Chapter 16, Problem 66PS

### 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 when 50.0mL of 0.40M NH 3 is mixed with 50.0mL of 0.40 M HCl.

Interpretation Introduction

Interpretation:

The hydronium ion concentration and the pH has to be calculated when 50.0mL of 0.40M NH3 is mixed with 50.0mL of 0.40 M HCl.

Concept introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

Ka is acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb is base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

Ionic product constant for water  Kw= [H3O+][OH-]       =1.00×10-14 pH = -log[H3O+]pOH= -log[OH-]Reltion between pH and pOH pH + pOH =14

Explanation

Given volume of 0.40MNH3 = 50mL

And it is mixed with 50mL of 0.40M HCl. Therefore, the hydronium ions from HCl neutralize ammonia a weak base to its conjugate acid. NH4+.

Hydronium ions reacts with ammonia to give NH4+.

The equilibrium chemical reaction is as follows.

NH3(aq)Â +H3O+(aq)Â â‡ŒNH4+(aq)Â +Â H2O(aq)

Letâ€™s calculate the amount of consumed HCl:

=Â (50mLÃ—1L1000mL)Â Ã—Â (0.40molL)=Â Â 0.02Â mol

Letâ€™s calculate the consumed amount of NH3

=Â (50mLÃ—1L1000mL)Â Ã—Â (0.40molL)=Â Â 0.02Â mol

Letâ€™s calculate the produced amount of NH4+ after completion of reaction.

=Â 0.02molÂ NH3Â Â Ã—Â 1molÂ NH4+1molÂ NH3=Â 0.02Â molÂ NH4+

TheÂ totalÂ volumeÂ ofÂ solutionÂ =Â 50.0mLÂ +Â 50.0mL=Â 100.0mLÂ Ã—Â 1L1000mL=Â 0.1L

ConcentrationÂ ofâ€‰NH4+â€‰=0.02Â mol0.1L=Â 0.2M

Therefore, 0.2M is the final volume of NH4+ after completion of the reaction.

The NH4+ ions react with water. The resulting solution is acidic in nature.

The equilibrium reaction is as follows.

NH4+(aq)Â +Â H2O(l)Â â‡ŒNH3(aq)Â +Â H3O+(aq)

The equilibrium expression:

Ka=Â [NH3][H3O+][NH4+]

KaÂ ofÂ NH4+Â isÂ 5

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