Anand A.Mahajan (Asst. Prof. Pharm. Chem)

Methyl orange

Strong acid versus Strong base titration:

It involves neutralization reaction in which an acid is reacted with an equivalent amount of base.

Example: HCl + NaOH NaCl + H 2O

The H+ and OH- combine to form H2O, and Na+ and Cl- combine to form NaCl, so

the net result of neutralization is conversion of the HCL to a neutral solution of NaCl.

We have to construct titration curve in order to detect the end point; the end point signals the completion of reaction. A titration curve is constructed by plotting the pH of the solution as a function of the volume of titrant added. The titrant is always strong acid or a strong base. The analyte may be either a strong base or acid or a weak base or acid.

Example:

Consider titration of 100 ml of 1 M HCl with 1 M NaOH.

The pH of 1 M HCl is 0 as activity coefficients of ions are zero.

Titration is performed and readings are noted in following table.

Table: pH during titration of 100 ml of HCl with NaOH of equal concentration

NaOH added (ml) 1 M solution (pH) 0.1 M solution (pH) 0.01 M solution (pH)

00.0 0.0 1.0 2.0

50.0 0.5 1.5 2.5

75.0 0.8 1.8 2.8

90.0 1.3 2.3 3.3

98.0 2.0 3.0 4.0

99.0 2.3 3.3 4.3

99.5 2.6 3.6 4.6

99.8 3.0 4.0 5.0

99.9 3.3 4.3 5.3

100.0 7.0 7.0 7.0

100.1 10.7 9.7 8.7

100.2 110. 10.0 9.0

100.5 11.4 10.4 9.4

101.0 11.7 10.7 9.7

We start with 1 M x 100 ml = 100 mmol H+ ion concentration

After adding 50 ml of NaOH 1 M x 50 ml = 50 mmol

What is pH ??????

At start mmol of H+ = 100 mmol H+

After addition of 50 ml of NaOH = 50 mmol OH-

H+ left after addition = 50 mmol H+ in 150 ml

[H+] = 50 mmol / 150 ml = 0.333 M

pH = - log 0.333 = 0.48

Calculations

Formula [H+] = (Macid x Vacid – M base x V base) / V total

[OH] = (M base x V base – M acid x V acid) / V total

Following calculation will show change in H+ ion concentration and calculation of pH after every addition of NaOH from burette.

For 50 ml of NaOH, [H+] 50 x 1/150 = 3.33 x 10-1 pH = 0.48

For 75 ml of NaOH, [H+] 25 x 1/175 = 1.43 x 10-1 pH = 0.84

For 90 ml of NaOH, [H+] 10 x 1/190 = 5.26 x 10-2 pH = 1.30

For 98 ml of NaOH, [H+] 02 x 1/198 = 1.01 x 10-2 pH = 2.00

For 99 ml of NaOH, [H+] 01 x 1/199 = 5.03 x 10-3 pH = 2.30

For 99.99 ml of NaOH, [H+] 0.1 x 1/199.9 = 5.00 x 10-4 pH = 3.30

Upon adding 100 ml of base, the pH will change sharply to 7 i.e. theoretical equivalence point. The resulting solution is simply of NaCl. Any NaOH addedbeyond this will be in excess of that needed for neutralization

For 100.1 ml of base [OH-] = 0.1/200.1 = 5.00 x 10-4 pOH = 3.3 and pH = 10.7

For 101.0 ml of base [OH-] = 01/201 = 5.00 x 10-3 pOH = 2.3 and pH = 11.7

15_327

01.0

Vol NaOH added (mL)

50.0

7.0

13.0pH

100.0

Equivalencepoint

15_328

Vol 1.0 M HCl added

7.0

14.0

50.0 mL

Equivalencepoint

pH

NaOH v/s HCl

HCl v/s NaoH

Graph constructed for strong acid strong base titration

100

1 M

0.1 M

0.01M

0.01M

0.1 M

1 M

These results shows that, as the titration proceeds, initially pH rises slowly, but between the addition of 99.9 ml and 100.1 ml of alkali, the pH of solution rises from 3.3 to 10.7 i.e. in the vicinity of the equivalence point there is a very rapid rate of pH change

Selection of Indicator for Strong acid v / s Strong base titration

Conc. of solutions Effective pH range Indicator

1 M 3 – 10.5

0.1 M 4.5 – 9.5 Phenolphthalein Methyl orange

0.001 M 5.5 – 8.5 Methyl red, Bromothymol blue, Phenol red

pH range of acid base indicator

Detection of end point: Indicators

A measurement of end point is done in such a way that end point coincides with or very close to equivalence point

It is more convenient to add an indicator to the solution and visually detect a color change rather than doing titration potentiometrically.

