chapter 17 part 2 solubility equilibria - miami dade college · factors affecting solubility -...

Solubility Equilibria

Chapter 17 – Part 2

Solubility Equilbria

• Ksp

• Calculating Solubility

• Factors that affect solubility

• Precipitation

• Qualitative Analysis


Quantitative predictions about how much of a given ionic compound will dissolve in water is possible with the solubility product constant, Ksp.

AgCl(s) ⇌ Ag+(aq) + Cl–(aq)

Ksp = [Ag+][Cl–]

Compound Dissolution Equilibrium Ksp

Aluminum hydroxide Al(OH)3(s) ⇌ Al3+(aq) + 3OH–(aq) 1.8 x 10–33

Calcium fluoride CaF2(s) ⇌ Ca2+(aq) + 2F–(aq) 4.0 x 10–11

Silver bromide AgBr(s) ⇌ Ag+(aq) + Br–(aq) 7.7 x 10–13

Silver chloride AgCl(s) ⇌ Ag+(aq) + Cl–(aq) 1.6 x 10–6

Zinc sulfide ZnS(s) ⇌ Zn2+(aq) + S2–(aq) 3.0 x 10–23


Molar solubility is the number of moles of solute in 1 L of a saturated solution (mol/L)

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

To calculate a compound’s molar solubility:

1) Construct an equilibrium table.

2) Fill in what is known.

3) Determine the unknowns.

Practice Problem - Solubility Equilibria

The Ksp of silver bromide is 7.7 x 10–13. Calculate the molar solubility.

Practice Problem - Solubility Equilibria

The Ksp for CaF2 is 3.9 x 10-11 at 25oC. Assuming that CaF2 dissociates completely upon dissolving, calculate the solubility of CaF2.

Factors Affecting Solubility

Several factors exist that affect the solubility of ionic compounds:

1. The Common Ion Effect

2. pH

3. The formation of complex ions

Factors Affecting Solubility - Common Ion effect

The common ion effect:

The presence of either Ag+ (aq) or Cl-(aq) in a solutions reduces the solubility of AgCl, shifting the equilibrium to the left.

AgCl(s) ⇌ Ag+(aq) + Cl–(aq)

Practice Problem

The common ion effect:

Calculate the molar solubility of calcium fluoride that is (a) 0.010 M in Ca(NO3)2

(b) 0.010 M in NaF

Factors Affecting Solubility - pH

The pH of a solution affects the solubility of any substance whose anion is basic. Consider the following solubility equilibrium:

Ksp = [Mg2+][OH–]2 = 1.2 x 10–11

(s)(2s)2 = 4s3 = 1.2 x 10–11 s = 1.4 x 10–4 M

At equilibrium:

[OH–] = 2(1.4 x 10–4 M ) = 2.8 x 10–4 MpOH = –log(2.8 x 10–4) = 3.55

pH = 14.00 – 3.55 = 10.45

Mg(OH)2(s) ⇌ Mg2+ (aq) + 2OH–(aq) Ksp = 1.2 x 10-11


If the pH of the medium were higher than 10.45 [OH–] would be higher and the solubility of Mg(OH)2 would decrease.In a solution with a pH of less than 10.45, the solubility of Mg(OH)2 increases.

Mg(OH)2(s) ⇌ Mg2+ (aq) + 2OH–(aq)

Mg(OH)2(s) + 2H+ (aq) ⇌ Mg2+ (aq) + 2H2O(l)

2H+(aq) + 2OH–(aq) → 2H2O(l)

Overall:


As the concentration of F – decreases, the concentration of Ba2+ must increase so satisfy the equality:

Ksp = [Ba2+][F–]2

The solubilities of salts containing anions that do not hydrolyze are unaffected by pH:

Cl– , Br – , NO3–

BaF2(s) ⇌ Ba2+ (aq) + 2F–(aq)

BaF2(s) + 2H+ (aq) ⇌ Ba2+ (aq) + 2HF(aq)

2H+(aq) + 2F–(aq) → 2HF(aq)

Overall:

Practice Problems

Which of these substances are more soluble in acidic solution than in a basic solution?

(a) Ni(OH)2 (s)

(b) CaCO3 (s)

(c) BaF2 (s)

(d) AgCl (s)

Factors Affecting Solubility - Complex Ion Formation

A characteristic property of metal ions is their ability to act as Lewis acids toward water molecules, which acts as a Lewis base.Lewis bases other than water can also interact with metal ions, particularly transition metal-ions.

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

A solution of CoCl2 is pink because of the presence of Co(H2O)6

2+ ions.


When HCl is added to a CoCl2 solution, there is a color change from pink to blue.

Co2+(aq) + 4Cl– (aq) ⇌ CoCl42–(aq)

This solution is blue because of the presence of presence of CoCl42– ions.


Cu(OH)2(s) + 4NH3(aq) ⇌ Cu(NH3)42+ (aq) + 2OH–(aq)

The formation of the Cu(NH3)42+ ion can be expressed as

Cu2+ (aq) + 2OH–(aq) → Cu(OH)2(s)

Cu2+(aq) + 4NH3(aq) ⇌ Cu(NH3)42+ (aq)

Formation Constants


For the dissociation of an ionic solid in water, the following conditions may exist:

1) The solution is unsaturated

2) The solution is saturated

3) The solution is supersaturated

The following relationships are useful in making predictions on when a precipitate might form.

Q < Ksp; no precipitate forms (solid dissolves)

Q = Ksp; no precipitate forms (equilibrium exists- saturated solutions)

Q > Ksp; a precipitate forms

Practice Problems - Solubility Equilibria

Predict whether a precipitate will form the following combination:

650 mL of 0.0080 M K2SO4 with 175 mL of 0.15 M AgNO3

Separation of Ions Using Differences in SolubilitySome compounds can be separated based on fractional precipitation.

Fractional precipitation is the separation of mixture based upon the components’ solubilities.

Compound Ksp

AgCl 1.6 x 10–10

AgBr 7.7 x 10–13

AgI 8.3 x 10–17

Qualitative analysis involves the principle of

selective precipitation and can be used to identify the types of ions present in a

solution.

Objectives

• Given either Ksp, molar solubility, or solubility for a substances, be able to calculate the other two quantities.

• Calculate the molar solubility in the presence of a common ion.

• Predict the effect of pH on solubility.

• Predict whether a precipitate will form when solutions are mixed by comparing Q and Ksp.

• Explain the effect of complex-ion formation on solubility.

chapter 17 part 2 solubility equilibria - miami dade college · factors affecting solubility -...

Documents