applications of aqueous equilibria common ions when we dissolve acetic acid in water, the following...

Applications of Aqueous Equilibria

“Common Ions”• When we dissolve acetic acid in water, the

following equilibrium is established:– CH3COOH CH3COO- + H+

• If sodium acetate were dissolved in solution, which way would this equilibrium shift?

“Common Ions”• When we dissolve acetic acid in water, the

following equilibrium is established:– CH3COOH CH3COO- + H+

• If sodium acetate were dissolved in solution, which way would this equilibrium shift?– Adding acetate ions from a strong electrolyte would

shift this equilibrium left (Le Chatelier’s principle).

Common Ion Effect

• Whenever a weak electrolyte (acetic acid) and a strong electrolyte (sodium acetate) share a common ion, the weak electrolyte ionizes less than it would if it were alone. (Le Chatelier’s)

• This is called the common-ion effect.

Steps for Common-Ion Problems

• 1. Consider which solutes are strong electrolyte and weak electrolytes.

• 2. Identify the important equilibrium (weak) that is the source of H+ and therefore determines pH.

• 3. Create an ICE chart using the equilibrium and strong electrolyte concentrations.

• 4. Use the equilibrium constant expression to calculuate [H+] and pH.

• What is the concentration of silver and chromate in a solution with silver chromate in 0.1 M silver nitrate

Ag2CrO4(s) 2Ag+1 + CrO4-2

I 0.1 0 C +2x +x E 0.1 + 2x x

Ksp = 9.0 x 10-12 = [Ag+1]2 • [CrO4-2]

= 9.0 x 10-12 = [.1 +2x]2 • [x] 9.0 x 10-12 = 0.12 • x

[CrO4-2] = x = 9.0 x 10-10 M

[Ag+1] = .1 + 2x = .1M

[Ag2CrO4] = 9.0 x 10-10

What are the sources of Ag+1?

Why is this a plus sign?

Sample Problem• What is the pH of a solution made by adding

0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution?

• Ka for acetic acid = 1.8 x 10-5

pH = 4.74

Sample Problem

• Calculate the fluoride concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl

• Ka for HF = 6.8 x 10-4.

• [F-] = 1.2x10-3M

• pH = 1.00

BuffersSolutions that resist pH change when small amounts of

acid or base are added.

Two typesTwo types

weak acid and its salt

weak base and its salt

HX(aq) + H2O(l) H+(aq) + X-(aq)

Add OH- Add H+

shift to right shift to left

Based on the common ion effect.

ml HCl added

pH buffered

unbuffered

Buffers

Buffers and bloodControl of blood pHControl of blood pH

Oxygen is transported primarily by hemoglobin in the red blood cells.

CO2 is transported both in plasma and the red blood cells.

CO2 (aq) + H2O H2CO3 (aq)

H+(aq) + HCO3-(aq)

Composition and Action of Buffered SolutionsComposition and Action of Buffered Solutions

The Ka expression is

A buffer resists a change in pH when a small amount of OH- or H+ is added.

HX(aq) H+(aq) + X-(aq)

.]X[

[HX]]H[

.[HX]

]X][H[

-

-

a

a

K

K

Buffers

Buffered SolutionsBuffered SolutionsAddition of Strong Acids or Bases to BuffersAddition of Strong Acids or Bases to Buffers• With the concentrations of HX and X- (note the

change in volume of solution) we can calculate the pH from the Henderson-Hasselbalch equation

acid conjugatebase conjugate

logppH

[HX]]X[

logppH-

a

a

K

K

Equation not necessary, since you knowEquation not necessary, since you know Ka = [X-]

[H+] [HX]

Buffers

Action of Buffered SolutionsAction of Buffered Solutions

Buffers

Buffer Capacity and pHBuffer Capacity and pH

Buffer capacity is the amount of acid or base neutralized by the buffer before there is a significant change in pH.

The greater the amounts(molarity) of the conjugate acid-base pair, the greater the buffer capacity.

The pH of the buffer depends on Ka

Buffered SolutionsBuffered SolutionsAddition of Strong Acids or Bases to BuffersAddition of Strong Acids or Bases to Buffers

BufferBufferAddition of Strong Acids or Bases to BuffersAddition of Strong Acids or Bases to Buffers

Break the calculation into two parts: stoichiometric and equilibrium.

The amount of strong acid or base added results in a neutralization reaction:

X- + H+ HX + H2O

HX + OH- X- + H2O.

By knowing how much H+ or OH- was added (stoichiometry) we know how much HX or X- is formed.

