Question

(a) What are the concentration at equilibrium of a 0.205 M HA, where HA is a weak acid that dissociates to H+ and A- with a Kc = 4.7 x 10-9 . The equilibrium (NH4)2CO3 (s)  2 NH3 (g) + CO2 (g) + H2O

Answer 1

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

initial conc.            c               0         0

change                -cα            +cα      +cα

Equb. conc.         c(1-α)          cα    cα

Dissociation constant , K_a = cα x cα / ( c(1-α)

                                         = c α² / (1-α)

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

So K_a = cα²

==> α = √ ( K_a / c )

Given K_a = 4.7x10^-9

          c = concentration = 0.205 M

Plug the values we get α = 1.514x10^-4

[H⁺] = cα = 0.205 x1.514x10^-4 = 3.10x10^-5 M

[A^-]= cα = 0.205 x1.514x10^-4 = 3.10x10^-5 M

[HA] =  c(1-α) = 0.205x[1- (1.514x10^-4 )] = 0.205 M

pH = - log [H⁺]

    = - log(3.10x10^-5 )

    = 4.51