Quantitative Chemical Analysis 9e And Sapling Advanced Single Course For Analytical Chemistry (access Card)
Quantitative Chemical Analysis 9e And Sapling Advanced Single Course For Analytical Chemistry (access Card)
9th Edition
ISBN: 9781319090241
Author: Daniel C. Harris, Sapling Learning
Publisher: W. H. Freeman
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Chapter 15, Problem 15.29P

a)

Interpretation Introduction

Interpretation:

The ionic strength of the buffer is μ=0.100m has to be proved.

a)

Expert Solution
Check Mark

Explanation of Solution

Given:

pH is 6.865 at 250C for 0.0250mKH2PO4|0.0250mNa2HPO4 buffer.

The change in buffer species when mixed with the acid H2PO4- with its conjugate base, HPO42- are negligible. The ionic strength of 0.0250mKH2PO4 (1:1electrolyte) is 0.0250m and the ionic strength of 0.0250mNa2HPO4 (2:1eletrolyte) is 0.0750m so the total ionic strength is 0.100m .

Conclusion

The ionic strength of the buffer is μ=0.100m was proved.

b)

Interpretation Introduction

Interpretation:

The pH and K2 for phosphoric acid, the quotient of activity coefficients γHPO4-/γH2PO4-atμ=0.100m has to be calculated.

Concept Introduction:

Metal ion buffer:

A metal-ion buffer gives a controlled source of free metal ions in a way related to the regulation of hydrogen ion concentration by a pH buffer. A metal-ion buffer solution has the free (hydrated) metal ion along with a complex compound made by the association of the ion with a ligand in excess. The concentration of free metal ion mainly depends on the total concentration of each component (ligand and metal ion) and also on the stability constant of the complex. The concentration of the free metal ion depends upon on solution pH if the ligand can undergo protonation.

Concentration= [M][L][ML]

b)

Expert Solution
Check Mark

Answer to Problem 15.29P

The pH and K2 for phosphoric acid, the quotient of activity coefficients γHPO4-/γH2PO4-atμ=0.100m was calculated as K2=0.4645 .

Explanation of Solution

Given:

Concentration= [M][L][ML]

K2=[H+]γH+[HPO42-]γHPO42-[H2PO4-]γH2PO4-

Given K2=10-7.198 and [H+]γHI+=10-pH

Where pH= 6.685

[H+]γHI+=10-pH=10-6.865

Then,

γHPO42-γH2PO42-=K2[H2PO4-][H+]γH+[HPO42-]=10-7.168[0.0250]10-6.685[0.0250]=0.4645

Units will cancel each other since, both the numerator and denominator has concentrations.

Conclusion

The pH and K2 for phosphoric acid, the quotient of activity coefficients γHPO4-/γH2PO4-atμ=0.100m was calculated as K2=0.4645 .

c)

Interpretation Introduction

Interpretation:

The needed molalities of KH2PO4andNa2HPO4 should be mixed to obtain a pH of 7.000 and μ=0.100m has to be explained.

Concept Introduction:

Metal ion buffer:

A metal-ion buffer gives a controlled source of free metal ions in a way related to the regulation of hydrogen ion concentration by a pH buffer. A metal-ion buffer solution has the free (hydrated) metal ion along with a complex compound made by the association of the ion with a ligand in excess. The concentration of free metal ion mainly depends on the total concentration of each component (ligand and metal ion) and also on the stability constant of the complex. The concentration of the free metal ion depends upon on solution pH if the ligand can undergo protonation.

Concentration= [M][L][ML]

c)

Expert Solution
Check Mark

Answer to Problem 15.29P

The needed molalities of Na2HPO4andKH2PO4 should be mixed to obtain a pH of 7.000 and μ=0.100m was 0.0268m,0.0196m .

Explanation of Solution

The pH 7.000 can be obtained by decreasing the concentration of acid (H2PO4-) and increase the concentration of base (HPO4-) .

In order maintain a constant ionic strength we must decrease KH2PO4 three times as much as we increase Na2HPO4 because Na2HPO4 can contributes three times as much as KH2PO4 to the ionic strength.

So increasing the Na2HPO4 by x and decrease KH2PO4 by 3x .

Given:

Concentration= [M][L][ML]

K2=[H+]γH+[HPO42-]γHPO42-[H2PO4-]γH2PO4-

Given K2=10-7.198

10-7.198=10-7.000[0.0250+x][0.0250-3x](0.4645)

x=0.0018m

So, the new concentration are Na2HPO4=0.0268m and KH2PO4=0.0196m

Conclusion

The needed molalities of Na2HPO4andKH2PO4 should be mixed to obtain a pH of 7.000 and μ=0.100m was 0.0268m,0.0196m .

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