# What is the pH of a buffer solution that is 0.10 M NH 3 and 0.10 M NH 4 + ? What is the pH if 12 mL of 0.20 M hydrochloric acid is added to 125 mL of buffer?

### General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
Publisher: Cengage Learning
ISBN: 9781305580343

### General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
Publisher: Cengage Learning
ISBN: 9781305580343

#### Solutions

Chapter
Section
Chapter 16, Problem 16.77QP
Textbook Problem

## What is the pH of a buffer solution that is 0.10 M NH3 and 0.10 M NH4+? What is the pH if 12 mL of 0.20 M hydrochloric acid is added to 125 mL of buffer?

Expert Solution
Interpretation Introduction

Interpretation:

The pH of a buffer solution that is 0.10 M NH3 and 0.10 M NH4+ has to be calculated.

The pH of a solution resulting from addition of 12.0 mL of 0.20 M Hydrochloric acid to the 125 mL of the buffer solution has to be calculated.

Concept Information:

Base ionization constant, Kb :

The equilibrium expression for the weak base is given below.

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

Where,

Kb is base ionization constant,

[OH] is concentration of hydroxide ion

[BH+] is concentration of conjugate acid

[B] is concentration of the base

Relationship between pH and pOH:

pH + pOH = 14

### Explanation of Solution

To Calculate: The pH of a buffer solution that is 0.10 M NH3 and 0.10 M NH4+

Given data:

The concentration of NH3 = 0.10 M

The concentration of NH4+ = 0.10 M

pH Calculation:

Construct an equilibrium table with assumption of x as unknown concentration.

Consider the concentration of NH3 from 0.10 M NH3 solution and concentration of NH4+ from 0.10 M NH4+ solution

The hydrolysis reaction of NH3 is as follows,

 NH3     +     H2O  ⇌     NH4+   +   OH− Initial (M) 0.10 −x 0.10-x 0.10 0.00 Change (M) +x +x Equilibrium (M) 0.10+x x

The base ionization constant (Kb) for NH3 is 1.8×105

Substitute the equilibrium concentrations into the equilibrium-constant expression.

Kb =[NH4+][OH-][NH3] =(0.10+x)(x)(0.10x)   1.8×105 (0.10)(x)(0.10)      x =1.8×105 M

Here x gives the concentration of hydroxide ion, [OH] .

The pOH is calculated from the obtained [OH] as follows,

pOH = -log[OH] =log(1.8×105) =4.744

The pH is calculated from pOH as follows,

pH + pOH =14 pH =14 - 4.744 =9.26

Therefore, the pH of the buffer solution of 0.10 M NH3 and 0.10 M NH4+ (before addition of HCl ) is 9.26

To Calculate: The pH of a solution resulting from addition of 12.0 mL of 0.20 M Hydrochloric acid to the 125 mL of the buffer solution

Given data:

The concentration of Hydrochloric acid is 0.20 M

The volume of Hydrochloric acid is 12.0 mL

The volume of the buffer is 125 mL

Explanation:

12. mL of 0.20 M HCl is added to the 125 mL buffer, noting that the H3O+ ion reacts with NH3 to form NH4+

The total volume of the buffer is calculated as,

Total volume = 125 mL + 12 mL =0.125 L + 0.012 L =0.137 L

The moles of HCl added is calculated as:

mol of HCl =(0

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