Acid-Base Titration Things to learn : - strong acid – strong base titration - weak acid – strong base titration - strong acid – weak base titration - prediction of titration curve - acid-base indicator

Strong Acid - Strong Base Titration In strong acid – strong base titration, there are three regions of the titration curve that represent different kinds of calculations : - before equivalence point - at equivalence point - after equivalence point Example : Consider the titration of ml of M KOH with 0.10 M HBr KOH + HBrH 2 O + KBr Before titration : Moles of HO - present = (0.050 l)(0.020 mol/l) = moles

What happens when 3.00 ml of HBr is added ? No. of moles in 3.00 ml HBr = (0.003)(0.10) = Moles of HO - unconsumed = – = Since the volume in the flask in now 53 ml, [HO - ] in the flask = (0.007 mol)/(0.053 l) = M pH of solution =-log [H + ] = -log = KwKw [HO - ] What happens at the equivalence point ?

At the equivalence point, the H + is just sufficient to react with all the HO - to form water. The pH is determined by the dissociation of water. Since there is mol HO - in the flask, mol H + must be added to reach equivalence point Volume of H + added ==10 ml Let : x = [H + ] = [HO - ] K w = [H + ] [HO - ] = x 2 x = 1.00 x pH = mol 0.10 mol/l pH at the equivalence point in any strong acid – strong base titration will be 7.00 at 25 o C What happens after the equivalence point ?

When ml of HBr is added to the solution : moles of excess H + = ( l)(0.10 mol/l) = mol Concentration of excess H + = =8.26 x M pH = -log (8.26 x ) = mol l

Weak Acid - Strong Base Titration Strong + weakcomplete reaction In weak acid – strong base titration, the titration curve consists of four regions : - before any base is added HAH + + A - - before the equivalence point : solution consists of a mixture of unreacted HA and A - - at the equivalence point : all the HA has been converted to A - A - + H 2 O HA + HO - - beyond the equivalence point : excess strong base is added and the pH of the solution is determined by the amount of strong base KaKa KbKb

Example : Consider a 50.0 ml solution of M HA with pK a = 6.15 which is treated with 0.10 M NaOH Before the addition of NaOH : HAH + + A - K a = Let x = [H + ] =[A - ] K a === x = 1.19 x pH = -log (1.19 x ) = 3.93 [H + ][A - ] [HA] x2x x 2 What happens when 3.0 ml of 0.10 M NaOH is added ?

When NaOH is added, a mixture of HA and A - is created a buffer Moles of NaOH added = (0.003 l)(0.10 M) = Concentration of A - = = 5.66 x M Moles of HA left = (0.050 l)(0.020 M) – = Concentration of HA == M Using the Henderson-Hasselbalch equation pH = pK a + log = log = [A - ] [HA] 5.66 x What happens at the equivalence point ?

At the equivalence point, sufficient amount of NaOH has been added to react with all the HA Moles of HA present = (0.050 l)(0.020 M) = Volume of NaOH added = (0.0010)/(0.10 M) =0.01 l = 10 ml Concentration of A - == M HA + HO - H 2 O + A - HA + HO - Since K w = K a K b K b = =1.43 x Let x = [HA] = [HO - ] mol l KbKb KaKa KwKw KaKa

K b ===1.43 x x = 1.54 x M Using K w = [H + ][HO - ] = 1x [H + ] = pH = -log[H + ] = 9.18 The pH at the equivalence point in not In weak acid – strong base titration, the pH at the equivalence point is always higher than 7 [HO - ][HA] [A - ] x2x x 1x x What happens after the equivalence point ?

Now there is excess NaOH in the solution Since NaOH is a strong base, we can say that the pH is determined by the concentration of the excess NaOH When ml of NaOH is added there is an excess of 0.10 ml of NaOH. [HO - ] == 1.66 x M pH = -log[H + ] = -log = KwKw [HO - ] (0.10 M)( l) l

Weak Base - Strong Acid Titration B + H + BH + Since it is a strong acid, therefore reaction goes essentially to completion In weak base – strong acid titration, the titration curve consists of four regions : - before any acid is added: B + H 2 OBH + + HO - - before the equivalence point – solution is a buffer B + HABH + + A - BH + + H 2 OB + H 3 O + pH = pK a (for BH + ) + log KbKb KaKa [B] [BH + ]

- at the equivalence point B + HA BH + + A - BH + + H 2 O B + H 3 O + pH is obtained by considering the acid dissociation [BH + ]  original [B] because of dilution Since the solution at equivalence point contains BH +, thus the pH should be below 7 - after the equivalence point There is excess HA in the solution. Since HA is a strong acid, it determines the pH (contribution from the hydrolysis of BH + is comparatively small and can be neglected)

