Question

1.         What is the K_b for B^-, if the K_a for the acid (HB) is 2.5 x 10^-8?

2.         What is the hydronium ion concentration of a 3.50 x 10^-2 M solution of the monoprotic acid, HBrC₆H₄COO, for which K_a = 1.00 x 10^-4?

3.         What is the K_a value for HB, if a solution that is 0.25 M B^- and 0.55 M HB produces a pH of 4.95?

4.         Calculate the pH of a solution containing .225 moles of lactic acid (HC₃H₅O₃) in water to produce one liter of solution. (Ka = 1.4 x 10^-4)

Answer 1

(1) The relation between K_b , K_a & K_w is K_w = K_a x K_b

                                                          K_b = K_w / K_a

                                                              = (1.0x10^-14) / (2.5x10^-8)

                                                              = 4.0x10^-7

Therefore K_b for B^- 4.0x10^-7

(2) 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 a is very small so 1-a is taken as 1

So K_a = ca²

==> a = √ ( K_a / c )

Given K_a = 1.00x10^-4

          c = concentration = 3.50x10^-2 M

Plug the values we get a = 0.0534

So [H⁺] = ca = 3.50x10^-2 M x 0.0534 = 1.87x10^-3 M

Therefore the [H⁺] = 1.87x10^-3 M

(3) According to Henderson's equation, pH = pK_a + log ([B^-]/[HB])

                                                       pK_a = pH - log ([B^-]/[HB])

                                                             = 4.95 - log(0.25/0.55)

                                                             = 5.29

(4) 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 a is very small so 1-a is taken as 1

So K_a = ca²

==> a = √ ( K_a / c )

Given K_a = 1.4x10^-4

          c = concentration = number of moles / volume in L

             = 0.225 mol / 1.0 L

              = 0.225 M

Plug the values we get a = 0.0249

[H⁺] = ca = 0.225 M x 0.0249 = 5.612x10^-3 M

pH = - log[H+] = 2.25