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York University CHEM 1001 3.0 Solubility - 1

Solubility Equilibria

Reading: from chapter 19 of Petrucci, Harwood and Herring(8th edition):

Required: Sections 19-1 through 19-8.

Examples: 19-1 through 19-8, 19-11, 19-13.

Problem Set:

Chapter 19 questions: 4, 5, 7a-c, 10, 11a,b, 14, 47, 53, 58,62, 64.

Additional problems from Chapter 19: 65, 71.


Solubility C Some liquids can dissolve in each other in any amount.

Example: ethanol and water.

C Many liquids can only dissolve in each other to only alimited degree. Example: benzene and water.

C Solids will dissolve in liquids, but there is always a limit.

C Solubility is the amount of a substance that can dissolve ina given amount of another substance (usually a liquid).

C Units should be M but other units are often used (such asgrams per 100 mL).

C Important for separations and quantitative analysis.


Non-Dissociating SoluteWe can treat dissolution as a chemical reaction.

Example: sucrose(s) W sucrose(aq)

At 25 °C, KC = 1.97 and KC = [sucrose(aq)].

Y The solubility of sucrose in water is 1.97 M.

C If [sucrose(aq)] = 1.97 M, the solution is saturated;additional sucrose can not dissolve.

C If [sucrose(aq)] < 1.97 M, the solution is unsaturated;additional sucrose can dissolve.

C If [sucrose(aq)] > 1.97 M, the solution is supersaturated;if solid sucrose is present, sucrose will precipitate.


The Solubility Product Constant

When salts dissolve in water, they usually dissociate. Theequilibrium constant for this process is called the solubilityproduct constant, Ksp.

Example: AgCl(s) W Ag+(aq) + Cl-(aq)

Ksp = [Ag+][Cl-] = 1.8×10-10 at 25 °C.

Q: How much AgCl can dissolve in water at 25 °C?

From stoichiometry, [Ag+] = [Cl-] = x.

At equilibrium Ksp = x2 = 1.8×10-10 Y x = 1.3×10-5 M

The solubility of AgCl is 1.3×10-5 M.


Calculating Solubility from Ksp

For Mg3(AsO4)2, Ksp = 2.0×10-20. Find the solubility.

Solution: Mg3(AsO4)2(s) W 3Mg2+(aq) + 2AsO43-(aq)

Ksp = [Mg2+]3[AsO43-]2

Let S = solubility (moles of Mg3(AsO4)2 dissolved per liter).

[Mg2+] = 3S [AsO43-] = 2S

Y Ksp = (3S)3(2S)2 = 27S3 × 4S2 = 108S5 = 2.0×10-20

Y S = 4.5×10-5 M

The solubility of Mg3(AsO4)2 is 4.5×10-5 M.


Calculating Ksp from Solubility

The solubility of MgNH4PO3 is 5.8×10-5 M. Determine Ksp.

Solution:

MgNH4PO3 W Mg2+(aq) + NH4+(aq) + PO3

3-(aq)

KSP = [Mg2+][NH4+][PO3

3-]

Let S = solubility. [Mg2+] = [NH4+] = [PO3

3-] = S

Y Ksp = S×S×S = S3 = (5.8×10-5)3 = 2.0×10-13

For MgNH4PO3, Ksp = 2.0×10-13.


The Common Ion Effect

Adding a common ion to a solution of a slightly soluble saltsuppresses the solubility of the salt.

Examples:

AgCl(s) W Ag+ + Cl- Ksp = 1.8×10-10

Add: NaCl(s) W Na+ + Cl-

C Cl- is a common ion. By LeChâtelier's principle, adding itshifts the AgCl equilibrium to the left.

Add: AgNO3(s) W Ag+ + NO3-

C The common ion, Ag+, suppresses the solubility of AgCl.


The Common Ion Effect - Example

For PbI2, Ksp = 7.1×10-9. By how much does 0.10 MPb(NO3)2 reduce the solubility of PbI2?

Let C = [Pb(NO3)2(aq)] = [Pb2+] if no PbI2 dissolves.

Reaction PbI2(s) W Pb2+ + 2I-

Initial - C 0 M

Change - S 2S M

Equil. - C+S 2S M

In water (C=0), Ksp = [Pb2+][I-]2 = 4S3 Y S = 1.2×10-3 M

In 0.10 M Pb(NO3)2, C=0.10 o S Y Ksp . 4CS 2

Y S = 1.3×10-4 M Nearly a factor of 10 smaller.


Other Influences on Solubility

C Salt effect - Ions influence each other electrostatically. Thiscauses activities to be lower than concentrations andincreases solubility.

C Ion pairs - At high concentrations, ions tend to "clump".This usually increases activities and decreases solubility.

C Simultaneous reactions can change concentrations of ions.Example:

CaCO3(s) W Ca2+ + CO32-

CO32- + H2O(l) W OH- + HCO3

-

Reaction of CO32- with H2O increases solubility of CaCO3.


