1. Practical Analytical chemistry Section (3)

2. Choice of Acid- Base indicators: The Equivalence Point [Theoretical]  It is the point at which the added amount of the standard solution is chemically equivalent to the amount of the substance being determined.  Its real position can only be theoretically calculated.  It is expressed by the volume of the titrant added. The End Point [Experimental]  It is the point at which the visual change of the indicator takes place in the titration.  Its position can be determined experimentally.  It is expressed by the volume of the titrant added.

3. The Titration Curve of Acid-Base titrations:  It is the curve obtained by plotting the pH of the titrated solution (y-axis) against the volume of the titrant added (x-axis).  The titration curve is characterized by a Sigmoid shape and the midpoint of the vertical part of the curve corresponds to the equivalence point. pH mls added of the titrant Equivalence point

4. The pH-range of the indicator:  It is the pH-range ( pH-interval ) within which the indicator changes its color.  Ex.: - The pH-range of M.O. ≈ (3 – 4.5) approximately.  - The pH-range of ph.ph. ≈ (8.5 – 10) approximately.

5. Illustrative diagrams for the titration of NaOH ≠ standard HCl

6. Illustrative diagrams for the titration of Na 2 CO 3 ≠ standard HCl

7. Illustrative diagrams for the titration of NaHCO 3 ≠ standard HCl

8. Summary:  This section includes:  Determination of NaOH/Na 2 CO 3 mixture. HCO 3 - CO 3 -- OH - No E.P. ½ CO 3 -- all OH - E.P ph.ph. (1st flask): all HCO 3 - all CO 3 -- all OH - E.P M.O. (2nd flask):

9. Determination of NaOH & Na 2 CO 3 mixture  Principle:  According to the two-indicators method:  First step: (E.P ph.ph )  Titration of 10 ml of the mixture ≠ standard HCl using ph.ph. indicator.  E.P ph.ph ≡ OH - + 1/2 CO 3 --  Second step: (E.P M.O )  Titration of another 10 ml of the mixture ≠ standard HCl using M.O. ind.  E.P M.O ≡ OH - + CO 3 --

10. Reaction equations  For NaOH:  NaOH + HCl → NaCl + H 2 O  For Na 2 CO 3 :  Na 2 CO 3 + HCl → NaCl + NaHCO 3...................Half neutralization (pH ≈ 8.3)  NaHCO 3 + HCl → NaCl + CO 2 + H 2 O...............Complete neutralization  (pH ≈ 3.8).  Na 2 CO 3 + 2HCl → 2NaCl + CO 2 + H 2 O

11. Procedure: 1. E.P. ph.ph :-  Transfer 10 ml of the mixture into a clean conical flask.  Add 10 drops ph.ph. indicator.  Titrate ≠ 0.2 N HCl.  { Color change at E.P.: from Pink to Colorless } 2. E.P M.O.:  Transfer another 10 ml of the mixture into a clean conical flask.  Add 2 drops M.O. indicator.  Titrate ≠ 0.2 N HCl.  { Color change at E.P.: from Yellow to Orange }

12. Calculations : For NaOH :  NaOH + HCl NaCl + H20   1 HCl ≡ 1 NaOH Equivalence factor (F):  each ml of 0.2 N HCl ≡1 x M.W. of NaOH x 0.2≡.......... g 1 x 1000 Concentration:  Concn. of NaOH =[E.P2 - 2(E.P2 - E.P1)] x F x 1000=.......... g/L 10

13. Calculation:  For Na 2 CO 3 Na 2 CO 3 + 2HCl → 2NaCl + CO 2 + H 2 O   2 HCl ≡ 1 Na 2 CO 3   1 HCl ≡ 1/2 Na 2 CO 3  Equivalence factor (F): each ml of 0.2 N HCl ≡ 1/2xM.W. of Na 2 CO 3 x 0.2≡... g 1000  Concentration: Conc. of Na 2 CO 3 =2(E.P2 - E.P1) x F x 1000=.......... g/L 10