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

Answer:

Percentage dissociated = 0.41%

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

$K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}$

where ;

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

$3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}$

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 = $\frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100$

Percentage dissociated = $\frac{7.3729*10^{-3}}{1.8 }*100$

Percentage dissociated = 0.4096

Percentage dissociated = 0.41%     (to two significant digits)

Answer 2

Answer:

0.00091%

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%