Criteria for Precipitation

The ion product, Qsp, is used to decide if a salt may precipitate.

Example: PbI2(s) W Pb2+ + 2I-

Qsp = [Pb2+]init[I-]2

init Ksp = [Pb2+]eq[I-]2

eq

C If Qsp < Ksp, the solution is unsaturated. A precipitate willnot form and PbI2(s), if present, will dissolve.

C If Qsp = Ksp, the solution is saturated. It is in equilibriumwith PbI2(s).

C If Qsp > Ksp, the solution is supersaturated. A precipitatemay form. PbI2(s) will precipitate until solution becomessaturated. Precipitation may initially be slow.


Applying the Criteria for Precipitation

10.0 mL of 0.015 M Pb(NO3)2 are mixed with 20.0 mL of0.030 M KI. Will a precipitate form?

Solution: Pb(NO3)2, KI, and KNO3 are all very soluble.

PbI2(s) W Pb2+ + 2I- Ksp = 7.1×10-9

[Pb2+] = (0.015 M) (10.0 mL) /(10.0+20.0 mL) = 0.0050 M

[I-] = (0.030 M) (20.0 mL) /(10.0+20.0 mL) = 0.020 M

Qsp = [Pb2+]init[I-]2

init = (0.0050)(0.020)2 = 2.0×10-6

Qsp > Ksp Y A precipitate of PbI2 will form.


Completeness of Precipitation

The concentration of an ion in solution can be determined byprecipitating a salt of the ion and weighing the precipitate.

Example: SO42- can be determined by adding BaCl2:

BaCl2 6 Ba2+ + 2Cl-

BaSO4(s) W Ba2+ + SO42- Ksp = 1.1×10-10

It is impossible to precipitate all the SO42-. The analysis will

be valid if the fraction of SO42- left in solution is negligible.

If this is so, the precipitation is said to be complete.


Completeness of Precipitation - Example

60 mL of 0.040 M BaCl2(aq) is added to 140 mL of solutioncontaining about 2.2 mmol SO4

2-. What fraction of the SO42-

is left in solution?

Solution: BaSO4(s) W Ba2+ + SO42- Ksp = 1.1×10-10

Is Ba2+ in excess? nBa = (60 mL)(0.040 M) = 2.4 mmol

Y 2.2 mmol of BaSO4(s), 0.2 mmol Ba2+(aq)

[Ba2+]f = (0.2 mmol) / (60+140 mL) = 1×10-3 M

[SO42-]f = Ksp/[Ba2+]f = (1.1×10-10)/(1×10-3) = 1.1×10-7 M

amount in solution = (1.1×10-7 M) (200 ml) = 2.2×10-5 mmol

Y 0.001% is left in solution. Precipitation is complete.


Fractional Precipitation

Differing solubilities can be used to precipitate differentions one at a time. Can be used for

C separation and/or purification

C precipitation titrations

Concept: To separate two ions, find a counterion that formssalts of differing solubility. Gradually add the counterion tothe solution.

Example: Separate Br- and CrO42- using Ag+ (counterion).

Ag2CrO4(s) W 2Ag+ + CrO42- Ksp = 1.1×10-12

AgBr(s) W Ag+ + Br- Ksp = 5.0×10-13


Fractional Precipitation - Example

Slowly add AgNO3 to 0.010 M [Br-] and 0.010 M [CrO42-].

Which precipitates first?

For AgBr:

Ksp = [Ag+][Br-]

5.0×10-13 = [Ag+](0.010)

[Ag+] = 5.0×10-11 M

For Ag2CrO4:

Ksp = [Ag+]2[CrO42-] Y [Ag+] = (Ksp/[CrO4

2-])½ = 1.0×10-5 M

Y AgBr precipitates first.


Fractional Precipitation - continued

How much Br- is left in solution when Ag2CrO4 starts toprecipitate?

[Ag+] = 1.0×10-5 M (from previous slide)

Using Ksp for AgBr: [Br-] = Ksp / [Ag+] = 5.0×10-8 M

Y 0.0005% of Br- is left in solution

Conclusion: Precipitation of AgBr is complete beforeprecipitation of Ag2CrO4 begins.

Also, red-brown Ag2CrO4 can be used as an indicator intitrating Cl- or Br- with Ag+.


Influence of pH on Solubility

If a salt contains the anion of a weak acid or the cation ofa weak base, then its solubility will depend on pH.

Example: CaCO3(s) W Ca2+ + CO32- Ksp = 2.8×10-9

Hydrolysis: CO32- + H2O(l) W OH- + HCO3

- pKb = 3.67

Consequences:

C Unless pH is high,solubility is greater thanimplied by Ksp.

C Solubility increases aspH decreases.