BuffersThe final [HX] and [X-] after the neutralization reaction

are used as the initial concentrations for the equilibrium reaction.

HX H+ X-

Initial conc., M

Change, M

Eq. Conc., M

Then the equilibrium constant expression is used to find [H+] and pH = - log [H+]

Ka = [H+] [X-]

[HX]

Buffer ExampleDetermine the initial pH of a buffer solution that has 0.10 M benzoic acid and 0.20 M sodium benzoate at 25 oC. Ka = 6.5 x 10-5

HBz(aq) + H2O(l) H+(aq) + Bz-(aq)

HBz H+ Bz-

Initial conc., M 0.10 0.00 0.20

Change, DM -x +x +x

Eq. Conc., M 0.10 - x x 0. 20 + x

Buffer ExampleSolve the equilibrium equation in terms of Solve the equilibrium equation in terms of xx

Ka = 6.5 x 10-5 =

x = (6.5 x 10-5 )(0.10) / (0. 20)

(assuming x<<0.10)

= 3.2 x 10-5 M H+

pH = - log (3.2 x 10-5 M) = 4.5

initial pH

x (0. 20 + x)0.10 - x

Buffer ExampleDetermine the pH of a buffer solution that has 0.10 M benzoic acid and 0.20 M sodium benzoate at 25 oC after 0.05 moles of HCl is added.

HBz H+ Bz-

Initial conc., M 0.15 0.00 0.15Change, DM -x +x +xEq. Conc., M 0.15 - x x 0. 15 + x

The 0.05 mol HCl reacts completely with 0.05 mol Bz-

(aq) to form 0.05 mol HBz(aq)Then equilibrium will be re-established based on the new initial concentrations of 0.15 M HBz(aq) and 0.15 M Bz-(aq).

First, find the concentrations of First, find the concentrations of the HBz and Bz- after HCl is added.


Ka = 6.5 x 10-5 =

x = (6.5 x 10-5 )(0.15) / (0. 15)

(assuming x<<0.15)

= 6.5 x 10-5 M H+

pH = - log (6.5 x 10-5 M) = 4.2

after 0.05 mole HCl added

x (0. 15 + x)0.15 - x

Buffer ExampleDetermine the pH of a buffer solution that has 0.10 M benzoic acid and 0.20 M sodium benzoate at 25 oC after 0.05 moles of NaOH is

added.

HBz H+ Bz-

Initial conc., M 0.05 0.00 0.25Change, DM -x +x +xEq. Conc., M 0.05 - x x 0. 25 + x

The 0.05 mol NaOH reacts completely with 0.05 mol HBz(aq) to form 0.05 mol Bz-(aq)

Then equilibrium will be re-established based on the new initial concentrations of 0.05 M HBz(aq) and 0.25 M Bz-(aq).

First, find the concentrations of First, find the concentrations of HBz and Bz- after NaOH is added.


Ka = 6.5 x 10-5 =

x = (6.5 x 10-5 )(0.05) / (0. 25)

(assuming x<<0.05)

= 1.3 x 10-5 M H+

pH = - log (1.3 x 10-5 M) = 4.9

after 0.05 mole NaOH added

x (0. 25 + x)0.05 - x

Acid-Base TitrationsAcid-Base Titrations Strong Acid-Base TitrationsStrong Acid-Base Titrations

• The plot of pH versus volume during a titration is a titration curve.

Remember the Acid-Base Titration Curves?Remember the Acid-Base Titration Curves?

HCl

HC2H3O2

Buffer Buffer ZoneZone

For this experiment, you will prepare a buffer that contains acetic acid and its conjugate base, the acetate ions. The equilibrium equation for the reaction is shown below:

HC2H3O2(aq) + H2O(l) <=> H+ (aq)+ C2H3O2- (aq)

The equilibrium expression for this reaction, Ka, has a value of 1.8 x 10-5 at 25ºC.

Acetic Acid/Acetate Ion Buffer LabAcetic Acid/Acetate Ion Buffer Lab

Acetic Acid/Acetate Ion Buffer LabAcetic Acid/Acetate Ion Buffer Lab The ratio between the molarity of the acetate ions to the

molarity of the acetic acid in your buffer must equal the ratio between the Ka value and 10- assigned pH.

This ratio should be reduced , so that either the [HC2H3O2] or [C2H3O2

- ] has a concentration of 0.10 M, and the concentration of the other component must fall within a range from 0.10 M to 1.00 M.

Complete the calculations onlyonly that are needed to prepare 100.0 mL of your assigned buffer solution that has these specific concentrations. Can you predict the final pH when a strong acid or base is added to the buffer solution?

