c h a p t e rc h a p t e r c h a p t e rc h a p t e r 16 applications of aqueous equilibria...

C h a p t e rC h a p t e r 1616Applications of Aqueous

EquilibriaApplications of Aqueous

Equilibria

Chemistry 4th EditionMcMurry/Fay

Chemistry 4th EditionMcMurry/Fay

Buffer Solutions 01Buffer Solutions 01

A Buffer Solution: is a solution of a weak acid and a weak base (usually conjugate pair); both components must be present.

A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.

Buffers are very important to biological systems!


Base is neutralized

by the weak acid.

Acid is neutralized

by the weak base.


Buffer solutions must contain relatively high concentrations of weak acid and weak base components to provide a high “buffering capacity”.

The acid and base components must not neutralize each other.

The simplest buffer is prepared from equal concentrations of an acid and its conjugate base.


a) Calculate the pH of a buffer system containing 1.0 M CH3COOH and 1.0 M CH3COONa.

b) What is the pH of the system after the addition of 0.10 mole of HCl to 1.0 L of buffer solution?

a) pH of a buffer system containing 1.0 M CH3COOH and 1.0 M CH3COO– Na+ is: 4.74 (see earlier slide)


b) What is the pH of the system after the addition of 0.10 mole of HCl to 1.0 L of buffer solution?

CH3CO2H(aq) + H2O(aq) CH3CO2–(aq) + H3O+(aq)

I 1.0 1.0 0 add 0.10 mol HCl

+ 0.10 – 0.10 C -x -x +x

E 1.1 – x 0.9 + x x

Ka = 1.8 x 10-5 = =(0.9+x)(x) (0.9)x

(1.1 – x) (1.1)

x = (1.1/0.9) 1.8 x 10-5

x = 2.2 x 10-5

pH = 4.66

reacts w/

Water – No BufferWater – No Buffer

What is the pH after the addition of 0.10 mole of HCl to 1.0 L of pure water?

HCl – strong acid, completely ionized.

H+ concentration will be 0.10 molar.

pH will be –log(0.10) = 1.0

Adding acid to water:

ΔpH = 7.0 - 1.0 = 6.0 pH units

Adding acid to buffer:

ΔpH = 4.74 - 4.66 = 0.09 pH units!!

The power of buffers!The power of buffers!

Titration: a procedure for determining the concentration of a solution using another solution of known concentration.

Titrations involving strong acids or strong bases are straightforward, and give clear endpoints.

Titration of a weak acid and a weak base may be difficult and give endpoints that are less well defined.

Acid–Base Titrations 01Acid–Base Titrations 01

The equivalence point of a titration is the point at which equimolar amounts of acid and base have reacted. (The acid and base have neutralized each other.)

For a strong acid/strong base titration, the equivalence point should be at pH 7.


HH++ ((aqaq) ) + OH + OH–– ((aqaq) ) → → HH22OO ( (ll))

neutral

The equivalence point of a titration is the point at which equimolar amounts of acid and base have reacted. (The acid and base have neutralized each other.)

Titration of a weak acid with a strong base gives an equivalence point with pH > 7.


HAHA ((aqaq)) + + OHOH– – ((aqaq) ) →→ HH22OO ((ll) + ) + AA–– ((aqaq))

basic

Acid–Base Titration Examples 03Acid–Base Titration Examples 03

Titration curve for strong acid–strong base:

Note the very sharp endpoint(vertical line) seen with strong acid – strong base titrations.

The pH is changingvery rapidly in this region.

add one drop ofbase: get a BIG change in pH

Titration of 0.10 M HCl 03Titration of 0.10 M HCl 03

Volume of Added NaOH

1. Starting pH (no NaOH added) 1.00

2. 20.0 mL (total) of 0.10 M NaOH. 1.48

3. 30.0 mL (total) of 0.10 M NaOH. 1.85

4. 39.0 mL (total) of 0.10 M NaOH. 2.90

5. 39.9 mL (total) of 0.10 M NaOH. 3.90

6. 40.0 mL (total) of 0.10 M NaOH. 7.00

7. 40.1 mL (total) of 0.10 M NaOH. 10.10

8. 41.0 mL (total) of 0.10 M NaOH. 11.08

9. 50.0 mL (total) of 0.10 M NaOH 12.05

pH


pH at the equivalence point will always be >7 w/ weak acid/strong base

Titration curve for weak acid–strong base:The endpoint(vertical line) is less sharp with weak acid – strong base titrations.

strong acid

weak acid

strong acid equivalence point

weak acid equivalence point

Acid–Base Titration CurvesAcid–Base Titration Curves

weak acid

very weak acid

With a very weak acid, the endpoint may be difficult to detect.


Strong Acid–Weak Base:

The (conjugate) acid hydrolyzes to form weak base and H3O+.

At equivalence point only the (conjugate) acid is present.

pH at equivalence point will always be <7.

Aqueous Solubility Rules for Ionic Compounds

A compound is probably soluble if it contains the cations: a. Li+, Na+, K+, Rb+ (Group 1A on periodic table)

b. NH4+

A compound is probably soluble if it contains the anions:

a. NO3– (nitrate), CH3CO2– (acetate, also written C2H3O2

–)

b. Cl–, Br–, I– (halides) except Ag+, Hg22+, Pb2+ halides

c. SO42– (sulfate) except Ca2+, Sr2+, Ba2+, and Pb2+ sulfates

Other ionic compounds are probably insoluble.

Solubility Equilibria 01Solubility Equilibria 01

Old way to analyze solubility.Old way to analyze solubility.

Answer is either “yes” or “no”.Answer is either “yes” or “no”.

