1. Redox QUANTITATIVE ANALYTICAL CHEMISTRY Electrochemistry
1) Volumetric (titrimetric) analysis
Acid-Base
preciptimetry
Complexmetry
Redox
2) Gravimetric analysis
3) instrumental analysis
Electrochemistry
Spectrophotometry
Spectrofluorometry
Flame photometric methods

2. Statisticsد:محمد عبد الجليل
Redox theoryأ.د.حريه
Redox applicationأ.د.حسن عزقل
Complexometry أ.د.سميحه

3. STATISTICS

4. Error: deviation from the absolute value .
Absolute error (E) : the difference between an observed or measured value (O) and the true value (T) , with regard to the sign and it is reported in the same units as the measurement.
E = O – T
Mean error: the difference between the average of several measurements and the true value (T).
Relative Error: absolute or mean error (E) expressed as a percentage (%) of the true value (T)

5. Example,1:
If a 2.62 g sample of material analyzed to be 2.52 g.
The absolute error (E) = 2.52 – 2.62 = g.
Example,2:
In the titration of 10 ml of 0.1 N NaOH with laboratory prepared 0.1 N HCl, the true value is 9.9 ml, and we have:
10.1, 9.9, 9.8, 10.2, 10.1 observed values,
So, Mean = summation of observed values / their number
= ( ) / 5 = ml.
And mean error = – 9.9 = 0.12 ml.

6. Relative Error = (-0.10/2.62) x 100% = - 3.8 % In example,2:
Relative Mean Error = ( 0.12/9.9) x 100% = 1.21%

7. Types of Errors:
(A) Determinate or systemic (constant) errors:
can be determined, (can be avoided)
(B) Indeterminate (random, accidental or chance) errors:
cannot be determined or corrected.

8. Accuracy:agreement of a measurement with the true value.
Determination of accuracy:
Absolute method
Accuracy is determined from the relative error;
In example,1: Relative Error = (-0.10/2.62) x 100% = -3.8 %
And accuracy = – 3.8 = 96.2 %.
In example,2: Relative Mean Error = ( 0.12/9.9) x 100% = 1.21%
And accuracy = – 1.21 = %.

9. Example 3:
In practical exam of volumetric analysis, three students get the following results:

10. Precision :The agreement between several measurements of the same substance.
Mean (X):It is the arithmetic average of all measured values.
The range (w): the"spread":
It is the difference between the highest measurement and the lowest one.
The median: It is the measurement in the middle of the arranged measurements where the numbers of higher and lower measurements are equal.

11. standard deviation (s):
Variance =The square of the standard deviation = S2
Relative standard deviation (RSD)
Coefficient of variation (C.V.) :

12. Example:
Analysis of a sample of iron ore gave the following % values:
7.08, 7.21, 7.12, 7.09, 7.16, , 7.07, 7.14, 7.18,
Calculate the mean, standard deviation, the variance and coefficient of variation;
Find also the median and the range for these data.

14. Variance (S2) = 0.002
(b) The arranged data are:
7.07, 7.08, 7.09, 7.11, 7.12, 7,14, 7,14, 7.16, 7.18, 7.21
The median is : ( ) / 2 = 7.13
The range is : – 7.07 = 0.14

15. Rejection of a result (The Q test):
The Q test is used to determine if an “outlier” is due to a determinate error or due to indeterminate error.
If it is due to a determinate error, it should be rejected.
If it is not due to a determinate error, then it falls within the expected random error and should be retained.
The ratio Q is calculated by arranging the data in decreasing order of numbers.

16. The difference (a) between the suspect number (the outlier) and its nearest neighbour number is divided by:
the range (w), which is the difference between the highest number and the lowest number,

17. The ratio is compared with the tabulated values of Q (see the Table).
If Q measured is equal or greater than the tabulated value, the suspected observation can be rejected.
If it is smaller than the tabulated value, the suspected observation is retained

18. Example:
The following set of chloride analysis on separate aliquots of serum were reported; 103, 106, 107 and 114 meq/L. one value appears suspect.
Determine if it may be rejected or not.

19. Answer:
The suspected result is 114 meq/L.
It differs from the nearest neighbor by (a) : 114 – 107 = 7 meq/L.
The range (w) is : 114 – 103 = 11 meq/L.
Therefore, Q = a/w = 7/11 = 0.64
The tabulated Q value (4 observations, 95% confidence level) is: 0.829
Since the calculated Q value is less than the tabulated Q value,
the suspected no. (114 meq/L) retained.

20. Significant figures
‘digit’ = 0, 1, 2, ………..8,9
A significant figure = is a digit which denotes the amount of quantity in the place in which it stands.
The digit 0 is a significant figure except when it is the first figure in a number.
In g and g
the zero is significant,
but in the quantity kg
the zero is not significant, because kg = 2.5 g.

21. 1 g means that it is between 0.9 and 1.1 g
Take 10.0 ml of Zn2+ sample, add 10 ml of NH3-buffer
Weigh g of powdered drug sample, add 2 g of hexamine reagent …..
1 kg of tomato xxxxxxx kg of gold !!!
volume which is known to be between 20.5 ml and 20.7 ml should be written as 20.6 ml; but not as
20.60 ml indicates that the value lies between ml and ml.

22. at different confidence levels
Rejection quotient, Q,
at different confidence levels