   Chapter 17, Problem 26PS

Chapter
Section
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 1.8×105.

The pKb value is calculated as follows;

pKb=log(Kb)

Substitute,1.8×105 for Kb.

pKb=log(1.8×105)=4.74

Therefore, pKb value of NH3 is 4.74.

The initial concentration of NH3 is 0.169 molL1.

The initial moles of NH4Cl is 0.183 molL1.

The volume of buffer solution is 80 mL.

Unit conversion of 80 mL into L.

(80 mL)(1 L1000 mL)=0.080 L

Therefore, volume of the solution is 0.080 L.

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

Unit conversion of 20 mL into L.

(20 mL)(1 L1000 mL)=0.020 L

Therefore, volume of the NaOH solution is 0.020 L.

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

pOH=pKb+log[conjugateacid][base]

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

pOH=4.74+log(0.183)(0.169 )=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,

pH + pOH = 14

Rearrange for pH

pH  = 14pOH

Substitute 4.77 for pOH

pH  = 144.779.23

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)Na+(aq)+ OH(aq)

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

NH4+(aq)+ OH(aq)H2O(l) + NH3(aq)

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

OH(from added NaOH)      NH4+ (from buffer)       NH3 (from buffer)initial(mol)0.002000.01460.0135change(mol)0.002000.00200+0.00200after reaction(mol)00.01260.0155

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

Molarity=Number of molesvolume of solvent (2)

The total volume of the solution will increase by 0

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