Chapter 17, Problem 26PS

### 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

# What is the pH change when 20.0 mL of 0.100 M NaOH is added to 80.0 mL of a buffer solution consisting of 0.169 M NH3 and 0.183 M NH4Cl?

Interpretation Introduction

Interpretation:

The change in value of pH is to be calculated when 20 mL of 0.100 molL1NaOH is added to 80 mL of the buffer solution which contains 0.169 M NH3 and 0.183 M NH4Cl.

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 0.169Â molâ‹…Lâˆ’1.

The initial moles of NH4Cl is 0.183Â molâ‹…Lâˆ’1.

The volume of buffer solution is 80Â mL.

Unit conversion of 80Â mL into L.

(80Â mL)(1Â L1000Â mL)=0.080Â L

Therefore, volume of the solution is 0.080Â L.

The concentration of NaOH is 0.100Â M, and volume of NaOH added is 20Â mL.

Unit conversion of 20Â mL into L.

(20Â mL)(1Â L1000Â mL)=0.020Â L

Therefore, volume of the NaOH solution is 0.020Â L.

The value of pH for the buffer solution before addition of NaOH is calculated by using equation (1).

pOH=pKb+log[conjugateâ€‰acid][base]

Substitute 4.74 for pKb, 0.183 for [conjugateÂ acid], 0.169Â  for [base].

pOH=4.74+log(0.183)(0.169Â )=4.74+log(1.08)=4.74+0.033=4.77

Therefore, pOH value of the buffer solution is 4.77.

The pH value is calculated by using expression,

pHÂ +Â pOHÂ =Â 14

Rearrange for pH

pHÂ Â =Â 14âˆ’pOH

Substitute 4.77 for pOH

pHÂ Â =Â 14âˆ’4.77=Â 9.23

Therefore value of pH for the buffer solution is 9.23.

Calculation of pH value after the addition of NaOH is given below.

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

NaOH(aq)â†’Na+(aq)+Â OHâˆ’(aq)

The OHâˆ’ ions will react with ammonium ion and affects the equilibrium conditions.

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

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

Â Â Â Â Â Â Â Â Â Â Â OHâˆ’(fromÂ addedÂ NaOH)Â Â Â Â Â Â NH4+Â (fromÂ buffer)Â Â Â Â Â Â Â NH3Â (fromÂ buffer)initial(mol)0.002000.01460.0135change(mol)âˆ’0.00200âˆ’0.00200+0.00200afterÂ reaction(mol)00.01260.0155

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

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