Practical Analytical chemistry

Section (3)

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.

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 addedof the titrant

Equivalencepoint

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.

Illustrative diagrams for the titration of NaOH ≠ standard HCl

Illustrative diagrams for the titration ofNa2CO3 ≠ standard HCl

Illustrative diagrams for the titration ofNaHCO3 ≠ standard HCl

Summary:

This section includes: Determination of NaOH/Na2CO3 mixture.

OH-CO3-- HCO3

-

E.Pph.ph.

)1st flask:( all OH- ½ CO3

-- No E.P.

E.PM.O.

)2nd flask:( all OH- all CO3

--

all HCO3-

Determination of NaOH & Na2CO3 mixture

Principle: According to the two-indicators method: First step: (E.Pph.ph) Titration of 10 ml of the mixture ≠ standard HCl using

ph.ph. indicator. E.Pph.ph ≡ OH- + 1/2 CO3

--

Second step: (E.PM.O) Titration of another 10 ml of the mixture ≠ standard

HCl using M.O. ind. E.PM.O ≡ OH- + CO3

--

Reaction equations For NaOH: NaOH + HCl → NaCl + H2O For Na2CO3: Na2CO3 + HCl → NaCl +

NaHCO3 ...................Half neutralization (pH ≈ 8.3) NaHCO3 + HCl → NaCl + CO2 +

H2O ...............Complete neutralization (pH ≈ 3.8) . Na2CO3 + 2HCl → 2NaCl + CO2 + H2O

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.PM.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 }

Calculations:For NaOH : NaOH + HCl NaCl + H20 1 HCl ≡ 1 NaOHEquivalence 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

Calculation:

For Na2CO3

Na2CO3 + 2HCl → 2NaCl + CO2 + H2O

2 HCl ≡ 1 Na2CO3

1 HCl ≡ 1/2 Na2CO3

Equivalence factor (F): each ml of 0.2 N HCl ≡ 1/2xM.W. of Na2CO3 x 0.2≡ ... g

1000 Concentration:

Conc. of Na2CO3 =2(E.P2 - E.P1) x F x 1000= .......... g/L

10