Acetic Acid/Acetate Ion Buffer LabAcetic Acid/Acetate Ion Buffer LabBuffer 1

Given: 1.00 M acetic acid solid sodium acetate trihydrate

Problem: Prepare 100.0 mL of a buffer solutionwith a pH of 4.50 that has a concentration of eachcomponent within the range of 0.10 M to 1.00 M.Use the lowest possible concentrations for the acidand conjugate base in the buffer.

SHOW ALL CALCULATIONS USED!

Buffer 2Given: 1.00 M hydrochloric acid

solid sodium acetate trihydrateProblem: Prepare 100.0 mL of a buffer solutionwith a pH of 4.50 that has a concentration of eachcomponent within the range of 0.10 M to 1.00 M.Use the lowest possible concentrations for the acidand conjugate base in the buffer.

SHOW ALL CALCULATIONS USED!Buffer 3

Given: 1.00 M acetic acid 1.00 M sodium hydroxide



Buffer 4Given: 1.00 M acetic acid



Given: 1.00 M hydrochloric acid solid sodium acetate trihydrate




1.00 M sodium hydroxideProblem: Prepare 100.0 mL of a buffer solutionwith a pH of 5.00 that has a concentration of eachcomponent within the range of 0.10 M to 1.00 M.Use the lowest possible concentrations for the acidand conjugate base in the buffer.


Acetic Acid/Acetate Ion Buffer LabAcetic Acid/Acetate Ion Buffer LabBuffer 7

Given: 1.00 M acetic acid solid sodium acetate trihydrate



Buffer 8Given: 1.00 M hydrochloric acid



Given: 1.00 M acetic acid 1.00 M sodium hydroxide






Given: 1.00 M hydrochloric acid solid sodium acetate trihydrate



Making a Buffer CalculationsMaking a Buffer Calculations

Name Formula Ka1 Ka2 Ka3

acetic acid HC2H3O2 1.8 x 10-5 tartaric acid H2C4H4O6 1.0 x 10-3 4.6 x 10-5 citric acid H3C6H5O7 7.4 x 10-4 1.7 x 10-5 4.0 x 10-7

malonic acid H2C3H2O4 1.5 x 10-3 2.0 x 10-6 carbonic acid H2CO3 4.3 x 10-7 5.6 x 10-11

phosphoric acid H3PO4 7.5 x 10-3 6.2 x 10-8 4.2 x 10-13 ammonium NH 41+ 5.6 x 10-10

You want to prepare 500.0 mL of a buffer with a pH = 10.00, with both the acid and conjugate base having molarities between 0.10 M to 1.00 M. You may choose from any of the acids listed below:

You must select an acid with a Ka value close to 10- assigned pH. The only two options are ammonium or the hydrogen carbonate ions.

Ka = 5.6 x 10-10 =H+ NH3

NH4+

Ka2 = 5.6 x 10-11 =H+ CO3

2 HCO3

1-


Ka = 5.6 x 10-10 =H+ NH3

NH4+

Divide both [ ] by 10..... 0.56

0.1=

NH3 NH4

+

5.6 x 10-10

H+ =

NH3 NH4

+

5.6 x 10-10

10-10 =

NH3 NH4

+

5.6

1=

NH3 NH4

+

x g NH4Cl = 500 mL 0.1 mol NH4

+

1000 mL

53.5 g NH4Cl

1 mol NH4+

= 2.68 g NH4Cl

x mL conc NH3 = 500 mL buffer 0.56 mol NH3

1000 mL buff

1000 mL conc NH3

14.8 mol NH3

=18.9 mL 14.8 M NH3


x g NH4Cl = 500 mL buffer 0.56 mol NH3

1000 mL buff

1 mol NH 4Cl

1 mol NH 3

58.5 g NH4Cl

1 mol NH4Cl

=16.38 g NH4Cl

Place ~250 mL of distilled water in a 500 mL volumetric flask. Add 2.68 g NH4Cl and dissolve. Then mix in 18.9 mL of 14.8 M NH3. Fill with distilled water to the 500 mL mark on the flask.

If there is no concentrated NHIf there is no concentrated NH33 available, the NH available, the NH33 can be can be

produced by neutralizing additional NHproduced by neutralizing additional NH44Cl with 1.00 M NaOH. Cl with 1.00 M NaOH.

x mL NaOH = 500 mL buffer 0.56 mol NH3

1000 mL buff

1 mol NaOH

1 mol NH3

1000 mL NaOH

1.00 mol NaOH

= 280. mL NaOH

Dissolve 19.06 g NHDissolve 19.06 g NH44Cl (2.68 g + 16.38 g) in 280. mL of 1.00 M NaOH. Cl (2.68 g + 16.38 g) in 280. mL of 1.00 M NaOH.

