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 7, Problem 7.19P

(a)

Interpretation Introduction

Interpretation:

The pCa2+ value before equivalence point for the titration of Na2C24VsCa(NO3)2 has to be calculated.

Concept introduction:

pX = - log 10[X]pfunction[X]isconcentrationofX

Molarity can be defined as the moles of solute (in grams) to the volume of the solution (in litres). The molarity of a solution can be given by the formula,

Molarity(M)=Molesofsolute(ing)Volumeofsolution(inL)

(a)

Expert Solution
Check Mark

Answer to Problem 7.19P

The pCa2+ value before equivalence point for the titration of Na2C24VsCa(NO3)2 is 6.81 .

Explanation of Solution

To calculate: The pCa2+ value before equivalence point for the titration of Na2C24VsCa(NO3)2

Given,

25.00 mL of 0.03110 M Na2C2O410.00mLof0.02570M Ca(NO3)2

The titration reaction of Na2C24VsCa(NO3)2 is as follows

Ca2+ + C2O4-  CaC2O4(s)

Before equivalence point, more moles of oxalate ion present in solution than calcium ions.  The strength of unprecipitated oxalate ion is calculated as

Moles of C2O4- = original moles of C2O4- - moles of Ca2+ added=(0.025 L)(0.03110 mol/L) - (0.010 L)(0.02570 mol/L)=0.0005205molC2O4-

Total volume of both solution is 0.035L(25.00 mL + 10.00 mL)

The concentration of iodide ion is

[C2O4-]=0.0005205mol0.035L=0.01487M

The concentration of silver ion which is in equilibrium with iodide ion is

[Ca2+]=Ksp[C2O4-]=2.3×10-170.01487=1.5466×10-7

pCa2+ = - log 10[Ca2+]=log(1.5466×10-7)=6.81

(b)

Interpretation Introduction

Interpretation:

The pCa2+ value at equivalence point for the titration of Na2C24VsCa(NO3)2 has to be calculated.

Concept introduction:

pX = - log 10[X]pfunction[X]isconcentrationofX

Molarity can be defined as the moles of solute (in grams) to the volume of the solution (in litres). The molarity of a solution can be given by the formula,

Molarity(M)=Molesofsolute(ing)Volumeofsolution(inL)

(b)

Expert Solution
Check Mark

Answer to Problem 7.19P

The pCa2+ value at equivalence point for the titration of Na2C24VsCa(NO3)2 is 4.32 .

Explanation of Solution

To calculate: The pCa2+ value at equivalence point for the titration of Na2C24VsCa(NO3)2

Given,

25.00 mL of 0.03110 M Na2C2O40.02570M Ca(NO3)2

The titration reaction of Na2C24VsCa(NO3)2 is as follows

Ca2+ + C2O4-  CaC2O4(s)

At equivalence point, moles of oxalate present in solution is equal to calcium ions.

The concentration of silver ion and iodide ion is calculated as

[Ca2+]=[ C2O4-][Ca2+][C2O4-]=Ksp(x)(x)=2.3×10-9x=2.3×10-9=4.795×10-5

pCa2+ = - log 10[Ca2+]=log(4.795×10-5)=4.32

(c)

Interpretation Introduction

Interpretation:

The pCa2+ value after equivalence point for the titration of Na2C24VsCa(NO3)2 has to be calculated.

Concept introduction:

pX = - log 10[X]pfunction[X]isconcentrationofX

Molarity can be defined as the moles of solute (in grams) to the volume of the solution (in litres). The molarity of a solution can be given by the formula,

Molarity(M)=Molesofsolute(ing)Volumeofsolution(inL)

(c)

Expert Solution
Check Mark

Answer to Problem 7.19P

The pCa2+ value after equivalence point for the titration of Na2C24VsCa(NO3)2 is 2.69

Explanation of Solution

To calculate: The pCa2+ value after equivalence point for the titration of Na2C24VsCa(NO3)2

Given,

25.00 mL of 0.03110 M Na2C2O40.02570M Ca(NO3)2

The titration reaction of Na2C24VsCa(NO3)2 is as follows

Ca2+ + C2O4-  CaC2O4(s)

The volume of calcium ions at equivalence point is

(0.025L)(0.03110M)=Ve×(0.02570)Ve=0.002050.05110=30.25mL

Most of oxalate ion are precipitated at equivalence point.  After equivalence point, more moles of calcium ions present in solution.  The volume of calcium ion after equivalence point is 35.00-30.25=4.75mL .

Moles of Ca2+ = (0.00475 L)(0.02570 mol/L)=0.000344molCa2+[Ca2+]=(0.0001221mol)/(0.060L)=0.0020346

pCa2+ = - log 10[Ca2+]=log(0.0020346)=2.69

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