Question

A certain weak acid, HA, has a Ka value of 2.0×10⁻⁷.

Part A: Calculate the percent ionization of HA in a 0.10 M solution.

Part B: Calculate the percent ionization of HA in a 0.010 M solution.

Answer 1

Part A :

Let a be the dissociation of the weak acid
                            HA <---> H ⁺ + A^-

initial conc.            c               0         0

change                -ca            +ca      +ca

Equb. conc.         c(1-a)          ca       ca

Dissociation constant , K_a = ca x ca / ( c(1-a)

                                         = c a² / (1-a)

In the case of weak acids α is very small so 1-a is taken as 1

So K_a = ca²

==> a = √ ( K_a / c )

Given K_a = 2.0x10^-7

          c = concentration = 0.10M

Plug the values we get a = 1.414x10^-3

∴ % dissociation =  1.414x10^-3 x 100 =  1.414x10^-1

Part B :

Let a be the dissociation of the weak acid
                            HA <---> H ⁺ + A^-

initial conc.            c               0         0

change                -ca            +ca      +ca

Equb. conc.         c(1-a)          ca       ca

Dissociation constant , K_a = ca x ca / ( c(1-a)

                                         = c a² / (1-a)

In the case of weak acids α is very small so 1-a is taken as 1

So K_a = ca²

==> a = √ ( K_a / c )

Given K_a = 2.0x10^-7

          c = concentration = 0.010M

Plug the values we get a = 4.472x10^-3

∴ % dissociation = 4.472x10^-3 x 100 = 4.472x10^-1