TITRATION CURVES.

TITRATION CURVES

TITRATION CURVES Titration Curves: a plot of pH as a function of volume of added titrant.

Typically the titrant is a strong (completely) dissociated acid or base. Such curves are useful for determining endpoints and dissociation constants of weak acids or bases. These curves can be calculated and can also determined experimentally. 1. The calculation of titration curves involves calculation of PH at various points during titration by using various equations. 2. Experimental determination is done potentiometrically by using PH meter.

Titration curves for titration of
Strong Acid –Strong Base Weak Acid- Strong Base Weak Base- Strong Acid Weak Acid- Weak Base Important regions for calculating PH during any of the above type of titration are PH before titration begins PH Before the end point i.e. during tiration PH At the end point i.e. equivalance point PH After the end point

Titration curves for titration of Strong Acid –Strong Base
The pH at starts is low, reflecting the high [H3O+] of the strong acid. The pH increases gradually as acid is neutralized by the added base. Suddenly the pH rises steeply. This occurs in the immediate vicinity of the equivalence point. For this type of titration the pH is 7.0 at the equivalence point. Beyond this steep portion, the pH increases slowly as more base is added.

Problem 1. Consider the titration of 40. 0 mL of 0. 100 M HCl with 0
Problem 1. Consider the titration of 40.0 mL of M HCl with M NaOH. Region 1-PH before titration begins is the PH of 0.100M HCl Therefore [H+] = 0.100M As PH = -log [H+] Therefore PH = - log 0.100 PH = 1 Mmoles of acid = Vol ˣ Molarity = 40 ˣ 0.1=4.0mmoles of HCl

Region 2- Before the equivalence point,
1)after adding 1.0 mL of titrant i.e M NaOH, mmoles of base added = Vol added ˣ Molarity of NaOH = 1 ˣ 0.1= 0.1 mmoles of NaOH When 1 ml of NaOH added to acid the total volume in flask becomes 41.0 ml. and the solution contains = 3.9 mmoles of HCl in 41 ml of solution. Now the [H+] = 3.9/41 = M Therefore PH = -log of = 2) after adding 10.0 mL of titrant i.e M NaOH mmoles of base added = = 10 ˣ 0.1= 1 mmoles of NaOH When 10 ml of NaOH added to acid the total volume in flask becomes 51.0 ml. and the solution contains = 3 mmoles of HCl in 51 ml of solution. Now the [H+] = 3/51 = M Therefore PH = -log of = 1.23 At 20 mi PH= 1.52 At 30 mi PH= 2.0

Region 3- at equivalence point
after adding 40.0 mL of titrant i.e M NaOH, mmoles of base added = = 40 ˣ 0.1= 4 mmoles of NaOH When 40 ml of NaOH added to acid the total volume in flask becomes ml. and the solution contains = 0 mmoles of HCl in 80 ml of solution. Now the [H+] = 0/00 = 0 M also [OH-] = 0/00 = 0 M Therefore PH will be of water i.e. 7.

Region 4- After the equivalence point,
When 41 ml of NaOH added to acid the total volume in flask becomes ml. and the solution contains 1 ML excess or 1 ˣ 0.1= 1 mmoles of NaOH in 81 ml of solution. Then the [OH-] = 1/81= M P OH = -log 0f [OH-] = -log 0f =2.91 Therefore PH = P Kw - P OH = = 11.09 After adding 45 ml PH =11.77 After adding 45 ml PH =12.05

2. Dependence on the initial concentrations (e.g. [HCl]):
PH change pattern will be different but the equivalence point will be at PH 7. [HCl]0 ↓ 0% 50% 90% 99% 99.9% 100% 100.1% 101% 110% 1 N 0,3 1 2 3 7 11 12 13 0,1 N 1,3 4 10 0,01 N 2,3 5 9 0,001 N 3,3 6 8 pH change around the end point ΔpH 3 – 11 4 – 10 5 – 9 6 – 8 100

Problem 1. Consider the titration of 50. 0 mL of 0. 100 M HCl with 0
Problem 1. Consider the titration of 50.0 mL of M HCl with M NaOH. Region 1-PH before titration begins is the PH of 0.100M HCl Therefore [H+] = 0.100M As PH = -log [H+] Therefore PH = - log 0.100 PH = 1 Mmoles of acid = Vol ˣ Molarity = 50 ˣ 0.1=5.0mmoles of HCl

