \ \ \ \to \ \ \ \ C_3H_6ClCO_2^-_((aq)) \ \ + \ \ \ \ H^+_((aq))`

Initial (M) 1.8 0 0

Change (M) - x + x + x

Equilibrium (M) (1.8 -x) x x

`K_a = ([C_3H_6ClCO^-_2][H^+])/([C_3H_6ClCO_2H])`

where ;

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

`3.02*10^(-5) = ((x)(x))/((1.8-x))`

Since the value for
`K_a` is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

`3.02*10^(-5) *(1.8) = {(x)(x)}`

`5.436*10^(-5)= {(x^2)`

`x = \sqrt{5.436*10^(-5)}`

`x = 0.0073729 \\ \\ x = 7.3729*10^(-3) \ M`

Dissociated form of 4-chlorobutanoic acid =
`C_3H_6ClCO_2^- = x= 7.3729*10^(-3) \ M`

Percentage dissociated =
`(C_3H_6ClCO^-_2)/(C_3H_6ClCO_2H) *100`

Percentage dissociated =
`(7.3729*10^(-3))/(1.8 )*100`

Percentage dissociated = 0.4096

Percentage dissociated = 0.41% (to two significant digits)