Then dilute with distilled water and fill to the 500 mL mark on the flask.Then dilute with distilled water and fill to the 500 mL mark on the flask.


x mL NH 3 = 500 mL buffer 0.10 mol NH4

1000 mL buff

1 mol NH3

1 mol NH4

1000 mL NH3

14.8 mol NH3

= 3.38 mL NH 3

If there is no NHIf there is no NH44Cl available, the NHCl available, the NH44++ can be produced by can be produced by

neutralizing additional NHneutralizing additional NH33 with 1.00 M HCl. with 1.00 M HCl.

x mL HCl = 500 mL buffer 0.10 mol NH4

1000 mL buff

1 mol HCl

1 mol NH4

1000 mL HCl

1.00 mol HCl

= 50.0 mL HCl

Place ~250 mL distilled water in volumetric flask. Add 50.0 mL of 1.00 M Place ~250 mL distilled water in volumetric flask. Add 50.0 mL of 1.00 M HCl and mix. Then add 22.3 mL of concentrated NHHCl and mix. Then add 22.3 mL of concentrated NH33 (18.9 mL + 3.4 mL). (18.9 mL + 3.4 mL).

Mix and fill with distilled water to the 500 mL mark on the flask. Mix and fill with distilled water to the 500 mL mark on the flask.

Place ~250 mL of distilled water in a 500 mL volumetric flask. Add 2.68 g NH4Cl and dissolve. Then mix in 18.9 mL of 14.8 M NH3. Fill with distilled water to the 500 mL mark on the flask.


Ka2 = 5.6 x 10-11 =H+ CO3

2 HCO3

1-

5.6 x 10-11

H+ =

CO32

HCO31-

5.6 x 10-11

10-10 =

CO32

HCO31-

0.56

1=

CO32

HCO31-

Multiply both [ ] by 0.179..... 0.100

0.179=

CO32-

HCO31-

Prepare 500. mL of the buffer that has Prepare 500. mL of the buffer that has [CO[CO33

2-2-] = 0.100 M and [HCO] = 0.100 M and [HCO331-1-] = 0.179 M. ] = 0.179 M.


x g Na2CO3 = 500 mL 0.100 mol CO3

2-

1000 mL

106 g Na2CO3

1 mol CO32-

= 5.30 g Na2CO3

Prepare 500. mL of the buffer that has [COPrepare 500. mL of the buffer that has [CO332-2-] = 0.100 M ] = 0.100 M

and [HCOand [HCO331-1-] = 0.179 M. ] = 0.179 M.

x g NaHCO 3 = 500 mL 0.179 mol HCO 3

1-

1000 mL

84.0 g NaHCO 3

1 mol HCO 31-

= 7.52 g NaHCO 3

Place ~250 mL distilled water in a 500 mL volumetric flask. Place ~250 mL distilled water in a 500 mL volumetric flask. Add 5.30 g NaAdd 5.30 g Na22COCO33 and 7.52 g NaHCO and 7.52 g NaHCO33 and dissolve. Fill with and dissolve. Fill with

distilled water to the 500 mL mark on the flask.distilled water to the 500 mL mark on the flask.


• The plot of pH versus volume during a titration is a titration curve.

Indicator examples• Acid-base indicators are weak acids that undergo a

color change at a known pH.

phenolphthalein

pH

Indicator examples

methyl red

bromthymol blue

Select the indicator that undergoes a color change closest to the pH at the equivalence point, where all of the acid has been neutralized by the base.

Acid-Base TitrationsAcid-Base Titrations Strong Acid-Base TitrationsStrong Acid-Base Titrations• Consider adding a strong base (e.g. NaOH) to a

solution of a strong acid (e.g. HCl).– Before any base is added, the pH is given by the strong acid

solution. Therefore, pH < 7.

– When base is added, before the equivalence point, the pH is given by the amount of strong acid in excess. Therefore, pH < 7.

– At equivalence point, the amount of base added is stoichiometrically equivalent to the amount of acid originally present. Therefore, the pH is determined by the salt solution. Therefore, pH = 7.

Acid-Base TitrationsAcid-Base Titrations Strong Acid-Base TitrationsStrong Acid-Base Titrations• Consider adding a strong base (e.g. NaOH) to a

solution of a strong acid (e.g. HCl).

