Chapter 17, Problem 81GQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Calculate the hydronium ion concentration and the pH of the solution that results when 50.0 mL of 0.40 M NH3 is mixed with 25.0 mL of 0.20 M HCl.

Interpretation Introduction

Interpretation:

The hydronium ion concentration and pH for the resultant solution has to be calculated when 50 mL of 0.40 M NH3 is mixed with 25 mL of 0.20 M NaOH.

Concept introduction:

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

pOH=pKb+log[conjugateacid][base] (1)

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

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

Explanation

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

Given:

Refer to table 16.2 in the textbook for the value of Kb.

The value of Kb for NH3 is 1.8Ã—10âˆ’5.

The pKb value is calculated as follows;

pKb=âˆ’log(Kb)

Substitute, 1.8Ã—10âˆ’5 for Kb.

pKb=âˆ’log(1.8Ã—10âˆ’5)=4.74

Therefore, pKb value of NH3 is 4.74.

The initial concentration of NH3 is 0.40Â molâ‹…Lâˆ’1.

The volume of buffer solution is 50Â mL.

Unit conversion of 50Â mL into L.

(50Â mL)(1Â L1000Â mL)=0.050Â L

Therefore, volume of the solution is 0.050Â L.

The concentration of HCl is 0.20Â M, and volume of HCl added is 25Â mL.

Unit conversion of 25Â mL into L.

(25Â mL)(1Â L1000Â mL)=0.025Â L

Therefore, volume of the HCl solution is 0.025Â L.

As, HCl is a strong base it ionize completely into its ions in aqueous solution.

HCl(aq)Â +Â H2O(l)â†’H3O+(aq)+Â Clâˆ’(aq)

The H3O+ ions will react with ammonia.

NH3(aq)+Â H3O+(aq)â†’H2O(l)Â +Â NH4+(aq)

Table gives the relationship of the concentration for the reaction between H3O+, NH3 and NH4+.

Â Â Â Â Â Â Â Â Â H3O+(fromÂ addedÂ HCl)NH3Â Â Â Â Â Â Â Â Â Â Â Â Â NH4+Â Â initial(mol)0.0050.020.0change(mol)âˆ’0.005âˆ’0.005+0.005afterÂ reaction(mol)00.0150.005

The concentration of ammonia and ammonium ion after the reaction is calculated by using equation (2);

Molarity=NumberÂ ofÂ molesvolumeÂ ofÂ solvent (2)

The total volume of the solution will increase by 0.025Â L after reaction.

Therefore, total volume of the solution is calculated as,

0.050Â L+Â 0.025Â LÂ =Â 0.075Â L

Total volume of the solution is Â 0.075Â L.

Substitute, 0.015Â mol for NumberÂ ofÂ moles and Â 0.075Â L for Â volumeÂ ofÂ solvent to calculate the concentration of ammonia.

Molarity=0.015Â mol0.075Â L=0.2Â molâ‹…Lâˆ’1

Therefore, concentration of ammonia ion left after reaction is 0.2Â molâ‹…Lâˆ’1.

Substitute, 0.005Â mol for NumberÂ ofÂ moles and Â 0.075Â L for Â volumeÂ ofÂ solvent to calculate the concentration of ammonia in equation (2)

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