Question

write the appropriate equilibrium reaction for each problem. Use the quadratic equation if the 5% rule does not apply. Show all work.

1. Calculate the Ka of a .48 M weak acid HA if the pH of the solution is 3.45

2. A weak base has a Kb of 1.050E-03. Calculate the pH of a .225 M solution of the weak base.

Answer 1

(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 c = concentration = 0.48 M

pH = 3.45

- log [H⁺] = 3.45

[H⁺] = 10^-3.45 = 3.55x10^-4 M

[H⁺] = cα = 3.55x10^-4 M

α = (3.55x10^-4 )/ c = (3.55x10^-4 )/0.48 = 7.39x10^-4 M

K_a = cα² = 2.62x10^-7

(2)

Let α be the dissociation of the weak base
                            BOH <---> B ⁺ + OH^-

initial conc.            c               0         0

change               -cα            +cα      +cα

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

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

                                          = c α² / (1-α)

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

So K_b = cα²

==> α = √ ( K_b / c )

Given K_b = 1.050x10^-3

          c = concentration = 0.225 M

Plug the values we get α = 0.0683
So the concentration of [OH^-] = cα

= 0.225 x 0.0683

= 0.0153 M

pOH = - log [OH^-]

        = - log 0.0153

   = 1.81

So pH = 14 - 1.81 = 1.19