Acid-Base TitrationsAcid-Base Titrations Strong Acid-Base TitrationsStrong Acid-Base Titrations• We know the pH at equivalent point is 7.00. • To detect the equivalent point, we use an indicator

that changes color somewhere near 7.00.– Usually, we use phenolphthalein that changes color between

pH 8.3 to 10.0.– In acid, phenolphthalein is colorless.– As NaOH is added, there is a slight pink color at the

addition point.– When the flask is swirled and the reagents mixed, the pink

color disappears.– At the end point, the solution is light pink.– If more base is added, the solution turns darker pink.

Acid-Base TitrationsAcid-Base Titrations Strong Acid-Base TitrationsStrong Acid-Base Titrations• The equivalence point in a titration is the point at

which the acid and base are present in stoichiometric quantities.

• The end point in a titration is the observed point.• The difference between equivalence point and end

point is called the titration error.• The shape of a strong base-strong acid titration curve

is very similar to a strong acid-strong base titration curve.

Acid-Base TitrationsAcid-Base Titrations Strong Acid-Base TitrationsStrong Acid-Base Titrations• Initially, the strong base is in excess, so the pH > 7.• As acid is added, the pH decreases but is still greater

than 7.• At equivalence point, the pH is given by the salt

solution (i.e. pH = 7).• After equivalence point, the pH is given by the strong

acid in excess, so pH < 7.

Acid-Base TitrationsAcid-Base Titrations Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations• Consider the titration of acetic acid, HC2H3O2 and

NaOH.• Before any base is added, the solution contains only

weak acid. Therefore, pH is given by the equilibrium calculation.

• As strong base is added, the strong base consumes a stoichiometric quantity of weak acid:

HC2H3O2(aq) + NaOH(aq) C2H3O2-(aq) + H2O(l)

Acid-Base TitrationsAcid-Base Titrations Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations

Acid-Base TitrationsAcid-Base Titrations Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations• There is an excess of acetic acid before the

equivalence point. • Therefore, we have a mixture of weak acid and its

conjugate base.– The pH is given by the buffer calculation.

• First the amount of C2H3O2- generated is calculated,

as well as the amount of HC2H3O2 consumed.

(Stoichiometry.)• Then the pH is calculated using equilibrium

conditions. (Henderson-Hasselbalch.)

Acid-Base TitrationsAcid-Base Titrations Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations• At the equivalence point, all the acetic acid has been

consumed and all the NaOH has been consumed. However, C2H3O2

- has been generated.

– Therefore, the pH is given by the C2H3O2- solution.

– This means pH > 7.• More importantly, pH 7 for a weak acid-strong base titration.

• After the equivalence point, the pH is given by the strong base in excess.

Acid-Base TitrationsAcid-Base Titrations Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations• For a strong acid-strong base titration, the pH begins

at less than 7 and gradually increases as base is added.

• Near the equivalence point, the pH increases dramatically.

• For a weak acid-strong base titration, the initial pH rise is more steep than the strong acid-strong base case.

• However, then there is a leveling off due to buffer effects.

Acid-Base TitrationsAcid-Base Titrations Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations• The inflection point is not as steep for a weak acid-

strong base titration.• The shape of the two curves after equivalence point is

the same because pH is determined by the strong base in excess.

• Two features of titration curves are affected by the strength of the acid:– the amount of the initial rise in pH, and

– the length of the inflection point at equivalence.

Acid-Base TitrationsAcid-Base Titrations Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations

• The weaker the acid, the smaller the equivalence point inflection.

• For very weak acids, it is impossible to detect the equivalence point.

Acid-Base TitrationsAcid-Base Titrations Weak Acid-Strong Base TitrationsWeak Acid-Strong Base Titrations• Titration of weak bases with strong acids have similar

features to weak acid-strong base titrations.

Acid-Base TitrationsAcid-Base Titrations Titrations of Polyprotic AcidsTitrations of Polyprotic Acids• In polyprotic acids, each ionizable proton dissociates

in steps.• Therefore, in a titration there are n equivalence

points corresponding to each ionizable proton.

• In the titration of Na2CO3 with HCl there are two equivalence points:– one for the formation of HCO3

-

– one for the formation of H2CO3.

Acid-Base TitrationsAcid-Base Titrations Titrations of Polyprotic AcidsTitrations of Polyprotic Acids

Acid-Base TitrationsAcid-Base Titrations Titrations of Polyprotic AcidsTitrations of Polyprotic AcidsKa = [H+] [X-]

[HX]At the equivalence point, [H+] = [OH-]At ½ the equivalence point, [X-] = [HX]

SOOOO…. At ½ equivalence point, Ka = [H+]

So pKa = pH

Solubility

• What happens when an ionic compound is dissolved in water?