Region 2- Before the equivalence point,
1)after adding 1.0 mL of titrant i.e M NaOH, mmoles of base added = Vol added ˣ Molarity of NaOH = 1 ˣ 0.5= 0.5 mmoles of NaOH When 1 ml of NaOH added to acid the total volume in flask becomes 51.0 ml. and the solution contains = 4.5 mmoles of HCl in 51 mi of solution. Now the [H+] = 4.5/51 = M Therefore PH = -log of = =1.08 In this manner the calculations are made for several points throughout the titration. Few results are At 2 ml- PH = 1.11 At 4 ml- PH = 1.26 At 6 ml- PH = 1.45 At 8 ml- PH = 1.76 At 9ml- PH = 2.07

At the Equivalence (Stoichiometric)Point of
Region 3- at equivalence point after adding 10.0 mL of titrant i.e M NaOH, mmoles of base added = = 10 ˣ 0.5= 5 mmoles of NaOH When 10 ml of NaOH added to acid the total volume in flask becomes ml. and the solution contains = 0 mmoles of HCl in 80 ml of solution. At this point all the HCl has been neutralised by NaOH. THE SALT FORMED FROM A SA-SB TITRATION IS ALWAYS NEUTRAL. Since there is no SA, no SB, and just H2O and a NEUTRAL salt, the pH of the solution formed will be 7.00. At the Equivalence (Stoichiometric)Point of a SA—SB Titration, the pH is always = 7.00 Now the [H+] = 0/00 = 0 M also [OH-] = 0/00 = 0 M Therefore PH will be of water i.e. 7.

Region 4- After the equivalence point,
When 11 ml of NaOH added to acid the total volume in flask becomes ml. and the solution contains 1 ML excess or 1 ˣ 0.5= 0.5 mmoles of NaOH in 61 ml of solution. Then the [OH-] = 0.5/61= M P OH = -log 0f [OH-] = -log 0f =2.09 Therefore PH = P Kw - P OH = = 11.91

Strong-Weak Titrations
You might think that all titrations must have an equivalence point at pH 7 because that is the point at which concentrations of hydrogen ions and hydroxide ions are equal and the solution is neutral. This is not the case – some titrations have equivalence points at pH values less than 7, and some have equivalence points at pH values greater than 7. These differences occur because of reactions between the newly formed salts and water – salt hydrolysis. Some salts are basic (weak acid, strong base) and some salts are acidic (strong acid, weak base).

TITRATION CURVES II. Weak acid with strong base Weak base with strong acid e.g. Titration of CH3COOH with NaOH , Titration of NH4OH with HCl: I. At the start: pH Weak acid Weak base II. Before the end point: pH Buffer (acid / salt) Buffer (base / salt) III. At the end point: pH Hydrolysing salt (Brönsted base) Hydrolysing salt (Brönsted acid) IV. After the end point: pH Excess of strong base Excess of strong acid [OH–] = Cexcess base [H+] = Cexcess acid [OH–] = [NaOH]excess [H+] = [HClexcess

HPr = Propionic Acid

The four Major Differences Between a Strong Acid-Strong Base Titration Curve and a Weak Acid-Strong Base Titration Curve The initial pH is higher. A gradually rising portion of the curve, called the buffer region, appears before the steep rise to the equivalence point. The pH at the equivalence point is greater than 7.00. The steep rise interval is less pronounced.

WEAK ACID – STRONG BASE TITRATION CURVE

The titration is considered in four stages
1. Before titration begins: at this point the sample solution contains only the weak acid and the PH is calculated with the formula [H+ ] = √ (Ka .C) i.e [H+ ] = √ (Ka .[HA]) 2. During titration :Buffer region: after some strong base has been added, its reaction with the weak acid will produce an equivalent amount of conjugate weak base. The solution is now the mixture of weak acid and its conjugate base, in fact, it is a buffer solution and we apply Henderson- Hasselbalch equation, PKa = PH – log [salt] i.e. PH = PKa + log [salt] [Acid] [Acid]

3. at the equivalence point: now an amount of strong base has been added that is exactly equal to amount of weak acid initially present. The solution now contains only the conjugate base of weak acid, which will undergo hydrilysis, and its PH can be calculated by formula [OH-] =√ (Kb .C ) i.e. [OH-] =√ (Kb .A- ) 4. After the end point : An excess of titrant has now been added, and PH is determined by essentially by this excess , the appropriate dilution within the solution being taken into consideration. It is true that the conjugate weak base produced will also contribute to the solution POH, but its concentration is very small relative to effect of excess titrant. After the equivalence point, the solution contains excess of OH- which will repress the hydrolysis of salt. [OH-] = [Base]

Sample Calculation: Weak Acid-Strong Base Titration Curve
Problem 1. Consider the titration of mL of 0.1M acetic acid(Ka=1.82ˣ 10 -5) with M NaOH. PH of the solution at the equivalence point is calculated directly by the formula PH =1/2PKw +1/2PKa +1/2log C =1/2(14)+1/2(4.74) +1/2log 0.1 = /2(-1) = = 8.87

Region 1. The PH of solution of weak acid to be titrated, before any base is added is calculated from dissociation of acetic acid. CH3COOH ↔ CH3COO- + H+ Ka = [CH3COO- ] [H+ ] [CH3COOH ] As the [H+ ] = [CH3COO- ] , Then Ka = [H+ ][H+ ] = [H+ ]2 = Ka.[CH3COOH ] THEREFORE [H+ ] = √ (Ka . [CH3COOH ] ) = √ (1.82ˣ ˣ 0.1) = 1.35 ˣ 10 -6 Therefore PH = - log 1.35 ˣ 10 -3 PH = -(- 2.87) = 2.87

Region 2: PH before equivalence point:
During titration we have a mixture of a weak acid (CH3COOH) and the salt of its conjugate base (NaCH3COO). CH3COO- + NaOH ↔ CH3COONa A mixture of a weak acid and a weak base (the salt of its conjugate base) is called a BUFFER SOLUTION. As we know a buffer solution is a solution which maintains the pH at a fairly constant value. This causes the titration curve to decrease in slope during this stage. The area on the curve is called the “Buffer Region”. The salt and acid concentration at any time during the titration can be calculated from the volume of titrant added and then PH can be determined by the formula. PH = PKa + log [salt] [Acid]

When 10 ml of 0.1 M NaOH have been added, [salt] = vol of alkali added ˣ molarity = 10 ˣ 0.1= Total Volume 110 [acid] = vol of acid unreacted ˣ molarity = 90 ˣ 0.1= Now PH = PKa + log [salt] [Acid] Therefore PH = Log ( /0.0818)=4.74+(-0.95)= 3.78 When 20 ml of 0.1 M NaOH have been added, [salt] = vol of alkali added ˣ molarity = 20 ˣ 0.1= Total Volume 120 [acid] = vol of acid unreacted ˣ molarity = 80 ˣ 0.1=0.066 Therefore PH = Log( /0.066)=4.74+(-0.599)= 4.14

When 50 ml of 0.1 M Naoh have been added, [salt] = vol of alkali added ˣ molarity = 50 ˣ 0.1= Total Volume 150 [acid]= vol of alkali added ˣ molarity = 50 ˣ 0.1= Now PH = PKa + log [salt] [Acid] Therefore PH = LOG (0.033/ 0.033)=4.74+(0)= 4.74 Region 3:Equivalence region: when 100 ml of 0.1 M NaOH added, it is clearly shows equivalence point. Now an amount of strong base has been added that is exactly equal to amount of weak acid initially present. The solution now contains only the conjugate base of weak acid CH3COONa→CH3COO - + Na+ and its PH can be calculated by formula

Therefore Kw=Kb. Ka = Kb=Kw /Ka =1 ˣ 10-14 /1. 82 ˣ 10-5 =5
Therefore Kw=Kb.Ka = Kb=Kw /Ka =1 ˣ /1.82 ˣ 10-5 =5.49 ˣ & [salt] = vol of alkali added ˣ molarity = 100 ˣ 0.1= 0.05M Total Volume 200 Therefore [OH-]=√ (5.49 ˣ ˣ 0.05) = √( 0.274)ˣ 10-5 [OH-] = ˣ log [OH-] = -log (0.523 ˣ 10-5 ) POH=-(-5.2) = 5.2 There fore PH = =8.8

Region 4: After the equivalence point
Looking at the Balanced equation: NaOH + CH3COOH  H2O + NaCH3COO Once NaOH is in excess, you will have some STRONG BASE (NaOH) and some WEAK BASE (CH3COO-) in the resulting mixture. The OH- contributed by the weak base ( CH3COO-)was significant when there was no other base present (EP), but once a strong base (NaOH) is present, the OH- contributed by the weak base is insignificant compared to that produced by the NaOH. So, the titration curve past the EP for a WA/SB Titration is the same as it is for a SA/SB Titration (where NaOH is in excess). After the equivalence point has been passed, the solution contains excess of OH- ions, which will repress the hydrolysis of salt, hence we can assume the PH is due to excess of base present, hence PH can be determined by formula POH = [Excess of alkali]

When 101 ml of NaOH added to acid the total volume in flask becomes ml. and the solution contains 1 ML excess or 1 ˣ 0.1= 0.1 mmoles of NaOH in 201 ml of solution. Then the [OH-] =1ˣ 0.1/201= -log [OH-] = -log =-(-3.3)=3.3 PH = PKw – POH = = 10.7 Similarly, After addition of 110 ml PH = 11.7 After addition of 125 ml PH = 12.0 After addition of 150 ml PH = 12.3 After addition of 200 ml PH = 12.5

INDICATOR SELECTION The equivalence point is at PH 8.87, so it requires an indicator with the colour change PH interval slightly at alkaline side eg. Phenolphthalein, thymolphthalein or thymol blue (8-9.6) In general weak acids with Ka ˃ 5ˣ10-6 should be titrated with Phenolphthalein, thymolphthalein or thymol blue as indicators

The four Major Differences Between a Weak Acid-Strong Base Titration Curve and a Weak Base-Strong Acid Titration Curve The initial pH is above 7.00. A gradually decreasing portion of the curve, called the buffer region, appears before a steep fall to the equivalence point. The pH at the equivalence point is less than 7.00. Thereafter, the pH decreases slowly as excess strong acid is added.

The titration is considered in four stages
1. Before tiration begins: at this point the sample solution contains only the weak base and the PH is calculated with the formula [OH- ] = √ (Kb .C) i.e. [OH- ] = √ (Kb .[BOH]) 2. During titration :Buffer region: after some strong Acid has been added, its reaction with the weak base will produce an equivalent amount of conjugate weak acid. The solution is now the mixture of weak base and its conjugate acid, in fact, it is a buffer solution and we apply Henderson- Hasselbalch equation, PH = PKw-1/2PKb + log [salt] [Base]

3. at the equivalence point: now an amount of strong Acid has been added that is exactly equal to amount of weak base initially present. The solution now contains only the conjugate acid of weak base and its PH can be calculated by formula [H+] =√ (Ka .C ) i.e. [H+] =√ (Ka .B+ ) 4. After the end point : An excess of titrant has now been added, and PH is determined by essentially by this excess , the appropriate dilution within the solution being taken into consideration. It is true that the conjugate weak acid produced will also contribute to the solution PH, but its concentration is very small relative to effect of excess titrant. After the equivalence point, the solution contains excess of H+ which will repress the hydrolysis of salt. [H+] = [Acid]

Sample Calculation: Weak base-Strong acid Titration Curve
Problem 1. Consider the titration of mL of 0.1M ammonia (Kb=1.8ˣ 10 -5) with M HCl. PH of the solution at the equivalence point is calculated directly by the formula PH =1/2PKw -1/2PKb - 1/2log C = /2(log 0.1) = /2(-0.5) = = 5.13

Region 1. The PH of solution of weak base to be titrated, before any acid is added is calculated from dissociation of acetic acid. NH3+ H2O ↔ NH4 + + OH- Kb = [NH4 + ] [OH- ] [NH3] As the [NH4 + ] = [OH- ] , Then Ka = [OH-][OH-] = [OH-]2 = Kb.[NH3] [NH3 ] THEREFORE [OH-] = √ Kb . [NH3 ] = √ (1.8ˣ ˣ 0.1) = 1.34 ˣ 10 -3 Therefore POH = - log 1.35 ˣ 10 -3 POH = 2.87 PH = PKw - POH = =11.12

Region 2: PH before equivalence point:
During titration When HCl is added but the NH3 is still in excess, we will have a mixture of NH3 (a weak base) and NH4+ (a weak acid) which is a buffer. NH3 + HCl ↔ NH4 + + Cl- A mixture of a weak base and a weak acid (the salt of its conjugate base) is called a BUFFER SOLUTION. As we know a buffer solution is a solution which maintains the pH at a fairly constant value. This causes the titration curve to decrease in slope during this stage. The area on the curve is called the “Buffer Region”. The salt and base concentration at any time during the titration can be calculated from the volume of titrant added and then PH can be determined by the formula. PH = PKw- PKb - log [salt] [Base]

When 10 ml of 0.1 M HCl have been added, [salt] = vol of acid added ˣ molarity = 10 ˣ 0.1= Total Volume 110 [base] = vol of base unreacted ˣ molarity = 90 ˣ 0.1= Now PH = PKw- PKb - log [salt] [Base] Therefore PH = Log ( /0.0818)= (-0.95)= 8.31 When 20 ml of 0.1 M HCl have been added, [salt] = vol of acid added ˣ molarity = 20 ˣ 0.1= Total Volume 120 [acid] = vol of base unreacted ˣ molarity = 80 ˣ 0.1=0.066 Now PH = PKw- PKb - log [salt] Therefore PH = Log( /0.066)=4.74+(-0.599)= 8.67

When 50 ml of 0.1 M Naoh have been added, [salt] = vol of acid added ˣ molarity = 50 ˣ 0.1= Total Volume 150 [acid] vol of base unreacted ˣ molarity = 50 ˣ 0.1= Now PH = PKw- PKb - log [salt] [Base] Therefore PH = LOG (0.033/ 0.033)= (0)= 9.53

Region 3:Equivalence region: At the equivalence point in this titration, all of the HCl and NH3 will be gone and only NH4+ (a weak acid) and Cl- (a neutral spectator) will remain. Because there is a WEAK ACID (NH4+) present, the pH will be LESS THAN 7. (but not really low). when 100 ml of 0.1 M HCl added, it is clearly shows equivalence point. Now an amount of strong acid has been added that is exactly equal to amount of weak base initially present. The solution now contains only the conjugate acid of weak base. NH3 + HCl ↔ NH4 + + Cl- and its PH can be calculated by formula

Therefore Kw=Kb. Ka = Ka=Kw /Kb =1 ˣ 10-14 /1. 8 ˣ 10-5 =5
Therefore Kw=Kb.Ka = Ka=Kw /Kb =1 ˣ /1.8 ˣ 10-5 =5.49 ˣ & [salt] = vol of acid added ˣ molarity = 100 ˣ 0.1= 0.05M Total Volume 200 Therefore [H+]=√ (5.49 ˣ ˣ 0.05) = √ (0.274) ˣ 10-5 [H+] = ˣ log [H+] = -log (0.523 ˣ 10-5 ) PH=-(-5.2) = 5.2

Region 4: After the equivalence point
After the equivalence point has been passed, the solution contains excess of H+ ions, which will repress the hydrolysis of salt, hence we can assume the PH is due to excess of acid present, hence PH can be determined by formula PH = [Excess of Acid] When 101 ml of HCl added to acid the total volume in flask becomes ml. and the solution contains 1 ML excess or 1 ˣ 0.1= 0.1 mmoles of HCl in 201 ml of solution. Then the [H+] =1ˣ 0.1/201= -log [H+] = -log =-(-3.30)=3.30 PH = 3.3 Similarly, After addition of 110 ml PH = 2.32 After addition of 125 ml PH = 1.95 After addition of 150 ml PH = 1.69 After addition of 200 ml PH = 1.52

INDICATOR SELECTION The equivalence point is at PH 5.13, so it requires an indicator with the colour change PH interval slightly at aCIDIC side eg. Methyl red, Methyl orange, bromophenol blue(3-6.5) In general weak acids with Kb ˃ 5ˣ10-6 should be titrated with Methyl red, Methyl orange, bromophenol blueas indicators

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