An indicator for an acid – base titration is a weak acid or weak base that is highly colored

An indicator for an acid – base titration is a weak acid or weak base that ishighly colored.

The color of ionized form is markedly different from that of nonionized form.

One form may be colorless, but the other must be colored.

These substances are usually composed of highly conjugated organic constituent that gives rise to color

Assume the indicator is a weak acid, designated HIn, and assume that the nonionized form is red while the ionized form is blue

HIn H+ + In-

We can write a Henderson - Hasselbalch equation for this, just as for other weak acid

pH = pKIn + log [In-] / [HIn]

The indicator changes color over a pH range

The transition range depends on the ability of the observer to detect small color change.

With indicators in which both forms are colored, generally only one color is observed if the ratio of the concentration of the two form is 10:1; only color of more concentrated for is seen

Now we can calculate the pH transition range required to go from one color to other

When only the color of the nonionized form is seen

[In-] / [HIn] = 1 / 10Therefore

pH = pKa + log 1/10 = pKa - 1

When only the color of the ionized form is observed

[In-] / [HIn] = 10 / 1

pH = pKa + log 10/1 = pKa + 1

So the pH change is of two unit when indicator changes form one form to anotheri.e. pKa – 1 to pKa + 1

So indicators require transition range of two pH unit

Midway in the transition, the concentration of the two form are equal, and the pH = pKa .So the pKa of indicator should be close to the pH of the equivalence point.

Calculations similar to these can be made for weak base indicators, and they

revel the same transition range; the pOH midway in the transition is equal to pKb and hence the pH equals 14 – pKb. Hence , a weak –base indicator should be

selected such that pH = 14 – pKb.

Since an indicator is a weak acid or weak base, the amount added should be kept minimum so that it does not contribute appreciably to the pH and so that only a small amount of titrant will be requite to cause color change. Of course, sufficient indicator must be added to Impart an easily discernible color to the solution. Generally few tenth percent solution (wt / vol) of the indicator is prepared and two or three drops are added to the solution to be titrated.

Weak acid versus strong base titration

Consider titration of weak acid (acetic acid) 50 ml of 0.1 M with 0.1 M strong base (NaOH), the neutralization reaction is

CH3COOH + NaOH CH3COONa + H2O

The acetic acid, which is only a few percent ionized, depending upon the concentration, is neutralized to water an equivalent amount of the salt, sodium acetate

At startWhen we have 50 ml of 0.1M acetic acid then the pH is calculated for weak acid by following formula

CH3COOH CH3COO- + H+

The concentrations of the various species are as follows

[CH3COOH] [H+] [CH3COO-] Initial 0.1 0 0 Change -x + x + x Equilibrium 0.1 – x x x

From law of mass action (x) (x) / 0.1 – x = 1.75 x 10-5

[H+] = x = 1.32 x 10-3

pH = 2.88

As soon as the titration is started, some of CH3COOH is converted to CH3COONaand buffer system is set up. As the titration proceeds, the pH slowly increases as the ratio changes [CH3COO-] / [CH3COOH]

After addition of 10 ml of 0.1 M NaOH

At the start (0.1 M x 50.0 ml = 5.0 mmol CH3COOH) = 5.00 mmol After addition of 10.0 ml NaOH (0.1 M x 10.0 ml ) = 1.00 mmol

Amount of CH3COOH left = 4.00 mmol

As this is buffer system

pH = pKa + log [CH3OO-] / [CH3COOH] pH = 4.76 + log [1.00] / [4.00] = 4.16

At the mid point of titration [CH3COONa] = [CH3COOH] and pH is equal to pKa

After addition of 25 ml of 0.1 M NaOH

At the start (0.1 M x 50.0 ml = 5.0 mmol CH3COOH) = 5.00 mmol After addition of 25.0 ml NaOH (0.1 M x 25.0 ml ) = 2.50 mmol

Amount of CH3COOH left = 2.50 mmol

As this is buffer system pH = pKa + log [CH3OO-] / [CH3COOH]

pH = 4.76 + log [2.50] / [2.50] = 4.76

At the equivalence pointWe have solution of CH3COONa. Since this is a Bronsted base (it hydrolyzes), the pH at the equivalence point will be alkaline. The pH will depend on the concentration of CH3COONa. The greater the concentration , the higher the pH.

