Chapter 17: Applications of

Aqueous Equilibria

By: Ms. Buroker

The Common Ion EffectSuppose we had a solution that contained the weak acid HF (Ka= 7.2 x 10-4) and its salt NaF.

What would be the major species in solution?HF, Na+, F-, H2O

The common ion is this solution is F- because it comes from both the HF and the NaF!!!

So … what effect does the presence of the NaF have on the dissociation of HF?Let’s compare:

1.0M HF with (1.0M HF + 1.0M NaF)

According to LeChatelier’s principle we would expect the dissociation equilibrium for HF to be:

HF(aq) ↔ H+(aq) + F-

(aq)

Equilibrium shifts away from the

added component.

Added F- ions from NaF

Buffered Solutions

A solution that resists a change in pH when either hydroxide ions or protons are added.

Buffered solutions contain either: – A weak acid and its salt – A weak base and its salt

Acid/Salt Buffering Pairs

The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH)

Weak Acid Acid Formula

Example of salt of the weak acid

Hydrofluoric acid

HF KF

Hydrocyanic acid

HCN NaCN

Acetic Acid HC2H3O2 Mg(C2H3O2)2

Base/Salt Buffering Pairs

The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO3)

Weak Base Base Formula

Example of salt of the weak base

Ammonia NH3 NH4ClMethylamine CH3NH2 CH3NH2ClEthylamine C2H5NH2 C2H5NH3NO3

Example #1What is the pH of a buffer solution composed

of .100M propanic acid (HC3H5O2; Ka=1.3x10-

5) and .100M sodium propanate?

To decide pH, we need to figure out how the presence of the salt affects the acid’s equilibrium:

HC3H5O2 <=> H+ + C3H5O2-

Example #1 (continued)

HC3H5O2 <=> H+ + C3H5O2-

ICE

.100M 0 0.100M

-x +x +x

.100-x x 0.100 + x

Example #1 (continued)

K =[HC3H5O2]

[H+] [C3H5O2-]

K =[.100-x]

[x] [.100+x]

we will assume that the effect of “x” can be ignored (we will double check later)

K = [x] = 1.3 x 10-5 pH = -log[x] = 4.88

How do buffers work?

• The presence of the common ion from the conjugate helps moderate the equilibrium

• More of the common ion shifts the equilibrium toward the non-dissociated side of the equilibrium

• Adding more acid or base has a lesser effect on pH because of this

Example #2How does the addition of 0.010 moles of

solid NaOH affect the pH of 1 L of the buffer from the last example?

HC3H5O2 + OH- C3H5O2- + H2O

• Stoichiometry Point of View

• 0.010 moles of the base will react with the same amount of the acid, and produce the same amount of the conjugate

• Therefore we have:

• 0.100 - .010 = .090 moles of the acid remaining

• 0.100 + .010 = .110 moles of the conjugate present

Example #2 (continued)

HC3H5O2 H+ + C3H5O2-

ICE

.090M 0 .110M -x +x +x

.090-x x .110 + x

Example #2 (continued)

K =[HC3H5O2]

[H+] [C3H5O2-] K =

[.090-x][x] [.110+x]

we will assume that the effect of “x” can be ignored (we will double check later)

1.3x10-5 =[.090]

[x] [.110]

x = 1.06 x 10-5

pH = -log[x] = 4.97

Another Way to Solve

The stability of a buffer can be explained another way:

Ka =[HA]

[H+] [A-]

[H+] = [HA]

[A-]

Ka

can be re-arranged to:

Another Way:Henderson- Hasselbalch

Taking negative log of both sides gives:

-log[H+] = -log(Ka) - log[HA]

[A-]

pH = pKa + log

[HA]

[A-]

pH = pKa + log

[acid]

[base]Henderson-Hasselbalch Equation

Example #2 (again)

pH = pKa + log[acid]

[base]

pH = p(1.3x10-5) + log (.090)

(.110)

pH = 4.97

Example #3

A buffered solution contains .50M NH3 (Kb=1.8x10-5)and .30M NH4Cl. What is its pH?

Kw = (Ka)(Kb)

1x10-14 = (Ka)(1.8x10-5)5.6x10-10 = (Ka)

To solve H-H:

Example #3 (continued)

pH = pKa + log[acid]

[base]

pH = p(5.6x10-10) + log [.30]

[.50]

pH = 9.26 + (-.22) = 9.04

Acid- Base Titrations

Titration: Quantitatively determining the concentration of a solution by reacting it with another solution of precisely known concentration.Terminology– Titrant: The solution of known concentration used to titrate a

volume of the solution of unknown concentration.– Standardized (standard) solution: Solution of precisely known

concentration– Equivalence point: The point at which stoichiometrically

equivalent amounts of reactants have reacted– Endpoint: The point where an indicator (such as pH) changes

color.

