Question

1- If the Ka of a monoprotic weak acid is 6.5 × 10-6, what is the pH of a 0.43 M solution of this acid?

2- If the Kb of a weak base is 7.1 × 10-6, what is the pH of a 0.19 M solution of this base?

Answer 1

1) ka=6.5*10^-6

Let the acid be represented by HA

It partially ionizes in solution,

HA +H2O↔A- + H3O+

Ka=[ A-] [H3O+]/[HA]

ICE table

	[HA]	[A-]	[H3O+]
Initial	0.43 M	0	0
change	-x	+x	+x
equilibrium	0.43-x	x	x

Ka=6.5*10^-6 =[ A-] [H3O+]/[HA]=x^2/(0.43-x)

6.5*10^-6 =x^2/(0.43-x)

Ignore x with respect to 0.43M as ka is small ,so very little ionization of weak acid takes place.

6.5*10^-6 =x^2/(0.43)

Or, 6.5*10^-6 *0.43 =x^2

Or, x^2=2.795*10^-6

X=1.672*10^-3=[H3O+]

pH=-log[H3O+]=-log (1.672*10^-3)=-0.223+3=2.8

pH=2.8

2)

Kb=7.1*10^-6

Let B be the weak base,[B]=0.19M

B +H2O↔OH-+ BH

Kb=[OH-][BH]/[B]

	[B]	[BH]	[OH-]
initial	0.19M	0	0
change	-x	+x	+x
equilibrium	0.19-x	x	x

Kb=7.1*10^-6 =[ A-] [H3O+]/[HA]=x^2/(0.19-x)

7.1*10^-6 =x^2/(0.19-x)

Ignore x with respect to 0.19M as kb is small ,so very little ionization of weak base takes place.

7.1*10^-6 =x^2/(0.19)

Or, 7.1*10^-6 *0.19 =x^2

Or, x^2=1.349*10^-6

X=1.16*10^-3=[OH-]

pOH=-log[OH-]=-log (1.16*10^-3)=-0.06+3=2.94

pH=14-pOH=14-2.94=11.06

pH=11.06