   # A buffer solution is prepared by adding 0.125 mol of ammonium chloride to 5.00 × 10 2 mL of 0.500 M solution of ammonia. (a) What is the pH of the buffer? (b) If 0.0100 mol of HCl gas is bubbled into 5.00 × 10 2 mL of the buffer, what is the new pH of the solution? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 17, Problem 25PS
Textbook Problem
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## A buffer solution is prepared by adding 0.125 mol of ammonium chloride to 5.00 × 102 mL of 0.500 M solution of ammonia. (a) What is the pH of the buffer? (b) If 0.0100 mol of HCl gas is bubbled into 5.00 × 102 mL of the buffer, what is the new pH of the solution?

(a)

Interpretation Introduction

Interpretation:

For the buffer solution which is prepared by adding 0.125 mol of ammonium chloride in 5.0×102mL of 0.500 M NH3,pHof the buffer solution has to be calculated.

Concept introduction:

The Henderson-Hasselbalch equation relates pHof a buffer withpKaof acid, concentration of conjugate base and concentration of acid. The expression is written as,

pH=pKa+logconjugatebaseacid                                                  (1)

This equation shows that pHof buffer solution is controlled by two major factors. First, strength of the acid can be expressed on terms of pKa and second, the relative concentration of acid and its conjugate base at equilibrium.

The pH value is calculated by expression, pH + pOH = 14.

### Explanation of Solution

The calculation of pH is done by using Henderson-Hasselbalch equation.

The value of Kb for NH3 is 1.8×105.

The pKb value is calculated as follows;

pKb=log(Kb)

Substitute, 1.8×105for Kb.

pKb=log(1.8×105)=4.74

Therefore, pKbvalue of NH3 is 4.74.

The initial concentration of NH3is 0.500 molL1.

The initial moles of NH4Cl is 0.125 mol.

The volume of the buffer solution is 5×102mL.

Unit conversion of 5×102mLinto L.

5×102 mL1 L1000 mL=0.500 L

Therefore, volume of the solution is0.500 L.

The concentration of NH4Cl is calculated as follows;

Molarity=Number of molesvolume of solvent                                   (2)

Substitute 0.125 molfor number of moles and0.500 L for volume of solvent in equation (2).

Molarity=0

(b)

Interpretation Introduction

Interpretation:

For the buffer solution which is prepared by adding 0.125 mol of ammonium chloride in 5.0×102mL of 0.500 M NH3. pHof the solution when 0.0100 mol of HCl is added to 5.0×102 mLof buffer solution has to be calculated.

Concept Introduction:

Refer to (a)

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