   Chapter 17, Problem 90GQ

Chapter
Section
Textbook Problem

What volume of 0.200 M HCl must be added to 500.0 mL of 0.250 M NH3 to have a buffer with a pH of 9.00?

Interpretation Introduction

Interpretation:

The volume of 0.200 M HCl added to 500 mL of  0.250 M NH3 has to be calculated when pH of the buffer solution is 9.0.

Concept introduction:

For weak base-strong acid titration the pH value can be calculated at various points before and after equivalence point.

The equilibrium established during the titration of NH3 with HCl. The equilibrium can be represented as,

NH3(aq)+ H3O+(aq)H2O(l)+NH4+(aq)

Explanation

The calculation of volume 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 ammonia 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 is 4.74.

The initial concentration of NH3 is 0.250molL1.

The initial concentration of HCl is 0.200molL1.

The volume of NH3 solution is 500 mL.

Let the volume of HCl is x mL used to get the pH of buffer equals to 9.0.

The pOH value of buffer solution is 5.0.

Therefore total volume of the buffer solution is (500 + x )mL or (0.500 + 0.00x )L.

The calculation of moles is done by using the expression,

Numberof moles=concentration(molL1)volume(L)

The ICE table (1) for the reaction between NH3 and HCl is given below,

EquationNH3(aq)+H3O+(aq)H2O(l)+NH4+(aq)Initial(mol)0.1250.00200x0Change(mol)0.00200x0.00200x+0.00200xAfterreaction(mol)(0.1250.00200x)00.00200x

From ICE table (1),

Number of moles of ammonia left after reaction are (0.1250.00200x). The value of 0.00200x is very small as comparison to 0.125 therefore it can be neglected with respect to 0.125. So numbers of moles of ammonia left after reaction are 0.125 mol.

Number of moles of H3O+ left after reaction are 0mol

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