Answer to Question #106246 in Chemistry for kajal

Question

A buffer is prepared by adding 0.7387 g of NH4Cl (MM = 53.4915 g/mol) to 1.00 L of 0.2098 M NH3. Kb (NH3) = 1.8 x 10—5 . Calculate the pH.

Answer 1

Solution:

m(NH₄Cl) = 0.7387 g;

n(NH₄Cl) = m(NH₄Cl) / M(NH₄Cl) = (0.7387 g) / (53.4915 g/mol) = 0.0138 mol;

c(NH₄Cl) = n(NH₄Cl) / V(NH₄Cl) = (0.0138 mol) / (1.00 L) = 0.0138 mol / L = 0.0138 M.

c(NH₄⁺) = 0.0138 M

c(NH₃) = 0.2098 M

NH₃ is a weak base: NH₃ + H2O = NH₄⁺ + OH^-;

NH₄Cl is a salt: NH₄Cl → NH₄⁺ + Cl^-;

thus NH₄⁺ is a "common ion":

NH₃ + H₂O = NH₄⁺ + OH^- 

Let's use an ICE (Initial, Change, Equilibrium) table:

The equlibrium constant (K_b) is expressed as:

K_b = [OH^-] * [NH₄⁺] / [NH₃];

K_b = [x] * [0.0138 + x] / [0.2098 - x].

Approximation: ignore -x, +x terms: 

K_b = [x] * [0.0138] / [0.2098] = 0.0138x / 0.2098 = 0.06578x.

0.06578x = K_b = 1.8*10^-5;

x = (1.8*10^-5) / 0.06578 = 0.00027365 = 2.7365*10^-4;

x= [OH-] = 2.7365*10^-4.

pH = 14 - pOH;

pH = 14 - (-log[OH^-]) = 14 - (log[2.7365*10^-4]) = 14 - (3.56) = 10.44

pH = 10.44.

Answer: pH = 10.44