   Chapter 17.3, Problem 17.7CYU

Chapter
Section
Textbook Problem

Calculate the pH after 75.0 mL of 0.100 M HO has been added to 100.0 mL of 0.100 M NH3 (Figure 17.7).

Interpretation Introduction

Interpretation:

The value of pH has to be calculated at a point when 100mL, 0.100 MNH3 is titrated against 75mL, 0.100MHCl.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base. There are four types of acid-base titrations.

(1) Strong acid-Strong base, in this type of titration a strong acid is titrated against a strong base for example, HCl is titrated against NaOH.

(2) Strong acid-Weak base, in this type of titration a strong acid is titrated against a weak base for example, HCl is titrated against NH4OH.

(3) Weak acid-Strong base, in this type of titration a weak acid is titrated against a strong base for example, CH3COOH is titrated against NaOH.

(4) Weak acid-Weak base, in this type of titration a weak acid is titrated against a weak base for example, CH3COOH is titrated against NH4OH.

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)

Here, the H3O+ ion are generated from HCl, and the reaction is written as,

HCl(aq)+H2O(l)H3O+(aq)+Cl(aq)

At equivalence point all the base will be neutralized, and there will be only NH4+ ion and H3O+. The H3O+ will be produced due to the hydrolysis of NH4+ at equivalence point. The hydrolysis equilibrium is represented as,

NH4+(aq)+H2O(l) H3O+(aq)+NH3(aq)

By using the value of Ka for the NH4+ concentration of H3O+ can be calculated.

Ka=[H3O+][H2O][NH4+] (1)

As H3O+ ions are produced due to hydrolysis of NH4+, the value of pH will be lower than 7 at equivalence point for the weak base-strong acid titrations.

Explanation

The pH calculation just before the equivalence point is done by using Henderson-Hasselbalch equation.

As the addition of HCl is done there will be formation of buffer solution NH3/NH4+. The pOH calculation for buffer solution is done by using Henderson-Hesselbalch equation.

pOH=pKb+log[conjugateacid][base] (2)

The pH value will be calculated by using expression, pH + pOH = 14.

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 0.100molL1.

The initial concentration of HCl is 0.100molL1.

The volume of NH3 is 100mL.

Conversion of 100mL into L.

(100mL)(1L1000mL)=0.1L

The volume of HCl is 75mL

Conversion of 75mL into L.

(75mL)(1L1000mL)=0.075L

The total volume after the reaction is calculated as,

totalvolume=volumeofNH3(L) + volume of HCl(L)

totalvolume = 0.1(L)+0.075(L)=0.175L

Therefore, total volume after reaction is 0.175L.

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)0.01000.00750Change(mol)0.00750.00750.0075Afterreaction(mol)0.002500.0075

From ICE table (1),

Number of moles of ammonia left after reaction are 0.0025mol.

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

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