• What forces does water have to overcome?

• Are all ionic compounds soluble?– Remember the solubility rules?– It’s time to review them!

Solubility Rulesfor Common Ionic Compounds in Water

1. Most nitrates are soluble.2. Most salts containing alkali metal ions and ammonium are soluble

Li+1, Na+1, K+1, Cs+1, NH4+1

3. Most chloride, bromide, and iodide salts are solubleNotable exceptions are Hg2

+2, Ag+1, Pb+2

4. Most sulfate (SO4) salts are solubleNotable exceptions are Ca +2, Sr+2, Ba+2, Pb+2, Hg2

+2

5. Most hydroxides are only slightly soluble. Na+ and K+are soluble. Ba +2, Sr +2 and Ca +2 are marginally soluble

6. Most sulfide (S-2), carbonate (CO3-2), chromate (CrO4

-2) and phosphate (PO4

-3) are only slightly soluble.

Properties of aqueous solutionsThere are two general classes of solutes.

ElectrolyticElectrolyticionic compounds in polar solvents

dissociate in solution to make ions

conduct electricity

may be strong (100% dissociation) or weak (less than 10%)

NonelectrolyticNonelectrolyticdo not conduct electricity

solute is dispersed but does not dissociate

Heterogeneous Equilibria

A(s) + H2O B(aq) + C (aq)

Ksp = [B] • [C]

Why is this not divided by [A]?

Why is [H2O] not included?

Known as the Solubility Product = Ksp

Solubility Products, KSP

KSP expressions are used for ionic materials that are only slightly soluble in water.

Their only means of dissolving is by dissociation.

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

KSP =[ Ag+] [ Cl-]


At equilibrium, the system is a saturated solution of silver and chloride ions.

The only way to know that it is saturated it to observe some AgCl at the bottom of the solution.

As such, [AgCl] is a constant and KSP expressions do not include the solid form in the equilibrium expression. The [H2O] for solvation process is also excluded from the KSP expression.


Determine the solubility of AgCl in water at 20 oC in terms of grams / 100 mL

KSP = [Ag+] [Cl-] = 1.7 x 10-10

At equilibrium, [Ag+] = [Cl-] so

1.7 x 10-10 = [x] 2

[Ag+] = 1.3 x 10-5 M

g AgCl = 1.3 x 10-5 mol/L * 0.10 L * 143.32 g/mol

Solubility = 1.9 x 10-4 g / 100 mL

Calculating Ksp

Calculate the Ksp for Bismuth Sulfide which has a solubility of 1.0 x 10-15 .

Bi2S3 → 2Bi+3 + 3S-2

Ksp =

For every Bi2S3 dissolved

2 Bi+3 and 3S-2 are formed

Ksp = [2 x 1.0 x 10-15 ]2 • [3 x 1.0 x 10-15 ]3

= 1.1 x 10-73

[Bi+3]2 • [S-2]3

Calculating SolubilityCalculate the concentration of ions for copper (II) iodate.

ksp = 1.4 x 10-7 at 25 °C

Cu(IO3)2(s) Cu+2 + 2 IO3-1

1.4 x 10-7 = [x] • [ 2x]2 = 4x3

x = 3.3 x 10-3 mol/L = [Cu+2]

2x = 6.6 x 10-3 mol/L = [IO3-1]

3.3 x 10-3 mol/L = [Cu(IO3)2]

Solubility Products, KSPAnother exampleAnother example

Calculate the silver ion concentration when excess silver chromate is added to a 0.010 M sodium chromate solution.

KSP Ag2CrO4 = 1.1 x 10-12

Ag2CrO4 (s) 2Ag+ + CrO42-

Solubility Products, KSPKSP = 1.1 x 10-12 = [ Ag+ ]2 [ CrO4

2- ]

[CrO42-] = [CrO4

2-]Ag2CrO4 + [CrO42-]Na2CrO4

With such a small value for KSP, we can assume that the [CrO4

2-] from Ag2CrO4 is negligible.

If we’re wrong, our silver concentration will be significant (>1% of the chromate concentration.)

Then you’d use the quadratic approach.


KSP = 1.1 x 10-12 = [ Ag+ ]2 [ CrO42- ]

[CrO42-] = 0.010 M

[ Ag+ ] = ( KSP / [ CrO42- ] ) 1/2

= (1.1 x 10-12 / 0.010 M ) 1/2

= 1.1 x 10 -5 M

[ Ag+ ] << [ CrO42- ]

so our assumption was valid.

