TUTORIAL 5

LEARNING OUTCOMES By the end of this session the student should be able to: Compare between different titration curves. Select an Indicator for a certain titration. Apply double indicators. Applications

pH=pK a2 + log (base/acid) pH= log ( 0.1/ 0.15) pH = 7.02 Calculate the pH of the buffer system with 0.1M Na 2 HPO 4 / 0.15M KH 2 PO 4. (K a1 = 7.5*10 -3,K a2 = 6.2*10 -8 ) QUESTION:

NORMALITY VS MOLARITY N = Mn n is the number of acidic H in case of an acid. n is the number of OH in case of a base. n is the (number of cations x valency of cation) in case of a salt. n is the number of electrons lost or gained incase we are dealing with a reducing agent or oxidizing agent respectively. so you can convert from molarity to normality simply by knowing the n.

(C x V) titrant = (C x V) sample (N x V) titrant = (N x V) sample ( M x V x n) titrant = ( M x V x n) sample We also know that NORMALITY VS MOLARITY

ACID –BASE INDICATORS Example: Phenolphthalein HIn is colourless In - is pink Indicators are weak acids or weak bases. HIn H + + In - Acidic form basic form Colour Different colour

When we choose an indicator for a titration: ACID –BASE INDICATORS For a typical acid-base indicator with dissociation constant K a, the colour transition occurs over a range of pH values given by: pk a  1 we choose the indicator according to the pH at the endpoint It should lie within the pH range of the indicator and the best is at the midpoint of the pH range.

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DOUBLE INDICATOR TITRATIONS For the determination of mixtures of two monobasic acidspolyfunctional acids or bases two monobasic acids or polyfunctional acids or bases Difference in pK a must be at least 4

Example mixture of HCl and CH 3 COOH Methyl orange e.p detects HCl. Phenolphthalein e.p detects total acids. Simply by subtraction you get an e.p for CH 3 COOH. DOUBLE INDICATOR TITRATIONS

How to detect the number of end points for any weak polyprotic acid? The difference in the pk a1 and Pk a2 must be at least 4. The pH at the end point is equal to the average between the 2 pk a values. The last pk a will be compared to the pk w of water (14).

pK a1 = 2.12 pK a2 = 7.21 pK a3 = 12.3 pK w = = 5.09 (>4), so an e.p can be detected pH of e.p = /2= = 5.09 (>4), so an e.p can be detected pH of e.p = /2= =1.70(<4), so No e.p can be detected i.e. H 3 PO 4 DOUBLE INDICATOR TITRATIONS How to detect the number of end points for any weak polyprotic acid?

Assignment 5 1.In the following polyprotic acids: a. Determine how many E.P? b. Determine the suitable indicators. c. What is the significance of the end point (equivalent to how much of the acid) d. Sketch the titration graph (showing how many breaks).

a) Alanine k a1 = 4.49X10 -3 k a2 = 1.36 X pka 1 = 2.34 pka 2 = 9.86 pk a2 -pk a1 =7.52 E.p.1: pH=(pka 2 +pka 1 )/2 = /2=6.10 (bromocresol, E.B.T) e.p = ½ acid pk w –pka 2 =4.14 E.p.2: pH=(pka2+pka1)/2 = /2=11.93(Alizarin) e.p = total of acid

b) Aspartic acid ka 1 =1.02X10 -2 ka 2 =1.26X10 -4 ka 3 =9.95X pk a1 = 1.99 pk a2 = 3.89 pk a3 = pk a2 -pk a1 =1.9 No End point pk a3 –pk a2 =6.12 E.p.1: /2=6.95 (Bromothymol Blue) e.p = 2/3 of acid pk w –pk a3 =3.99 No End point

Phosphoric acid Na 2 CO 3 + NaOH Na 2 CO 3 + NaHCO 3 Methyl orange1/3 of AcidTotal Phenolpthalin2/3 of Acid ½ Na 2 CO 3 + NaOH½ Na 2 CO 3 N.B. This is valid only if we are working on a new portion of sample Titration of polyprotic acid against strong base or mixture of bases against strong acid

2. 10 ml mixture of Na 2 CO 3 and NaHCO 3 requires 10 ml of 0.1N HCI using phenolphthalein as indicator and requires 41ml of the same acid on using methyl orange. Calculate the Normality of Na 2 CO 3 and NaHCO 3 in the mixture.

Ph.Ph. gets ½ CO 3 -2 M.O. gets total basicity NV CO3 -2 = NV titrant (N*10 ) = (0.1 * 10*2) N CO3 -2 = 2/10=0.2 N NV HCO3 -2 = Nv titrant N*10 = 0.1*(41-20) N HCO3 -2 = 2.1/10 =0.21 N Solution:

3.12 ml of 0.1N NaOH were required for the end point of titration against 10 ml phosphoric acid using phenolphthalein as indicator. Calculate the concentration of the acid. M.O. gets 1/3 H 3 PO 4 Ph.Ph. gets 2/3 H 3 PO 4 NV H 3 PO 4 = NV titrant N * 10 = 0.1 * (3/2*12) N H 3 PO 4 = 1.8/10= 0.18N Solution:

4.15ml mixture of Na 2 CO3, NaHCO3 Consumed 10ml of 1M HCl upon using phenolphthalein as indicator. On the same flask methyl orange was added and the titration was continued consuming another 35 mls of 1M HCl Calculate the Molarity of Na 2 CO3 & NaHCO3 in the mixture.

Ph.ph e.p = ½ Na 2 CO3 So E.P for Na 2 CO3 = 10 x 2 = 20 mls M.O e.p = ½ Na 2 CO3 + NaHCO3 So E.P for NaHCO 3 = = 25 mls (M x V x n) Na2CO3 = (M x V x n) HCl (M x 15 x 2) Na2CO3 = (1 x 20 x 1) HCl M of Na 2 CO3 = 0.66 M (M x V x n) NaHCO3 = (M x V x n) HCl (M x 15 x 1) NaHCO3 = (1 x 25 x 1) HCl M of NaHCO 3 = 1.66 M Solution: