Acid/Base Titrations: CHEM 191 Module 1 Lecture 8

• Acid/Base Titrations are covered in CHEM 191 Module 1, Lecture 8 of the Chemical Reactions in Aqueous Solution course

Learning Objectives:

• Recognize the shapes of acid-base titration curves
• Understand chemical reactions during titration
• Understand how amino acid chemistry affects the shape of its titration curve

Titrations

• Titration allows determination of the concentration of a solution through reaction with a solution of known concentration.
• Reaction stoichiometry is needed to know the mole ratio between reactants.
• Titrations are used extensively in chemistry in the determination of metal ions in body fluids, determination of CO2 in seawater, determination of contaminants in products and analysis of wastewater.

Acid/Base Titration Curves

• A buffer solution could be prepared by adding NaOH(aq) to a solution of CH3COOH(aq).

• pH can be monitored as volumes of NaOH(aq) are added, this process is acid/base titration.

• Weak and strong acids behave differently in titrations with NaOH(aq).

• The essential reaction is H3O+ + OH- 2H2O

• For 50.00 mL of 0.100 mol L-1 CH3COOH(aq) with 0.100 mol L-1 NaOH(aq):

• The pH change near the equivalence point is smaller than in strong acid–strong base titration due to the formation of a buffer solution.

• At the equivalence point, the moles of base equal moles of acid.

• pH > 7.0 at equivalence point because CH3COO- is a weak base.

• Added OH converts CH3COOH(aq) into CH3COO-(aq), forming buffer solution.

• Acid is completely neutralised by added base, resulting in CH3COONa(aq) salt solution.

• No acid is left to react with excess OH-.

The Regions of Titration Curves

• Ka = 10^-4.74 = 1.82×10-5
• Initial pH: [H30+] = √(Ka x [CH3COOH]) = √(1.82×10-5)(0.1) = 1.35×10-3 mol/L
• pH= -log[H3O+] , pH= -log(1.35x10^-3) = 2.87
• 0.100 mol L-1 solution of CH3COOH has a pKa = 4.74 can be calculated to pH = 2.87.

The Acidic Region

• In the acidic region where volume is less than the equivalent volume (v < ve)
• Reaction occurs between CH3COOH and NaOH to give CH3COO-
• pH is controlled by dissociation of CH3COOH in the presence of increasing amounts of CH3COO-.
• This region is a buffer solution.
• When exactly 25 mL of NaOH(aq) has been added, [CH3COOH] = [CH3COO-], and from the Henderson-Hasselbalch equation: pH = pKa + log([A]/[HA]) = pKa = 4.74

Buffer Region

• The equivalence point; nNaOHadded = nCH3COOHinitial, v = Ve
• The reaction between CH3COOH and NaOH has gone to completion, resulting in a solution of CH3COO¯ and Na+ at this point.
• The pH is basic (8.34) due to the reaction of the conjugate base with water gives OH-.
• CH3COO- + H2O CH3COOH + OH-

pH at the equivalence point

• Initially, n(CH3COOH) = 0.100 mol L-1 x 0.050 L = 0.005 mol
• 0.005 mol of CH3COO- is formed when the equivalence point is reached.
• The total volume of solution at the equivalence point will be 50 mL (initial acid solution) + 50 mL (added base solution) = 100 mL
• Therefore [CH3COO¯] = 0.005 mol / 0.100 L = 0.05 mol L-1
• Since CH3COOH and CH3COO¯ are a conjugate acid-base pair: pKa(CH3COOH) + pKb(CH3COO¯) = 14
• Where pKb(CH3COO¯) = 14 – 4.74 = 9.26, so Kb = 10^-9.26 = 5.5 × 10^-10
• The general equation to calculated that pH of a weak base solution is [OH-] = √(Kb) [CH3COO-];]
• √(5.5×10^-10 )(0.05) = √(2.75×10-11 = 5.244×10^-6
• pOH = -log[OH-] = 5.28
• And hence pH = 14 - 5.28 = 8.72
• Kb = [CH3COOH]e[OH-]e / [CH3COO-]e = [OH-]^2/ [CH3COO-]initial
• [OH-]e = √(K x [CH3COO-]initial = √ 5.5x10-10 x 0.050 = 5.2 × 10-6
• pOH = -log(5.2×10^-6) = 5.28
• pH = 14 - 5.28 = 8.72

The Alkaline Region

• For the alkaline region Volume is greater than the equivalent volume v > ve:
• pH is determined solely by the amount of excess OH- ions added to the solution after the equivalence point has been reached.

Strong Acid/Strong Base Titration Curve

• There is no equilibrium to deal with.
• HCl + NaOH → H2O + NaCl
• H3O+ + OH- → 2H2O
• A strong acid strong base solution is sensitive to dilution, therefore is not a buffer region.
• At the equivalent point, pH = 7.0, NaCl(aq) salt solution
• Only HCl(aq) is present before titration
• H+ consumed as OH- added, forming H2O (pH < 7.0)
• H+ is completely neutralised by OH- (pH = 7.0)
• No H+ is left to react with excess OH- (pH > 7.0)
• There is no buffer region as pH must not change as in a buffer solution

Diprotic Acids

• A monoprotic acid donates one proton
• Diprotic acids have two protons that can donate, then it will do so one at a time, e.g., H3PO3

Amino Acids

• Amino acids are biological molecules necessary for life; at least 20 naturally occurring amino acids are known
• All amino acids have both a weak acid and a weak base within the same molecule.
• They differ in the nature of the R group attached to the carbon atom.

Zwitterions

• Neutral amino acids exist primarily as zwitterions, meaning they contain both a positive and a negative charge, but the overall molecule is neutral
• This can be in terms of the acidic –COOH proton protonating the basic -NH2 centre
• Zwitterions behave like ionic salts rather than organic molecules – they are generally involatile solids, have high melting points, are water-soluble, and are insoluble in nonpolar solvents
• Under very acidic conditions, both the –COOH and the –NH3+ groups will be protonated, creating a cation form
• Under very basic conditions, both the –COO- and the –NH2 groups are deprotonated, creating an anion form

Titration Curve for Glycine

• In the titration curve for glycine, it is good to remember the pH number for amino acids

The regions of Glycine curve

• Initial pH: More acidic than a normal weak acid as it is easier to lose a proton from a positively-charged ion than from a neutral molecule.

Region Where Less Than One Mole of OH- Has Been Added

• In this region, the more acidic –COOH proton reacts (is being titrated), hence there is a mixture of both is present.
• When half an equivalent of OH- has been added, the pH = pKa (pKa1 = 2.35)
• With One mole of OH- has been added at this point, the pH = pl (6.06) and the solution contains the zwitterion
• pI is called the isoelectric point and is the point at which most of the molecules in the solution have a net charge of zero.

Region where more than one mole of OH- has been added

• In this region, the less acidic –NH3+ proton now reacts (is titrated), hence there is a mixture of both.
• When 1.5 equivalents of OH- have been added, the pH = pKa (pKa2 = 9.78)
• The titration is complete when two equivalents of OH- have been added and contains a single structure.
• pI = (pkai + pkaz)/2