Relative Solubility

• Be careful when comparing solubility products (Ksp)

• Compare – AgI Ksp = 1.5 x 10-16

– CuI Ksp = 5.0 x 10-12

• Which is more soluble?

Copper (I) iodide is more soluble. Each compound produces the same number ions

Relative Solubility

• CuS Ksp = 8.5 x 10-45

• Ag2S Ksp = 1.6 x 10-49

• Bi2S3 Ksp = 1.1 x 10-73

Which is more soluble?Each has a different number of ions, so calculate the solubility of each.

Estimate (guess)three groups and calculate the solubility

Factors influencing solubilityCommon ion and salt effects.Common ion and salt effects.

As with other equilibria we’ve discussed, adding a ‘common’ ion will result in a shift of a solubility equilibrium.

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

KSP = [Ag+] [Cl-]

Adding either Ag+ or Cl- to our equilibrium system will result in driving it to the left.

Factors That Affect SolubilityFactors That Affect Solubility Common-Ion EffectCommon-Ion Effect

Factors influencing solubilityComplex ion formation.Complex ion formation.

The solubility of slightly soluble salts can be increased by complex ion formation.

Example. Addition of excess ClExample. Addition of excess Cl --

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

+ 2Cl-(aq)

AgCl2-(aq)

A large excess of chloride results in the formation of the complex.

More AgCl will dissolve as a result.

A large excess of chloride results in the formation of the complex.

More AgCl will dissolve as a result.

Factors That Affect SolubilityFactors That Affect Solubility Formation of Complex IonsFormation of Complex Ions

Factors That Affect SolubilityFactors That Affect Solubility Formation of Complex IonsFormation of Complex Ions• Consider the addition of ammonia to AgCl (white

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

Ag+(aq) + 2NH3(aq) Ag(NH3)2(aq)

AgCl(s) + 2NH3(aq) Ag(NH3)2(aq) + Cl-(aq)• The overall reaction is

• Effectively, the Ag+(aq) has been removed from solution.• By Le Châtelier’s principle, the forward reaction (the

dissolving of AgCl) is favored.

• Knet = Ksp•Kf = (1.8 x 10-10)(1.7 x 107) = 3.1 x 10-3 at 25ºC

+

+

Factors That Affect SolubilityFactors That Affect Solubility AmphoterismAmphoterism• Amphoteric oxides will dissolve in either a strong acid

or a strong base.• Examples: hydroxides and oxides of Al3+, Cr3+, Zn2+,

and Sn2+.• The hydroxides generally form complex ions with

four hydroxide ligands attached to the metal:

• Hydrated metal ions act as weak acids. Thus, the amphoterism is interrupted:

Al(OH3)(s) + OH-(aq) Al(OH)4-(aq)

Factors That Affect SolubilityFactors That Affect Solubility AmphoterismAmphoterism• Hydrated metal ions act as weak acids. Thus, the

amphoterism is interrupted:Al(H2O)6

3+(aq) + OH-(aq) Al(H2O)5(OH)2+(aq) + H2O(l)

Al(H2O)5(OH)2+(aq) + OH-(aq) Al(H2O)4(OH)2+(aq) + H2O(l)

Al(H2O)4(OH)+(aq) + OH-(aq) Al(H2O)3(OH)3(s) + H2O(l)

Al(H2O)3(OH)3(s) + OH-(aq) Al(H2O)2(OH)4-(aq) + H2O(l)

Factors influencing solubilityHydrolysis.Hydrolysis.

If the anion of a weak acid, or cation of a weak base is part of a KSP, solubilities are greater than expected.

AgCN(s) Ag+(aq) + CN-(aq)

+ H2O(l)

HCN(aq) + OH-(aq)

This competingequilibrium causes the CN- to be lower than expected.

More AgCN will dissolve as a result.

This competingequilibrium causes the CN- to be lower than expected.

More AgCN will dissolve as a result.

Factors That Affect SolubilityFactors That Affect Solubility Solubility and pHSolubility and pH• Again we apply Le Châtelier’s principle:

– If the F- is removed, then the equilibrium shifts towards the decrease and CaF2 dissolves.

– F- can be removed by adding a strong acid:

– As pH decreases, [H+] increases and solubility increases.

• The effect of pH on solubility is dramatic.

CaF2(s) Ca2+(aq) + 2F-(aq)

F-(aq) + H+(aq) HF(aq)

Factors That Affect SolubilityFactors That Affect Solubility Solubility and pHSolubility and pH

Solubility EquilibriaSolubility Equilibria Solubility-Product Constant, Solubility-Product Constant, KKspsp

Consider

for which

Ksp is the solubility product. (BaSO4 is ignored because it is a pure solid so its concentration is constant.)

