Question

A .053M solution of a weak base has a pH of 9.87. What is the Kb of this weak monprotic acid?

Answer 1

Molarity of base = 0.053 M

pH = 9.87

We know ka x kb = 1.0E -14

We have to find ka of this acid.

Lets find kb first and then get ka

Assume B is the weak base.

[B] = 0.053 M

ICE

            B (aq) + H2O (l)   --- > BH+ + OH- (aq))

I           0.053                           0          0

C          -x                                 +x        +x

E          (0.053-x)                     x          x

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

Once we get x then we can get the value of kb

[OH-]= x

[OH-] = Antilog ( pOH )

pOH – 14 – pH

pH = 9.87

Lets put the value of pH in above equation

pOH = 14 -9.87 = 4.13

[OH-]= Antilog ( - 4.13 )

= 7.41 E-5

Therefore x = 7.41 E-5

[B]= 0.053- x = 0.05291

Kb = (7.41 E-5)² / ( 0.05291)

Kb = 1.04 E-7

Value of ka = 1.0 E-14 / 1.04 E-7

= 9.63 E-8

Ka of this acid is 9.63 E-8