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Calculate the pH after 0.010 mole of gaseous HCl is added to 250.0 mL of each of the following buffered solutions. a. 0.050 M NH 3 /0.15 M NH 4 Cl b. 0.50 M NH 3 /1.50 M NH 4 Cl Do the two original buffered solutions differ in their pH or their capacity? What advantage is there in having a buffer with a greater capacity?

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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Section
BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 14, Problem 39E
Textbook Problem
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Calculate the pH after 0.010 mole of gaseous HCl is added to 250.0 mL of each of the following buffered solutions.

a. 0.050 M NH3/0.15 M NH4Cl

b. 0.50 M NH3/1.50 M NH4Cl

Do the two original buffered solutions differ in their pH or their capacity? What advantage is there in having a buffer with a greater capacity?

(I)

Interpretation Introduction

Interpretation: The value of pH for the given buffer solutions on addition of 0.010mole of HCl; the and the advantage of having a buffer with greater capacity.

Concept introduction: The solution that have tendency to resist the change in pH is called a buffer solution.

Explanation of Solution

Explanation

Given

The moles of HCl are 0.010.

The concentration of NH3 is 0.050M.

The concentration of NH4Cl is 0.15M.

The volume of buffer solution is 250mL.

The Henderson-Hasselbalch equation is represented as,

pOH=pKb+log[Salt][Base] (1)

Where,

  • pOH is the negative logarithm of OH ions concentration in the solution.
  • pKb is the negative logarithm of dissociation constant of the base.
  • [Salt] is the concentration of conjugate base of the given weak base.
  • [Base] is the concentration of given weak base.

The value of Kb for Ammonia is 1.8×105.

The relationship between pKb and Kb is given as,

pKb=logKb

Where,

  • Kb is the dissociation constant of a base.

Substitute the value of Kb for Ammonia in above equation.

pKb=logKb=log(1.8×105)=4.74

Substitute the value of concentration of NH3 and NH4Cl and pKb in the equation (1).

pOH=pKb+log[Salt][Base]=4.74+log0.150.05=5.21

The value of pOH for the given buffer solution is 5.21.

The relationship between pOH is given as,

pH+pOH=14 (2)

Substitute the value of pOH in the above equation.

pH+pOH=14pH+5.21=14pH=8.79

The pH of given buffer solution is 8.79.

The conversion of mL into L is done as,

1mL=0.001L

Hence the conversion of 250.0mL into L is done as,

250.0mL=250.0×0.001L=0.25L

The concentration of any species is given as,

Concentration=NumberofmolesVolumeofsolutioninlitres (3)

Rearrange the above equation to obtain the value of number of moles.

Numberofmoles=Concentration×Volumeofsolutioninlitres (4)

Substitute the value of number of moles of HCl and volume of solution in the equation (3).

Concentration=NumberofmolesVolumeofsolutioninlitres=0.010moles0.25L=0.04M

Make the table for the formation of Ammonium ion.

H++NH3NH4+Initialconcentration (M)          0.040.050.15Change (M)                              0.040.04+0.04Equilibriumconcentration (M)00

(II)

Interpretation Introduction

Interpretation: The value of pH for the given buffer solutions on addition of 0.010mole of HCl; the and the advantage of having a buffer with greater capacity.

Concept introduction: The solution that have tendency to resist the change in pH is called a buffer solution.

To determine: If the given original solutions differ in their pH or their capacity

(III)

Interpretation Introduction

Interpretation: The value of pH for the given buffer solutions on addition of 0.010mole of HCl; the and the advantage of having a buffer with greater capacity.

Concept introduction: The solution that have tendency to resist the change in pH is called a buffer solution.

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Chapter 14 Solutions

Chemistry: An Atoms First Approach
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