Quantitative Chemical Analysis
Quantitative Chemical Analysis
9th Edition
ISBN: 9781464135385
Author: Daniel C. Harris
Publisher: W. H. Freeman
Question
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Chapter 5, Problem 5.BE

a)

Interpretation Introduction

Interpretation:

The expression for the final concentration [Ni]f after adding 25.0mL of unknown to 0.500mL of standard has to be expressed.

Concept Introduction:-

Standard addition,

Standard addition are the addition of known quantities of analyte to unknown solution, with the increase in linearity we can deduce that how much analyte present in the original unknown solution.

Standardadditionequation=[X]i[S]f+[X]f=IXIS+X...(1)

Where,

[X]i = The concentration of analyte in initial solution.

[S]f+[X]f = The concentration of analyte plus standard in final solution.

IX = Signal from the initial solution

IS+X =signal from the final solution.

For an initial volume V0 of unknown and added volume Vs of standard with concentration [S]i ,

The total volume is

V=V0+Vs

And the concentration of equation 1 can be written as

[X]f=[X]i(V0V)[S]f=[S]i(VSV)

a)

Expert Solution
Check Mark

Explanation of Solution

Given,

Initial volume ( Vi ) = 25.0mL

Final volume ( Vf ) =25.5mL

Final concentration [Ni2+]f = [Ni2+]i(ViVf)

[Ni2+]f = [Ni2+]i(25.025.5) = 0.984[Ni2+]i

Conclusion

The expression for the final concentration [Ni]f after adding 25.0mL of unknown to 0.500mL of standard has to be expressed.

b)

Interpretation Introduction

Interpretation:

The final concentration of added standard [Ni2+] and designated as [S]f has to be expressed.

Concept Introduction:-

Standard addition,

Standard addition are the addition of known quantities of analyte to unknown solution, with the increase in linearity we can deduce that how much analyte present in the original unknown solution.

Standardadditionequation=[X]i[S]f+[X]f=IXIS+X...(1)

Where,

[X]i = The concentration of analyte in initial solution.

[S]f+[X]f = The concentration of analyte plus standard in final solution.

IX = Signal from the initial solution

IS+X =signal from the final solution.

For an initial volume V0 of unknown and added volume Vs of standard with concentration [S]i ,

The total volume is

V=V0+Vs

And the concentration of equation 1 can be written as

[X]f=[X]i(V0V)[S]f=[S]i(VSV)

b)

Expert Solution
Check Mark

Explanation of Solution

Given,

[S]f=[S]i(VSV)

Where,

Vs=0.500V=25.5[S]i=0.0287M

Substituting the above values we will get

[S]f=[S]i(VSV)

[S]f=0.0287M(0.50025.5)=0.005627M

Conclusion

The final concentration of added standard [Ni2+] and designated as [S]f was expressed as 0.005627M .

c)

Interpretation Introduction

Interpretation:

The concentration of [Ni2+]i in the unknown has to be calculated.

Concept Introduction:-

Standard addition,

Standard addition are the addition of known quantities of analyte to unknown solution, with the increase in linearity we can deduce that how much analyte present in the original unknown solution.

Standardadditionequation=[X]i[S]f+[X]f=IXIS+X...(1)

Where,

[X]i = The concentration of analyte in initial solution.

[S]f+[X]f = The concentration of analyte plus standard in final solution.

IX = Signal from the initial solution

IS+X =signal from the final solution.

For an initial volume V0 of unknown and added volume Vs of standard with concentration [S]i ,

The total volume is

V=V0+Vs

And the concentration of equation 1 can be written as

[X]f=[X]i(V0V)[S]f=[S]i(VSV)

c)

Expert Solution
Check Mark

Explanation of Solution

Given,

Standardadditionequation=[X]i[S]f+[X]f=IXIS+X...(1)

Where,

IX=2.36μΑIX+S=3.79μΑ[S]f=0.005627M[Ni2+]f=0.9804[Ni2+]i

Standardadditionequation=[Ni2+]i0.005627+0.9804[Ni2+]i=2.36μΑ3.79μΑ

[Ni2+]i=9.00×10-4M

Conclusion

The concentration of [Ni2+]i in the unknown was calculated as 9.00×10-4M .

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