Solubility Equilibria02Solubility Equilibria02

New method to measuremeasure solubility:Consider solution formation an equilibrium process:

MCl2(s) M2+(aq) + 2 Cl–(aq)

Give equilibrium expression Kc for this equation:

KC = [M2+][Cl–]2

This type of equilibrium constant Kc that measures

solubility is called: KKspsp


Solubility Product: is the product of the molar concentrations of the ions and provides a measure of a compound’s solubility.

MX2(s) M2+(aq) + 2 X–(aq)

Ksp = [M2+][X–]2

“Solubility Product Constant”

Al(OH)3 1.8 x 10–33

BaCO3 8.1 x 10–9

BaF2 1.7 x 10–6

BaSO4 1.1 x 10–10

Bi2S3 1.6 x 10–72

CdS 8.0 x 10–28

CaCO3 8.7 x 10–9

CaF2 4.0 x 10–11

Ca(OH)2 8.0 x 10–6

Ca3(PO4)2 1.2 x 10–26

Cr(OH)3 3.0 x 10–29

CoS 4.0 x 10–21

CuBr 4.2 x 10–8

CuI 5.1 x 10–12

Cu(OH)2 2.2 x 10–20

CuS 6.0 x 10–37

Fe(OH)2 1.6 x 10–14

Fe(OH)3 1.1 x 10–36

FeS 6.0 x 10–19

PbCO3 3.3 x 10–14

PbCl2 2.4 x 10–4

PbCrO4 2.0 x 10–14

PbF2 4.1 x 10–8

PbI2 1.4 x 10–8

PbS 3.4 x 10–28

MgCO3 4.0 x 10–5

Mg(OH)2 1.2 x 10–11

MnS 3.0 x 10–14

Hg2Cl2 3.5 x 10–18

HgS 4.0 x 10–54

NiS 1.4 x 10–24

AgBr 7.7 x 10–13

Ag2CO3 8.1 x 10–12

AgCl 1.6 x 10–10

Ag2SO4 1.4 x 10–5

Ag2S 6.0 x 10–51

SrCO3 1.6 x 10–9

SrSO4 3.8 x 10–7

SnS 1.0 x 10–26

Zn(OH)2 1.8 x 10–14

ZnS 3.0 x 10–23

Solubility Equilibria - Ksp Values 04Solubility Equilibria - Ksp Values 04


• The solubility of calcium sulfate (CaSO4) is found experimentally to be 0.67 g/L. Calculate the value of Ksp for calcium sulfate.

• The solubility of lead chromate (PbCrO4) is 4.5 x 10–5 g/L. Calculate the solubility product of this compound.

• Calculate the solubility of copper(II) hydroxide, Cu(OH)2, in g/L.

Equilibrium Constants - ReviewEquilibrium Constants - Review

The reaction quotient (Qc) is obtained by substituting initial concentrations into the equilibrium constant. Predicts reaction direction.

Qc < Kc System forms more products (right)

Qc = Kc System is at equilibrium

Qc > Kc System forms more reactants (left)

Equilibrium Constants - QcEquilibrium Constants - Qc

Predicting the direction of a reaction.

QQcc < K< Kcc

QQcc > > K Kcc


We use the reaction quotient (Qc) to determine if

a chemical reaction is at equilibrium:

compare Qc and Kc

Ksp values are also a type of equilibrium constant,

but are valid for saturated solutions only.

We can use “ion product” (IP) to determine whether a precipitate will form:

compare IP and Ksp


Ion Product (IP): solubility equivalent of reaction

quotient (Qc). It is used to determine whether a

precipitate will form.

IP < Ksp

IP = Ksp

IP > Ksp

Unsaturated (more solute can dissolve)

Saturated solution

Supersaturated; precipitate forms.


A BaCl2 solution (200 mL of 0.0040 M) is added to

600 mL of 0.0080 M K2SO4. Will precipitate form?

(Ksp for BaSO4 is 1.1 x 10-10)

0.200 L x 0.0040 mol/L = .00080 moles Ba2+

[Ba2+] = 0.00080 mol 0.800 L = 0.0010 M

0.600 L x 0.0080 mol/L = .00480 moles SO42–

[SO42–] = 0.00480 mol 0.800 L = 0.0060 M

[ Ba2+ ]

[ SO42– ]


IP = [ Ba2+ ]1 x [ SO42– ]1

= (0.0010) x (0.0060)= 6.00 x 10-6

A BaCl2 solution (200 mL of 0.0040 M) is added to

600 mL of 0.0080 M K2SO4. Will precipitate form?

(Ksp for BaSO4 is 1.1 x 10-10)

IP > Ksp, so ppt forms


Exactly 200 mL of 0.0040 M BaCl2 are added to

exactly 600 mL of 0.0080 M K2SO4. Will a

precipitate form?

If 2.00 mL of 0.200 M NaOH are added to 1.00 L

of 0.100 M CaCl2, will precipitation occur?

The solubility product (Ksp) is an equilibrium

constant; precipitation will occur when the ion

product (IP) exceeds the Ksp for a compound.

If AgNO3 is added to saturated AgCl, the increase

in [Ag+] will cause AgCl to precipitate.

IP = [Ag+]0 [Cl–]0 > Ksp

The Common-Ion Effect and SolubilityThe Common-Ion Effect and Solubility


MgF2

becomes less soluble as F- conc. increses


CaCO3 is more

soluble at low pH.

Calculate the solubility of silver chloride (in

g/L) in a 6.5 x 10–3 M silver chloride solution.

Calculate the solubility of AgBr (in g/L) in:

(a) pure water

(b) 0.0010 M NaBr


c h a p t e rc h a p t e r c h a p t e rc h a p t e r 16 applications of aqueous equilibria...

Documents

ph unitsadding acid

hcl strong acid

weak base components

conjugate base

drop of base

buffer system

x x e

equimolar amounts of