Titration Curve

The pH at the equivalence point of a strong acid-strong base titration is 7.

The pH of the equivalence point is taken as the mid-point in the vertical portion of the pH verses volume of titrant curve.

ExampleWhat is the pH after 25.0mL of 0.100M NaOH has been added to 50.0mL of 0.100M HCl? What is the pH after 50.50mL of NaOH has been added?

1.) The first thing you need to do is find out how much of the acid will be left once the base has been added … you know all 25mL of the base will react.

Answer to First Partmol acid left: (0.0500L)(0.100M) – (.0250L)(0.100M)

= .0025mol

New [H3O+] = .0025mol/.075L = .033M

pH = -log(.033) = 1.48Remember now that the base has been added to the acid … the volume has increased!! Draw a picture … it helps to keep things straight.

Now … back to the problem

What is the pH after 25.0mL of 0.100M NaOH has been added to 50.0mL of 0.100M HCl? What is the pH after 50.50mL of NaOH has been added?

2.) Now you will notice that all of the acid has reacted and you have excess base remaining. So now you just have to find the hydroxide ion concentration and go from there.

Answer to Second Part

At 50.50mL of NaOH all of the HCl is neutralized and an extra 0.50mL of .100M NaOH exists in solution.

[OH-] = (0.0005L)(0.100M) / .10050L = 4.975x10-4

pOH = -log(4.975x10-4) = 3.303pH = 14.0 – 3.303 = 10.7

Remember now that the base has been added to the acid … the volume has increased!! Draw a picture … it helps to keep things straight.

Weak Acid- Strong Base Titrations

*The pH before the titration begins is given by the equilibrium expression using Ka of the weak acid.

*The pH at the equivalence point is dominated by the conjugate base (in the salt that forms) from the weak acid that was titrated.*The pH at the halfway point (half-equivalence point) is given by:

pH = pKa + log([A-]/[HA])*at the halfway point [A-] = [HA] and the log(1) =

0 Therefore pH = pKa

Titration Curve

ExampleWhat is the pH of the solution when 35.0mL of 0.100M NaOH has been added to 100.0mL of 0.100M acetic acid?Answer:HC2H3O2 + NaOH →NaC2H3O2 + H2O

Moles NaOH = (.0350L)(0.100M) =.0035 molMoles HC2H3O2 = (.1000L)(0.100M) = .0100 mol

pH = pKa + log ([C2H3O2-]/[HC2H3O2])

pH = 4.74 + log (.0259M / 0.0481M)= 4.47 Moles of acid minus the moles of base it reacted with divided by liters of solution.

ExampleCalculate the pH after 75.0mL of 0.100M HCl has been added to 100.0mL of 0.100M NH3 (Kb of NH3 = 1.8x10-5)

Answer:mol HCl = (.0750L)(0.100M) = .00750mol HClinitial moles NH3 = (.100L)(0.100M) = .0100mol NH3

Final:mol NH4+ = .00750molmol NH3 = .0100 - .00750 = .0025mol

pOH = pKb + log([NH4+]/[NH3])

pOH = 4.74 + log(.00750/.0025) = 5.22pH = 14.00 – 5.22 = 8.78

Solubility of SaltsFor a given slightly soluble salt of the

form:• AxBy(s) ↔xAy

+(aq) + yBx

-(aq)

• Ksp= [Ay+]x[Bx

-]y

• Where Ksp is the solubility product constant (equilibrium constant) for the salt (in aqueous solution)

• Can be used to determine solubility (in mol/L or g/100mL H2O, etc.)Molar solubility (mol/L) is the number of moles of solute dissolved in 1 L of a

saturated solution.

Solubility (g/L) is the number of grams of solute dissolved in 1 L of a saturated solution.

Solubility Equilibria

AgCl (s) Ag+ (aq) + Cl- (aq)

Ksp = [Ag+][Cl-] Ksp is the solubility product constant

MgF2 (s) Mg2+ (aq) + 2F- (aq) Ksp = [Mg2+][F-]2

Ag2CO3 (s) 2Ag+ (aq) + CO32- (aq) Ksp = [Ag+]2[CO3

2-]

Ca3(PO4)2 (s) 3Ca2+ (aq) + 2PO43- (aq) Ksp = [Ca2+]3[PO3

3-]2

Dissolution of an ionic solid in aqueous solution:

Q = Ksp Saturated solution

Q < Ksp Unsaturated solution No precipitate

Q > Ksp Supersaturated solution Precipitate will form

What is the solubility of silver chloride in g/L ?