Change of Solubility with pH - Example

Given Ksp(CaF2) = 5.3×10-9 and pKa(HF) = 3.18, how doesthe solubility of CaF2 vary as pH changes from 5.0 to 1.0?Assume that the pH is controlled by using buffers.

Solution: CaF2(s) W Ca2+ + 2F-

HF + H2O(l) W H3O+ + F-

Significant species: Ca2+, F-, HF (H3O+ is fixed)

Electroneutrality: Not helpful in buffer solution.

Material balance: 2[Ca2+] = [F-] + [HF]

Equilibria: Ksp = [Ca2+] [F-]2

Ka [HF] = [H3O+] [F-]

3 equations, 3 unknowns


Change of Solubility with pH - continued

From equilibria: [F-] = (Ksp/[Ca2+])½

[HF] = [H3O+][F-]/Ka = ([H3O

+]/Ka) (Ksp/[Ca2+])½

Substitute into material balance:

2[Ca2+] = (Ksp/[Ca2+])½ (1 + [H3O+]/Ka)

Solve for [Ca2+]:

[Ca2+] = (Ksp/4)a (1 + [H3O+]/Ka)

b

At pH = 5.0, [H3O+] n Ka Y [Ca2+] = (Ksp/4)a = 1.1×10-3 M

At pH = 3.18, [H3O+] = Ka Y [Ca2+] = Ksp

a = 1.7×10-3 M

At pH = 1.0, [H3O+] o Ka Y [Ca2+] = (Ksp[H3O

+]2/(4Ka2))a

Y [Ca2+] = 3.1×10-2 M


Solubility at Low pH - Another Approach

If the pH is low enough that [HF] o [F-], the dissolutionreaction is

CaF2(s) + 2H3O+ W Ca2+ + 2HF + 2H2O(l) K = ?

This is equivalent to:

CaF2(s) W Ca2+ + 2F- K = Ksp

2H3O+ + 2F- W 2HF + 2H2O(l) K = (1/Ka)

2

CaF2(s) + 2H3O+ W Ca2+ + 2HF + 2H2O(l) K = Ksp/(Ka

2)

Let S / [Ca2+] Y K[H3O+]2 = [Ca2+][HF]2 = S(2S)2 = 4S3

Y S = (K[H3O+]2/4)a = (Ksp[H3O

+]2/(4Ka2))a

Same as result on previous slide for pH n pKa.


Complex Ions

A complex ion consists of:

C a central positive ion (usually a transition metal)

C a number of ligands

The ligands are anions or neutral molecules.

The complex ion may be positively or negatively charged.

Examples:

Ag+ + 2NH3 W [Ag(NH3)2]+

Cu+ + 4CN- W [Cu(CN)4]3-

Fe3+ + 2SCN- W [Fe(SCN)2]+


The Formation Constant

The equilibrium constant for the reaction producing thecomplex ion from its central ion and ligands is called theformation constant, Kf.

Co3+ + 6NH3 W [Co(NH3)6]3+

Pb2+ + 3Cl- W [PbCl3]-

Fe3+ + 6CN- W [Fe(CN)6]3-

Formation constants can be very large.


The Formation Constant - Example

A solution is prepared containing 0.010 M NaCN,0.010 M HCN, and 1.0×10-4 M AgNO3. Find 'free' [Ag+].

Solution: Ag+ + 2CN- W [Ag(CN)2]- Kf = 5.6×1018

Find the relative amounts of Ag+ and [Ag(CN)2]-:

Y

[CN-] = 0.010 M Y [Ag+] = 1.8×10-15[[Ag(CN)2]-]

Y [[Ag(CN)2]-] = 1.0×10-4 M, [Ag+] = 1.8×10-19 M


Complex Ions and SolubilityComplex ions can substantially increase solubility.

Example: Find the solubility of AgCl in 0.10 M HCl(aq).

AgCl(s) W Ag+ + Cl- Ksp = 1.8×10-10

[Ag+] = Ksp / [Cl-] = (1.8×10-10)/(0.10) = 1.8×10-9 M

But: Ag+ + 2Cl- W AgCl2- Kf = 1.1×105

[AgCl2-] = Kf[Ag+][Cl-]2 = (1.1×105)(1.8×10-9)(0.10)2

Y [AgCl2-] = 2.0×10-6 M 1000 times as large as [Ag+]

Y solubility = [Ag+] + [AgCl2-] = 2.0×10-6 M

Could also be done by the method of example 19-13.


Solubility Equilibria - Summary

C Solids dissolve in liquids to a finite extent.

C The solubility of salts in water is governed by thesolubility product constant.

C Solubility is reduced by the presence of common ions.

C Solubility can be enhanced by hydrolysis, so it maydepend on pH.

C Solubility can be enhanced by the formation of complexions, so it may depend on the presence of ligands.

C Precipitation of an ion is "complete" if only a negligibleamount remains in solution.

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