Titration in Diprotic Systems Treatment is an extension from the monoprotic system Example : Consider the titration of 10 ml of 0.10 M base, B, with 0.10 M HCl. The base is dibasic with pK b1 = 4.00 and pK b2 = Calculate the pH at each point along the titration curve Before an acid is added : B + H 2 OBH + + HO - Let x = [BH + ] = [HO - ]. Thus K b1 = == 1.00 x x = 3.11 x [H + ] =pH=11.49 K b1 [BH + ] [HO - ] [B] x2x x KwKw [HO - ]

When acid is added and before the first equivalence point, we have a buffer containing B and BH + B + H + BH + BH + + H + BH 2 2+ BH 2 2+ BH + + H + BH + B + H + If 1.5 ml of HCl has been added : Moles of HCl added = ( l)(0.10 M) = 1.5 x = moles of BH + formed [ BH + ] = = x M 1.5 x K b1 K b2 K a1 K a2

[B] [BH + ] Moles of B left =(0.010)(0.10) – (1.5 x ) = [B] == 7.39 x M Using the Henderson-Hasselbalch equation: pH = pK a2 + log = p( )+ log = log (5.667) = KwKw K b x x 10 -2

At the first equivalence point, all the B has been converted to BH + which is both an acid and a base Moles of B present = (0.010)(0.10) = Volume of acid used == 10 ml [BH + ] ==0.05 M Using : [H + ] = where K 1 = K a1 and K 2 = K a2 [H + ] =3.16 x10 -8 pH = 7.50 K 1 K 2 [BH + ] + K 1 K w K 1 +[BH + ] mol 0.10 M mol l

At regions between the first and second equivalence points, a buffer containing BH + and BH 2 + is formed : BH + + H + BH 2 + BH 2 + BH + + H + When 15 ml of HCl is added : Moles of BH 2 2+ = (0.005 l)(0.10 M) = [BH 2 2+ ] == 0.02 M Moles of BH + left = – = [BH + ] == 0.02 M K a mol l mol l

Using the Henderson-Hasselbalch equation: pH = pK a1 + log = log 1 = 5.00 [BH + ] [BH 2 2+ ] At the second equivalence point, all the BH + has been converted to BH 2 + BH + + H + BH 2 2+ BH 2 2+ BH + + H + pH of the solution is determined by the acid dissociation Moles of BH + present at first equivalence point = Volume of acid added between first and second equivalence point = (0.001 mol)/((0.10 M) = 10 ml K a1

Moles of BH 2 2+ = moles of BH + = [BH 2 2+ ] == M K a1 = = x = 5.72 x pH = -log (5.72 x ) = mol l [BH + ][H + ] [BH 2 2+ ] KwKw K b2 x2x x Beyond the second equivalence point, the pH is determined by the excess HCl If the total volume of HCl added is 25.0 ml, Moles of excess HCl = (0.005 l)(0.10 M) = [H + ]== 1.43 x M pH=

(a)Titration of 10.0 ml of M base (pK b1 = 4.00, pK b2 = 9.00) with M HCl (b) Titration of 10.0 ml of M nicotine (pK b1 = 6.15, pK b2 = 10.85) with M HCl

Finding the End Point Titrations are commonly performed to determine : - how much analyte is present - the equilibrium constants of the analyte How would one determine the end point ? (i) Autotitrator -pH is measured by electrodes immersed in the analyte solution  pH/  V and (  pH/  V)/  V are computed -when  pH/  V is maximum and (  pH/  V)/  V = 0 end point

(ii) Indicator An acid-base indicator is itself an acid or base whose various protonated species have different colours : eg phenolphthalein HlnIn - + H + pH = pK 1 + log The indicator changes color over a pH range Generally only one color is observed if the ratio of the concentrations of the two forms is 10:1 [In - ] [Hln] acid colorbase color

So the pH in going from one color to the other has changed from (pK 1 - 1) to (pK 1 + 1) most indicators require a transition range of about two pH units The pH range over which the color changes is called the transition range 0.7< pH< < pH< 9.6 When only the color of the unionized form is seen : = pH = pK 1 + log = pK When only the color of the ionized form is seen : = pH = pK [In - ] [Hln] 10 1 [In - ] [Hln]

Choosing an Indicator An indicator with a color change near pH 5.54 would be useful in determining the end point of the titration. The closer the point of color change is to pH 5.54 the more accurate will be the end point The difference between the observed end point (when there is a color change) and the true equivalence point is called the indicator error Use only a few drops of dilute indicator solution for each titration