1. 1 Example A 25.0  L sample was found to contain 26.7  g glucose. Express the concentration as ppm and mg/dL glucose. Solution A ppm is defined as  g/mL, therefore ppm = 26.7  g/(25.0x10 -3 mL) = 1.07x10 3 ppm

2. 2 Or one can use dimensional analysis considering always a ppm as mg/L as below ? mg/L glucose = (26.7  g/25.0  L) x (10 -3 mg/  g) x (10 6  L/L) = 1.07x10 3 ppm Now let us find mg glucose per deciliter ?mg glucose/dL = = (26.7  g/25.0  L) x (10 -3 mg/  g) x (10 6  L/L) x (L/10dL) = 107 mg/dL

3. 3 Example Find the molar concentration of a 1.00 ppm Li (at wt = 6.94 g/mol) and Pb (at wt = 207 g/mol). Solution A 1.00 ppm is 1.00 mg/L, therefore change this 1.00 mg into moles to obtain molarity. ? mol Li/L = (1.00x10 -3 g Li/L) x ( 1 mol Li/6.94 g Li) = 1.44x10 -4 M ? mol Pb/L = (1.00x10 -3 g Pb/L) x ( 1 mol Pb/207 g Li) = 4.83x10 -6 M

4. 4 Example Find the number of mg Na 2 CO 3 (FW = 106 g/mol) required to prepare 500 mL of 9.20 ppm Na solution. Solution The idea is to find the of mg sodium ( 23.0 mg/mmol) required and then get the mmoles sodium and relate it to mmoles sodium carbonate followed by calculation of the weight of sodium carbonate.

5. 5 ? mg Na = 9.20 mg/L x 0.5 L = 4.6 mg Na mmol Na = 4.60 mg Na/23.0 mg/mmol mmol Na 2 CO 3 = 1/2 mmol Na = 4.60/46.0 mmol = 0.100 ? mg Na 2 CO 3 = (4.60/46.0) mmol Na 2 CO 3 x (106 mg Na 2 CO 3 / mmol Na 2 CO 3 ) = 10.6 mg One can work such a problem in one step as below ? mg Na 2 CO 3 = (9.2 mg Na/1000mL) x 500 mL x (1mmol Na/23.0 mg Na) x (1 mmol Na 2 CO 3 /2 mmol Na) x (106 mg Na 2 CO 3 /1 mmol Na 2 CO 3 ) = 10.6 mg

6. 6 Stoichiometric Calculations: Volumetric Analysis In this section we look at calculations involved in titration processes as well as general quantitative reactions. In a volumetric titration, an analyte of unknown concentration is titrated with a standard in presence of a suitable indicator. For a reaction to be used in titration the following characteristics should be satisfied:

7. 7 1.The stoichiometry of the reaction should be exactly known. This means that we should know the number of moles of A reacting with 1 mole of B. 2.The reaction should be rapid and reaction between A and B should occur immediately and instantly after addition of each drop of titrant (the solution in the burette). 3.There should be no side reactions. A reacts with B only.

8. 8 4.The reaction should be quantitative. A reacts completely with B. 5.There should exist a suitable indicator which has distinct color change. 6.There should be very good agreement between the equivalence point (theoretical) and the end point (experimental). This means that Both points should occur at the same volume of titrant or at most a very close volume. Three reasons exist for the disagreement between the equivalence and end points. The first is whether the suitable indicator was selected, the second is related to concentration of reactants, and the third is related to the value of the equilibrium constant. These factors will be discussed in details later in the course.

9. 9 Standard solutions A standard solution is a solution of known and exactly defined concentration. Usually standards are classified as either primary standards or secondary standards. There are not too many secondary standards available to analysts and standardization of other substances is necessary to prepare secondary standards. A primary standard should have the following properties:

10. 10 1.Should have a purity of at least 99.98% 2.Stable to drying, a necessary step to expel adsorbed water molecules before weighing 3.Should have high formula weight as the uncertainty in weight is decreased when weight is increased 4.Should be non hygroscopic 5.Should possess the same properties as that required for a titration

11. 11 Remember!! NaOH and HCl are not primary standards and therefore should be standardized using a primary or secondary standard. NaOH absorbs CO 2 from air, highly hygroscopic, and usually of low purity. HCl and other acid in solution are not standards as the percentage written on the reagent bottle is a claimed value and should not be taken as guaranteed.

12. 12 Molarity Volumetric Calculations Volumetric calculations involving molarity are rather simple. The way this information is presented in the text is not very helpful. Therefore, disregard and forget about all equations and relations listed in rectangles in the text, you will not need it. What you really need is to use the stoichiometry of the reaction to find how many mmol of A as compared to the number of mmoles of B.

13. 13 Example A 0.4671 g sample containing NaHCO 3 (FW = 84.01 mg/mmol) was dissolved and titrated with 0.1067 M HCl requiring 40.72 mL. Find the percentage of bicarbonate in the sample. Solution We should write the equation in order to identify the stoichiometry

14. 14 NaHCO 3 + HCl  NaCl + H 2 CO 3 Now it is clear that the number of mmol of bicarbonate is equal to the number of mmol HCl mmol NaHCO 3 = mmol HCl mmol = M x V mL mmol NaHCO 3 = (0.1067 mmol/ml ) x 40.72 mL = 4.345 mmol Now get mg bicarbonate by multiplying mmol times FW Mg NaHCO 3 = 4.345 mmol x (84.01 mg/mmol) = 365.0 1 % NaHCO 3 = (365.0 1 x 10 -3 g/0.4671 g) x 100 = 78.14%

15. 15 We can use dimensional analysis to calculate the mg NaHCO 3 directly then get the percentage as above. ? mg NaHCO 3 = (0.1067 mmol HCl/ml) x 40.72 mL x (mmol NaHCO 3 /mmol HCl) x (84.01 mg NaHCO 3 / mmol NaHCO 3 ) = 365.0 mg