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Buffer Solutions
A buffer solution is comprised of a mixture of an acid (base) with its conjugate base (acid) that resists changes in pH when additional acid or base is added The Henderson-Hasselbalch Equation describes the pH in buffer solutions: Henderson-Hasselbalch Equation: [A-] = concentration of conjugate base [HA] = concentration of acid
Buffer Solutions
Example: Find the pH of a solution with [CH 3 COOH] = 0.250 M and [NaCH 3 COO] = 0.315 M [HA] = 0.250 M [A-] = 0.315 M Ka = 1.8 x 10-5^ ⇒ pKa = -log(1.8x10-5) = 4.
Buffer Solutions
Example: If we now add 10.0 mL of 0.100 M HCl to 50.0 mL of the buffer solution, determine the new pH of the solution. HCl, a strong acid, reacts with the acetate ion base to form water and acetic acid: H 3 O+(aq) + CH 3 COO-(aq) → H 2 O + CH 3 COOH(aq) We must first account for the dilution factors (C 1 V 1 =C 2 V 2 ): [H 3 O+] = (.100 M)(10.0 mL)/(60.0 mL) = 0.0167 M [CH 3 COO-] = (.315 M)(50.0 mL)/(60.0 mL) = 0.263 M [CH 3 COOH] = (.250 M)(50.0 mL)/(60.0 mL) = 0.208 M
Buffer Solutions
Example (con’t.): Now allow reaction between HCl and acetate to occur— since HCl is a strong acid, it will react completely as long as base remains in solution: H 3 O+(aq) + CH 3 COO-(aq) → H 2 O + CH 3 COOH(aq) .0167 – .0167 .263 – .0167 .208 +. [CH 3 COOH] = .208 + .0167 = .225 M [CH 3 COO-] = .263 - .0167 = .246 M If the same amount of acid had been added to pure water, the pH would change from 7.0 to
Polyprotic Acids and Bases
If an acid can donate more than one hydrogen ion, then it is called a diprotic (2 H+^ ions) or triprotic (3 H+ ions) acid If a base can accept more than one hydrogen atom, it is called a diprotic (2 H+^ ions) or triprotic (3 H+ ions) base Example: H 3 PO 4 H 3 PO 4 (aq) + H 2 O ↔ H 2 PO 4 - (aq) + H 3 O+(aq) Ka1 = 7.11 x 10- H 2 PO 4 - (aq) + H 2 O ↔ HPO 4 2-(aq) + H 3 O+(aq) Ka2 = 6.34 x 10- HPO 4 2-(aq) + H 2 O ↔ PO 4 3-(aq) + H 3 O+(aq) Ka3 = 4.22 x 10-
Polyprotic Acids and Bases
The resulting conjugate bases also can be polyprotic: H 2 PO 4 - (aq) + H 2 O ↔ H 3 PO 4 (aq) + OH-(aq) Ka1⋅Kb1 = Kw Kb1 = 1.41 x 10- HPO 4 2-(aq) + H 2 O ↔ H 2 PO 4 - (aq) + OH-(aq) Ka2⋅Kb2 = Kw Kb2 = 1.58 x 10- PO 4 3-(aq) + H 2 O ↔ HPO 4 2-(aq) + OH-(aq) Ka3⋅Kb3 = Kw Kb3 = 2.37 x 10- Species that can both accept and donate a hydrogen ion are called amphiprotic
Polyprotic Acids and Bases
Suppose we dissolve NaH 2 PO 4 in water to produce a solution that is 0.100 M NaH 2 PO 4. The resulting H 2 PO 4 - (aq) can behave either as an acid or a base. Which process will dominate and what will be the resulting pH? Acid: H 2 PO 4 - (aq) + H 2 O ↔ HPO 4 2-(aq) + H 3 O+(aq) Base: H 2 PO 4 - (aq) + H 2 O ↔ H 3 PO 4 (aq) + OH-(aq) We can use the requirements of charge and mass balance for chemical reactions to determine a solution to this problem (see section 8-4)
Polyprotic Acids and Bases
Find the pH of a 0.100 M NaH 2 PO 4 solution: Charge balance—the sum total of positive charges must equal the sum total of negative charges Positively charged species are H 3 O+^ and Na+ Negatively charged species are H 2 PO 4 - , HPO 4 2-, and OH- [H 3 O+] + [Na+] = [H 2 PO 4 - ] + 2[HPO 4 2-] + [OH-] We use 2[HPO 4 2-] because this ion has a -2 charge, so we need two singly charged cations to balance the negative charge from this anion
Polyprotic Acids and Bases
Find the pH of a 0.100 M NaH 2 PO 4 solution: Mass balance—the concentrations of all species containing PO 4 3-^ must equal the formal concentration (F) of the initial amount prepared in the solution: F = [H 3 PO 4 ] + [H 2 PO 4 - ] + [HPO 4 2-] In this particular example, [Na+] = F
Polyprotic Acids and Bases
Find the pH of a 0.100 M NaH 2 PO 4 solution: substitution of mass balance relationship into charge balance expression:
Polyprotic Acids and Bases
Find the pH of a 0.100 M NaH 2 PO 4 solution: [H 3 O+] + [H 3 PO 4 ] – [HPO 4 2-] – [OH-] = 0 Using the two acid dissociation reactions:
Polyprotic Acids and Bases
Find the pH of a 0.100 M NaH 2 PO 4 solution:
Titration of a Weak Base with a
Strong Acid
At equivalence point Addition of strong acid has converted all base to its conjugate acid HB+ This dissociates back to for B and H 3 O+ HB+^ + H 2 O ↔ B + H 3 O+^ Ka = Kw/Kb = [B][H 3 O+]/[HB+]
Titration of a Weak Base with a
Strong Acid
After the equivalence point Once the base has all reacted with strong acid, the only thing remaining in solution the weak conjugate acid, HB+, and hydronium from the excess strong acid added pH is determined solely from the added excess strong acid, taking into account the dilution of acid in solution [H 3 O+] = [HA](Va,xs – Va,eq pt)/Vtot
Titration of a Weak Base with a
Strong Acid
Example: let [HA] = 0.1000 M, [B]o = 0.0750 M, Vb = 25.00 mL, pKb = 5.75 ⇒ pKa = 8. Before titration begins: B + H 2 O ↔ HB+^ + OH-^ Kb = 1.78x10- .0750-x x x x^2 = (1.78x10-6)(.0750-x) [OH-] = 3.66 x 10-4^ M pH = 10.
Titration of a Weak Base with a
Strong Acid