At the addition of complete NaOH i.e. 50 ml

All of the CH3COOH has been converted to CH3COONa

It is 5.00 mmol in 100 ml or 0.05 M solution

[OH] = Kw /Ka x [CH3COONa]

[OH] = 1.0 x 10 -14 / 1.75 x 10 -5 x 0.05 = 5.35 x 10-6 M

pOH = 5.27 pH = 8.73

As excesses NaOH is added beyond the equivalence point, the ionization of base CH3COONa is suppressed to a negligible amount and the pH is determined only by the concentration of excess OH-. Therefore titration curve beyond equivalence point follows that for titration of a strong acid.

At 60.0 ml we have solution of CH3COONa. The hydrolysis of the acetate is negligible in the presence of added OH-. So the pH is determined by the conc. of excess

of OH-

mmol OH- = 0.100 x 10.0 ml = 1.00 mmol in 110 ml

[OH-] = 0.009009 M

pOH = -2.04 pH = 11.96

Vol of 0.1 m NaOH added

Slowly rising region before equivalence point is called as buffer region

Curve is flattest at mid point where the ratios[CH3COONa] / [CH3COOH]is unity so buffering capacity is greatest at this pH

The transition range of the indicator for this titration must fall within pH range of about 7 to 10. Phenolphthalein fits nicely

Vol of 0.1 m NaOH added

Weak base versus strong acid titration

The titration of weak base with strong acid is completely analogous to the titration of weak acid versus strong base titration, but the titration curves are reverse of those for weak acid versus strong base

Example: Consider neutralization reaction 50 ml of 0.1 M Hydrochloric acid versus 50 ml of 0.1 M aqueous Ammonia

NH4 OH + HCl NH4 + Cl- At startWhen we have 50 ml of 0.1M aqueous ammonia then the pH is calculated for weak base by following formula NH3 + H2O NH4

+ + OH-

The concentrations of the various species are as follows

[NH3] [NH4] [OH-] Initial 0.1 0 0 Change -x + x + x Equilibrium 0.1 – x x x

From law of mass action (x) (x) / 0.1 – x = 1.75 x 10-5

[OH- ] = x = 1.32 x 10-3 pOH = 2.88 pH = 11.12

As soon as the titration is started, some of NH3 is converted to NH4Cland buffer system is set up. As the titration proceeds, the pH slowly decreases as the ratio changes [NH3] / [NH4

+]

After addition of 10 ml of 0.1 M NaOH

At the start (0.1 M x 50.0 ml = 5.0 mmol NH3 ) = 5.00 mmol After addition of 10.0 ml HCl (0.1 M x 10.0 ml ) = 1.00 mmol

Amount of NH3 left = 4.00 mmol

As this is buffer system pH = ( pKw – pKb) - log [NH4+] / [NH3]

pH = (14 - 4.76) - log [4.00] / [1.00] = 9.84

At the mid point of titration [NH4Cl] = [NH3] and pH is equal to pKa

After addition of 25 ml of 0.1 M HCl

At the start (0.1 M x 50.0 ml = 5.0 mmol NH3) = 5.00 mmol After addition of 25.0 ml HCl (0.1 M x 25.0 ml ) = 2.50 mmol

Amount of NH3 left = 2.50 mmol

As this is buffer system pH = (pKw – pKa) - log [NH4

+] / [NH3] pH = (14 - 4.76) + log [2.50] / [2.50] = 9.24

At the equivalence pointWe have solution of NH4Cl. Since this is a Bronsted acid (it hydrolyzes) to give an acidic solution, the pH at the equivalence point will be acidic. The pH will depend on the concentration of NH4Cl. The greater the concentration , the lower the pH.

At the addition of complete HCl i.e. 50 ml

All of the NH3 has been converted to NH4Cl

It is 5.00 mmol in 100 ml or 0.05 M solution

[H+] = Kw /Kb x [NH4Cl]

[H+] = 1.0 x 10 -14 / 1.75 x 10 -5 x 0.05 = 5.35 x 10-6 M

pH = 5.27

As excesses HCl is added beyond the equivalence point, the ionization of acid NH4Cl is suppressed to a negligible amount and the pH is determined only by the concentration of excess H+. Therefore titration curve beyond equivalence point follows that for titration of a strong base.