Calculate the pH of each of the following solutions.(Kb= 1.1e-8).?

Question

a) 0.100 M HONH3Cl

b) A mixture containing 0.100 M HONH2 and 0.100 M HONH3Cl

Please help me in solving this problem. Thanx in advance.....

Answer 1

a) It is a salt derived from a weak base and a strong acid.
So it is an hydrolysis

Kh = Kw / Kb = 9.1 10^-7

Kh = (x)(x)/ 0.1-x = 9.1 10^-7

x = 0.000302 = concentration H+

pH = 3.52

b) it is a buffer solution

1.1 10^-8 = (0.1+x)(x)/ 0.1-x

x = 1.1 10^-8 = concentration OH-

pOH = 7.96

pH = 6.04

Answer 2

a) Ka = Kw/Kb = 9.09 * 10^-7
9.09 * 10^-7 = x2/0.100 - x
x <<< 0.100, so can be ignored in the denominator.
9.09 * 10^-8 = x2
x = 3.02 * 10^-4 = [H+]
pH = 3.52
b) Equal concentrations of acid and conjugate base means that by Henderson-Hasselbach, pH = pKa = 6.04

Answer 3

For a)
Kb tells you it behaves as a base,
HONH3Cl + H2O --> H2ONH3Cl + OH-

Kb = [H2ONH3Cl][OH] / [HONH3Cl]

1.1x10^-8 = [x][x]/.100
[.100][1.1x10^-8] = x^2
multiply, then take the square root of both sides. Now you have found OH, but to find pH:
1.0x10^-14 / x.
That gives you H+ then just take the -log(H+)

For b),
Disregard the Cl, it has nothing to do with what you want.

HONH2 + H2O --> HONH3 + OH

1.1x10^-8 = [.100x] / .100
Again, that gives you OH, so find H+ and take the negative log.