BaSO4(s) Ba2+(aq) + SO42-(aq)

]SO][Ba[ -24

2spK

Solubility EquilibriaSolubility Equilibria Solubility-Product Constant, Solubility-Product Constant, KKspsp

• In general: the solubility product is the molar concentration of ions raised to their stoichiometric powers.

• Solubility is the amount (grams) of substance that dissolves to form a saturated solution.

• Molar solubility is the number of moles of solute dissolving to form a liter of saturated solution.

Solubility EquilibriaSolubility Equilibria Solubility and Solubility and KKspsp

To convert solubility to Ksp

• solubility needs to be converted into molar solubility (via molar mass);

• molar solubility is converted into the molar concentration of ions at equilibrium (equilibrium calculation),

• Ksp is the product of equilibrium concentration of ions.

Solubility EquilibriaSolubility Equilibria Solubility and Solubility and KKspsp

Precipitation Reactions

• We will use “Q” and Ksp

• If Q is bigger than Ksp, what will happen?

– Smaller

• The Q in a solubilty problem is called the

Ion Product

A solution prepared by adding 750 mL of 4.00 x 10-

3 M Ce(NO3)3 to 300.0 mL of 2.00 x 10-2 M KIO3. Will Ce(IO3)3 precipitate? (Ksp = 1.9 x 10-10)

• Step one: Calculate the concentration of Ce+3 and IO3-1 before

a reaction occurs

[Ce+3]0 = (750.0 mL)(4.00x10-3 M) = 2.86 x 10-3 M(750. + 300.mL)

[IO3-1]0 = (300.0 mL)(2.00x10-3 M) = 5.71 x 10-3 M

(750. + 300.mL)

Q = (2.86 x 10-3 M) x (5.71 x 10-3 M)3 = 5.32 x 10-10

Q > Ksp therefore the precipitation reaction will occur

Precipitation and Separation of IonsPrecipitation and Separation of Ions

• At any instant in time, Q = [Ba2+][SO42-].

– If Q < Ksp, precipitation occurs until Q = Ksp.

– If Q = Ksp, equilibrium exists.

– If Q > Ksp, solid dissolves until Q = Ksp.

• Based on solubilities, ions can be selectively removed from solutions.

• Consider a mixture of Zn2+(aq) and Cu2+(aq). CuS (Ksp= 610-37) is less soluble than ZnS(Ksp=210-25), CuS will be removed from solution before ZnS.

BaSO4(s) Ba2+(aq) + SO42-(aq)

Precipitation and Separation of IonsPrecipitation and Separation of Ions• As H2S is added to the green solution, black CuS

forms in a colorless solution of Zn2+(aq).

• When more H2S is added, a second precipitate of white ZnS forms.

Selective Precipitation of IonsSelective Precipitation of Ions• Ions can be separated from each other based on their

salt solubilities.

• Example: if HCl is added to a solution containing Ag+ and Cu2+, the silver precipitates(Ksp for AgCl is 1.810-10) while the Cu2+ remains in solution, since CuCl2.

• Removal of one metal ion from a solution is called selective precipitation.

Qualitative Analysis for Metallic ElementsQualitative Analysis for Metallic Elements

• Qualitative analysis is designed to detect the presence of metal ions.

• Quantitative analysis is designed to determine how much metal ion is present.

Qualitative Analysis for Metallic ElementsQualitative Analysis for Metallic Elements• We can separate a complicated mixture of ions into

five groups:– Add 6 M HCl to precipitate insoluble chlorides (AgCl,

Hg2Cl2, and PbCl2).

– To the remaining mix of cations, add H2S in 0.2 M HCl to remove acid insoluble sulfides (e.g. CuS, Bi2S3, CdS, PbS, HgS, etc.).

– To the remaining mix, add (NH4)2S at pH 8 to remove base insoluble sulfides and hydroxides (e.g. Al(OH)3, Fe(OH)3, ZnS, NiS, CoS, etc.).

– To the remaining mixture add (NH4)2HPO4 to remove insoluble phosphates (Ba3(PO4)2, Ca3(PO4)2, MgNH4PO4).

– The final mixture contains alkali metal ions and NH4+.

applications of aqueous equilibria common ions when we dissolve acetic acid in water, the following...

Documents

x aq

x hx slide

x initial

oh x

terms of x

aq slide

ph change

ph buffer capacity