Question

Student Name Course/Section Date Instructor RELATED QUESTIONS 1. A buffer solution is prepared by making a solution 0.1M in NH, and 0.2M in NHACI (a) Write the equation for the ionization of NH, in water. (b) Write the expression for the ionization constant, K, for NHg. The value of K, at 25°C is 1.8 x 10 (c) Calculate the [OH-], the [H'), and the pH for this buffer solution. (d) Calculate the (OH), the [H), and the pH for 0.1M NH3.

Answer 1

Solution:

A) Ionization of NH3 in water:

NH3 + H2O = NH4+ + OH-

B) Kb = [NH4+] [OH-] / [NH3]

1.8 x 10^-5 = [NH4+] [OH-] / [NH3]

C) pH of the buffer solution is calculated by Hendersen equation,

pOH = pkb + log [NH4+] / [NH3+]

pOH = 4.74 + log (0.2 / 0.1)

pOH = 4.74 + log 2

pOH = 4.74 + 0.30 = 5.04

Thus,

[OH-] = 10^-pOH = 10^-5.04

[OH-] = 9.12 x 10^-6

Then,

[H+] = 1 x 10^-14 / [OH-]

= 1 x 10^-14 / 9.12 x 10^-6

[H+]= 1.1 x 10^-9

D) Calculation of OH- and H+:

NH3 + H2O = NH4+ + OH-

0.1 M -------------- 0 -------0 M (initial)

(0.1- X) ----------- X ------- X M (final)

Thus,

Kb = X . X / (0.1 - X)

Kb = X^2 / 0.1

(Since, X <<<< 0.1 because NH3 is a weak base, hence neglected from denominator)

1.8 x 10^-5 = X^2 / 0.1

X^2 = 1.8 x 10^-6

X = 1.34 x 10^-3

Therefore,

[OH-] = X = 1.34 x 10^-3 M

[H+] = 1 x 10^-14 / [OH-]

= 1 x 10^-14 / 1.34 x 10^-3

[H+] = 7.46 x 10^-12 M

pH = - log [H+]

= - log 7.46 x 10^-12

= 12 - log 7.46 = 12 - 0.87

pH = 11.13