Quantitative Chemical Analysis
Quantitative Chemical Analysis
9th Edition
ISBN: 9781464135385
Author: Daniel C. Harris
Publisher: W. H. Freeman
Question
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Chapter 4, Problem 4.33P

(a)

Interpretation Introduction

Interpretation:

For measurement giving y value of 2.58 , the corresponding x value and its standard uncertainty has to be found out.

Concept Introduction:

Uncertainty with a Calibration Curve:

Propagation for the equation y=mx+b (but not y=mx) gives,

The standard uncertainty in x is given as:

ux=sy|m|1k+1n+(yy¯)2m2(xix¯)2

Where,

ux is standard uncertainty for x

y is the corrected absorbance of unknown

xi is the mass of sample in standards

y¯ is the average of y values

x¯ is the average of x values

k is the number of replicate measurements

(a)

Expert Solution
Check Mark

Answer to Problem 4.33P

The corresponding x and its standard uncertainty value for y value of 2.58 is 2.0±0.3

Explanation of Solution

Given data:

Consider least-squares in Figure 4-11

A new single measurement gives a y value of 2.58

Finding of x value and its standard uncertainty:

The equation of straight line from figure 4-11 is y=mx+by=0.61538x+1.34615

On rearranging, we get,

x=ybm   =2.581.350.615   =2.00

The value of y¯ and x¯ is calculated as follows,

y¯=(2+3+4+5)/4   =3.5x¯=(1+3+4+6)/4   =3.5

The (xix¯)2 is found as follows,

(xix¯)2=(13.5)2+(33.5)2+(43.5)2+(63.5)2       =13.0

The standard uncertainty in x is given as,

ux=sy|m|1k+1n+(yy¯)2m2(xix¯)2    =0.19612|0.61538|11+14+(2.583.5)2(0.61538)2(13.0)    =0.38    =0.3 (rounded off)

Therefore, the value for x is 2.0±0.3

Conclusion

The corresponding x and its standard uncertainty value for y value of 2.58 is found out as 2.0±0.3

(b)

Interpretation Introduction

Interpretation:

The four times measurement of y gives an average value of 2.58 , the corresponding x value and its standard uncertainty based on four measurements has to be found out.

Concept Introduction:

Uncertainty with a Calibration Curve:

Propagation for the equation y=mx+b (but not y=mx) gives,

The standard uncertainty in x is given as:

ux=sy|m|1k+1n+(yy¯)2m2(xix¯)2

Where,

ux is standard uncertainty for x

y is the corrected absorbance of unknown

xi is the mass of sample in standards

y¯ is the average of y values

x¯ is the average of x values

k is the number of replicate measurements

(b)

Expert Solution
Check Mark

Answer to Problem 4.33P

The corresponding x and its standard uncertainty value based on four measurements is   2.0±0.2

Explanation of Solution

Given data:

Consider least-squares in Figure 4-11

The y is measured four times and its average value is 2.58

The numbers in subscript denotes insignificant figures.

Finding of x value and its standard uncertainty:

The equation of straight line from figure 4-11 is y=mx+by=0.61538x+1.34615

On rearranging, we get,

x=ybm   =2.581.350.615   =2.00

The value of y¯ and x¯ is calculated as follows,

y¯=(2+3+4+5)/4   =3.5x¯=(1+3+4+6)/4   =3.5

The (xix¯)2 is found as follows,

(xix¯)2=(13.5)2+(33.5)2+(43.5)2+(63.5)2       =13.0

Here, the number of replicate measurement is four. Hence, k=4

The standard uncertainty in x is given as,

ux=sy|m|1k+1n+(yy¯)2m2(xix¯)2    =0.19612|0.61538|14+14+(2.583.5)2(0.61538)2(13.0)    =0.26   =0.2 (rounded off)

Therefore, the value for x is 2.0±0.2

Conclusion

The corresponding x and its standard uncertainty value based on four measurements is found out as 2.0±0.2

(c)

Interpretation Introduction

Interpretation:

The 95% confidence intervals for (a) and (b) has to be found out.

Concept Introduction:

Confidence Intervals:

The confidence interval is given by the equation:

Confidence interval = x¯±tsn       =x¯±tux (since standard uncertainty(ux)=s/n

Where,

x¯ is mean

s is standard deviation

t is Student’s t

n is number of measurements

ux is standard uncertainty

To Find: The 95% confidence intervals for (a) and (b)

(c)

Expert Solution
Check Mark

Answer to Problem 4.33P

The 95% confidence intervals for (a) is 2.0±0.8

The 95% confidence intervals for (b) is 2.0±0.6

Explanation of Solution

Given data:

The results of the measurement in part (a) is 2.00±0.38

The results of the measurement in part (b) is 2.00±0.26

The numbers in the subscript denotes insignificant figures.

Calculation of Confidence intervals:

The t value corresponding to 90% confidence level for three degrees of freedom is t90%=2.353

The 90% confidence interval for part (a) is calculated as follows,

     confidence interval    = x¯±tux90% confidence interval =2.00(±2.353×0.38)    =2.00(±0.89)    =2.0(±0.8) (rounded off)

The 90% confidence interval for part (b) is calculated as follows,

     confidence interval    = x¯±tux90% confidence interval =2.00(±2.353×0.26)    =2.00(±0.61)    =2.0(±0.6) (rounded off)

Conclusion

The 95% confidence intervals for (a) is calculated as 2.0±0.8

The 95% confidence intervals for (b) is calculated as 2.0±0.6

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