AgCl (s) Ag+ (aq) + Cl- (aq)

Ksp = [Ag+][Cl-]Initial (M)

Change (M)

Equilibrium (M)

0.00

+x

0.00

+x

x x

Ksp = x2

x = Kspx = 1.3 x 10-5

[Ag+] = 1.3 x 10-5 M [Cl-] = 1.3 x 10-5 M

Solubility of AgCl = 1.3 x 10-5 mol AgCl

1 L soln

143.35 g AgCl

1 mol AgClx = 1.9 x 10-3 g/L

Ksp = 1.6 x 10-10

ExampleKnowing that the Ksp value for MgF2 is 5.2x10-11,

calculate the solubility of the salt in a) moles per liter and b) grams per literAnswer:5.2x10-11 = [X][2X]2

X = 2.35x10-4 = 2.4x10-4mol/L2.4x10-4mol/L(62.3g/mol) = 0.015gMgF2/L

Reaction:MgF2 ↔ Mg+2

(aq) + 2F-

(aq)

Ksp = [Mg+2][F-]2

If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl2, will a precipitate form?

The ions present in solution are Na+, OH-, Ca2+, Cl-.

Only possible precipitate is Ca(OH)2 (solubility rules).

Is Q > Ksp for Ca(OH)2?

[Ca2+]0 = 0.100 M [OH-]0 = 4.0 x 10-4 M

Ksp = [Ca2+][OH-]2 = 8.0 x 10-6

Q = [Ca2+]0[OH-]02 = 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8

Q < Ksp No precipitate will form

Solubility and the Common Ion Effect

The effect of having an ion already in solution that is common to one of the ions in a substance being added to the solution (with a given Ksp) is to reduce the solubility of that substance.

Example:Calculate the solubility of BaSO4 a) in pure water and b) in the presence of 0.010M Ba(NO3)2. Ksp for BaSO4 is 1.1x10-10.

Think LeChatelier!!!

Answera.) Ksp = [Ba2+][SO4

2-]

so … 1.1x10-10 = (x)(x)solving for x gives … x= 1.0x10-

5mol/L

b.) 1.1x10-10 = (x +.010)(x)solving for x gives … x = 1.1x10-

8mol/LThis comes from the common ion that was added to the solution of BaSO4.

Complex Ion Equilibria and Solubility

A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions.

Co2+ (aq) + 4Cl- (aq) CoCl4 (aq)2-

Co(H2O)62+ CoCl42-

16.10

Complex Ion Formation

• These are usually formed from a transition metal surrounded by ligands (polar molecules or negative ions).

• As a "rule of thumb" you place twice the number of ligands around an ion as the charge on the ion... example: the dark blue Cu(NH3)4

2+ (ammonia is used as a test for Cu2+ ions), and Ag(NH3)2

+.• Memorize the common ligands.

Common LigandsLigands Names used in the ion

H2O aqua

NH3 ammine

OH- hydroxyCl- chloroBr- bromoCN- cyanoSCN- thiocyanato (bonded through

sulfer) isothiocyanato (bonded through nitrogen)

Names• Names: ligand first, then cation

Examples:– tetraamminecopper(II) ion: Cu(NH3)4

2+

– diamminesilver(I) ion: Ag(NH3)2+.

– tetrahydroxyzinc(II) ion: Zn(OH)4 2-

• The charge is the sum of the parts (2+) + 4(-1)= -2.

When Complexes Form• Aluminum also forms complex ions as do some post

transitions metals. Ex: Al(H2O)63+

• Transitional metals, such as Iron, Zinc and Chromium, can form complex ions.

• The odd complex ion, FeSCN2+, shows up once in a while

• Acid-base reactions may change NH3 into NH4+ (or vice

versa) which will alter its ability to act as a ligand.• Visually, a precipitate may go back into solution as a

complex ion is formed. For example, Cu2+ + a little NH4OH will form the light blue precipitate, Cu(OH)2. With excess ammonia, the complex, Cu(NH3)4

2+, forms.• Keywords such as "excess" and "concentrated" of any

solution may indicate complex ions. AgNO3 + HCl forms the white precipitate, AgCl. With excess, concentrated HCl, the complex ion, AgCl2-, forms and the solution clears.

Coordination Number

• Total number of bonds from the ligands to the metal atom.

• Coordination numbers generally range between 2 and 12, with 4 (tetracoordinate) and 6 (hexacoordinate) being the most common.

Some Coordination Complexes

molecular formula

Lewis base/ligand

Lewis acid donor atom

coordination number

Ag(NH3)2+ NH3 Ag+ N 2

[Zn(CN)4]2- CN- Zn2+ C 4

[Ni(CN)4]2- CN- Ni2+ C 4

[PtCl6] 2- Cl- Pt4+ Cl 6

[Ni(NH3)6]2+ NH3 Ni2+ N 6