Chem 1310: Introduction to physical chemistry

Part 5: Buffers and solubility

Peter H.M. Budzelaar

Buffers

Water has a very small [H3O+] (10-7). Adding just a little bit of acid or base can change the pH drastically.

Add 0.001 M HCl: pH goes from 7 to 3!

For many applications this sensitivity is undesirable. One of the best ways to prevent pH swings is buffering: the use of a mixture of a weak acid and its conjugate base (which will be a weak base).

Buffers

Two important aspects:• What will be the resulting pH?• What will be the buffer capacity (how much

acid/base can be absorbed before the pH starts to change drastically)?

The pH of a buffer solution

Take a mixture of HOAc and NaOAc (both 0.1 M):

HOAc + H2O OAc⇋ - + H3O+

[HOAc]

]O][H[OAc 3

aK

HOAc OAc- H3O+

initial 0.1 0.1 (10-7)

change - x + x + x

equilibrium 0.1-x 0.1+x x

The pH of a buffer solution (2)

We usually assume x « buffer concentration, so

(Always check afterwards! If not valid,solve the full quadratic equation)

5108.11.0

)1.0(

x

xxKa

7.4pH

108.1108.11.0

1.0 55

xx

Ka

The pH of a buffer solution (3)

General formula (using the same x assumption):

where we replace the actual [HA], [A-] by the amounts weighed in (Henderson-Hasselbalch).

)logpK([HA]

][AlogpK

][A

[HA]loglog]O[HlogpH

][A

[HA]]O[H

[HA]

]O][H[A

aa

3

33

a

a

aa

K

K

KK

Buffer capacityHow much added acid or base can a buffer absorb?

At most until the buffer acid or its conjugate base is consumed.

If you have 1L of a buffer containing 0.2 M HOAc and 0.35 M NaOAc, this can absorb up to 0.35 moles of acid (all NaOAc consumed) or 0.2 moles of base (all HOAc consumed).

As long as you do not exceed the buffer capacity, you can calculate the new pH using Henderson-Hasselbalch.

Buffer capacity (2)

0.03 moles of HCl is added to 1L of a buffer of 0.1 M each of HOAc and NaOAc. What is the resulting pH?

New [HOAc] = 0.1+0.03 = 0.13,new [NaOAc] = 0.1-0.03=0.07:

4.5.13)log(0.07/04.7[HA]

][AlogpKpH a

Titration

• Slowly add acid of known concentration from a burette to a solution of base (or vv).

• Use an indicator to detect moment of fast pH change (happens at equivalence point).

• Strong acid, baselargest pH change, almost any indicator will work.

• Weak acid titrated with strong base:Solution will originally not be very acidic, but will go till very basic. Use indicator for pH > 7, e.g. phenolphtalein.

Titration

• Weak base titrated with strong acid:Solution will originally not be very basic, but will go till very acidic. Use indicator for pH < 7, e.g. methyl red.

• Do not titrate a weak acid with a weak base!No clear equivalence point.

Solubility in water

Just another equilibrium (see MSJ p839 for Ksp table):

AgCl(s) Ag⇋ +(aq) + Cl-

(aq)

KC = Ksp = [Ag+][Cl-]

The standard rules for writing equilibrium constants apply:

Mg3(PO4)2 (s) 3 Mg⇋ 2+(aq) + 2 PO4

3-(aq)

Ksp = [Mg2+]3[PO43-]2

No AgCl, becausethat is a pure solid.

Calculating the solubilityof a compound in pure water

Add excess AgCl to water; it starts to dissolve:

[Ag+][Cl-] = x2 = Ksp = 1.8·10-10

x 1.3·10-5 mol/L

Ag+ Cl-

initial 0 0

change + x + x

equilibrium x x

Calculating the solubilityof a compound in pure water (2)

Add excess PbCl2 to water; it starts to dissolve:

[Pb2+][Cl-]2 = 4x3 = Ksp = 1.7·10-5

x 0.015 mol/L

Pb2+ Cl-

initial 0 0

change + x + 2x

equilibrium x 2x

Calculating solubilityin the presence of "common ions"

Dissolve AgCl in a solution of 0.1 M NaCl:

[Ag+][Cl-] = x(0.1+x) = Ksp = 1.8·10-10

Ag+ Cl-

initial 0 0.1

change + x + x

equilibrium x x

Calculating solubilityin the presence of "common ions" (2)• Assume x « 0.1:

x(0.1+x) 0.1x = Ksp = 1.8·10-10

x = 1.8·10-9 mol/L(verify: x « 0.1!)

A lot less soluble than in pure water!

• Without assumption: solve the quadratic equation. This is often not a good idea!

Calculating whethera salt will precipitate

Calculate Qsp = [...][...](same formula as for Ksp)

• Qsp < Ksp: more could dissolve

• Qsp = Ksp: saturated solution

• Qsp > Ksp: super-saturated: salt will precipitate

(c.f. Q and K for other equilibria)

Solubility calculations are not always straightforward...

The real solution equilibrium might be more complicated:

PbCl2(s) PbCl⇋ + + Cl- Pb⇋ 2+ + 2 Cl-

The original Ksp expression is still valid, but we cannot assume all Pb in solution is present as Pb2+. There will also be some PbCl+, so the amount of Pb that goes into solution will be higher than expected.

Solubility calculations are not always straightforward...

Added reagents may complex with the solutes and reduce their concentrations, setting up new equilibria:

AgCl(s) Ag⇋ +(aq) + Cl-

(aq)

Ag+(aq) + 2 CN-

(aq) Ag(CN)⇋ 2-

(aq)

AgCl(s) + 2 CN-(aq) Ag(CN)⇋ 2

- (aq) + Cl-

(aq)

(Hess's law)

182

2

10

106.5]][CN[Ag

][Ag(CN)

108.1]][ClAg[

f

sp

K

K

92

2 100.1][CN

]][Ag(CN)[Cl

fsp KKK

Solubility and complexation

We have 1L of a solution 0.1 M in NaCN. Will it dissolve 0.01 moles of AgCl?

Assuming complete conversion to Ag(CN)2-:

[Cl-] = 0.01, [Ag(CN)2-] = 0.01, [CN-] = 0.008:

a)

b)

Either way: it will easily dissolve!

92

2 100.102.0][CN

]][Ag(CN)[Cl

KQ

102119

192

2

108.1108.201.0*108.2]][ClAg[

108.2]CN[

]Ag(CN)[]Ag[

spsp

f

KQ

K

Solubility and acid-base reactions

CaCO3(s) Ca⇋ 2+(aq) + CO3

2-(aq)

CO32- + H2O HCO⇋ 3

- + OH-

Part of CO32- removed via reaction with water

more will dissolved than you would calculate from Ksp.

With added acid:

CO32- + H3O+

HCO3- + H2O

HCO3- + H3O+

H2CO3 + H2O

H2CO3 CO2 + H2O

Solubility and acid-base reactions

Eventually, all CaCO3 dissolves in acid!

This happens with many poorly soluble salts of weak acids (S2-, CO3

2-, F-), except when Ksp is really very small (PbS, HgS, ...).