Question

(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conductance at infinite dilution (Λ°) of aqueous sodium chloride, sodium formate and hydrochloric acid are 1.264 × 10–2, 1.046 × 10–2 and 4.261 × 10–2 respectively at 25°C determine the acid dissociation constant and the pKa for the acid.

Answer 1

Explanation:

The given data is as follows.

   $\Lambda^{o}_{m}$(NaCl) = $1.264 \times 10^{-2}$

   $\Lambda^{o}_{m}$(H-O=C-ONO) = $1.046 \times 10^{-2}$

   $\Lambda^{o}_{m}$(HCl) = $4.261 \times 10^{-2}$

Conductivity of monobasic acid is $5.07 \times 10^{-2} S m^{-1}$

     Concentration = 0.01 $mol/dm^{3}$

Therefore, molar conductivity ($\Lambda_{m}$) of monobasic acid is calculated as follows.

                 $\Lambda_{m} = \frac{conductivity}{concentration}$

                                  = $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}$

                                 = $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}$

                                 = $5.07 \times 10^{-3} S m^{2} mol^{-1}$

Also, $\Lambda^{o}_{m}$ = $\Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}$

                            = $4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}$

                            = $4.043 \times 10^{-2} S m^{2} mol^{-1}$

Relation between degree of dissociation and molar conductivity is as follows.

               $\alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}$

                             = $\frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}$

                             = 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

                     K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

Putting the values into the above formula we get the following.

                     K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

                        = $\frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}$

                        = $0.017973 \times 10^{-2}$

                       = $1.7973 \times 10^{-4}$

Hence, the acid dissociation constant is $1.7973 \times 10^{-4}$.

Also, relation between $pK_{a}$ and $K_{a}$ is as follows.

                 $pK_{a} = -log K_{a}$

                              = $-log (1.7973 \times 10^{-4})$

                              = 3.7454

Therefore, value of $pK_{a}$ is 3.7454.