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

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

STATISTICS

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)

Example,1: If a 2.62 g sample of material analyzed to be 2.52 g. The absolute error (E) = 2.52 – 2.62 = -0.10 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 = (10.1 + 9.9 + 9.8 + 10.2 + 10.1) / 5 = 10.02 ml. And mean error = 10.02 – 9.9 = 0.12 ml.

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

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.

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 = 100.0 – 3.8 = 96.2 %. In example,2: Relative Mean Error = ( 0.12/9.9) x 100% = 1.21% And accuracy = 100.00 – 1.21 = 98.79 %.

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

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.

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

Example: Analysis of a sample of iron ore gave the following % values: 7.08, 7.21, 7.12, 7.09, 7.16, 7.14, 7.07, 7.14, 7.18, 7.11. Calculate the mean, standard deviation, the variance and coefficient of variation; Find also the median and the range for these data.

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 : (7.12 + 7.14) / 2 = 7.13 The range is : 7.21 – 7.07 = 0.14

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.

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,

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

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.

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.

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 1.2680 g and 1.0062 g 5 the zero is significant, but in the quantity 0.0025 kg 2 the zero is not significant, because 0.0025 kg = 2.5 g.

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 1.000 g of powdered drug sample, add 2 g of hexamine reagent ….. 1 kg of tomato xxxxxxx 1.000 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. 20.60 ml indicates that the value lies between 20.59 ml and 20.61 ml.

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