Solubility

Equilibria AP Chemistry

Ms. Grobsky

Introducing Solubility Equilibria

For the last couple of weeks, we have considered equilibria involving acids and bases

Homogeneous equilibria systems

All species have been in the same physical state

Now, we will consider the equilibria involved in the dissolution or precipitation of ionic compounds

Heterogeneous equilibria systems

Equilibrium of Dissolution and

Precipitation Reactions

A saturated solution is one in which the solution is in contact with undissolved solute

For example, when solid barium sulfate - an ionic compound that is a weak electrolyte - is added to water, the solid will dissolve and yield Ba2+ (aq) and SO4

2- (aq), readily establishing the equilibrium:

BaSO4 s ↔ Ba2+ aq + SO42−(aq)

As with any other equilibrium, the extent to which this dissolution reaction occurs is expressed by the magnitude of the equilibrium constant, Ksp

Ksp – The Solubility Product

Constant

Ksp represents the equilibrium associated with ionic

solids dissolving to form ions in aqueous solutions

sp stands for solubility product

This equilibrium constant indicates how soluble the

solid is in water

But wait…isn’t this just about using the solubility

rules?

The solubility rules only qualitatively predict whether

an ionic compound has a low or high solubility in

water

Solubility Equilibria

As it turns out, insoluble products according to the

solubility rules are actually a bit soluble after all

“Soluble” is often defined as 3 grams of solid

dissolving in 100 mL

Some ionic compounds do not dissolve completely

and they exist in a dynamic equilibrium state

Some of the solid dissolves and some remains in the

solid form

Ksp – The Solubility Constant

• No ions are

initially present

• As the solid dissolves, the concentration of ions increase until

equilibrium is established

• The solution is then saturated – no more solid forms

• The rate of solid dissolution is equal to solid formation (precipitation) – EQUILIBRIUM!

Let’s Take a Look At What’s

Happening at the Molecular

Level…

Solubility Animation

General form:

AaBb(s) aAb+ (aq) + bBa- (aq) Ksp = [Ab+]a [Ba-]b

Example: AgCl (s) Ag+ + Cl-

Ksp = [Ag+][Cl-] = 1.6 x 10-6 @ 25°C

Some notes on Ksp

Ksp is constant at a given temperature The greater the Ksp, the more soluble the solid is in H2O

The Solubility Product Expression

Solubility and Ksp

It is important to distinguish carefully between solubility and Ksp:

Solubility is the quantity of a substance in grams per liter of solution that dissolves to form a saturated solution

Molar solubility is the maximum number of moles of the solute that dissolves to form a liter of saturated solution

Solution is “saturated” and thus, equilibrium is established between solid and hydrated ions

Ksp(solubility product) is the equilibrium constant for the equilibrium between an ionic solid and its saturated solution

Unitless number and is a measure of how much of the solid dissolves to form a saturated solution

Difference is the units!

Solubility Product Calculations -

Determining Ksp Given Molar

Solubilities Plan:

Write the dissolution equation

Setup ICE table to calculate molar solubility

for ions

Substitute values into solubility product

constant expression and solve for Ksp

Practice!

# 1 on page

Solubility Product Calculations -

Determining Solubility Given Ksp

Plan:

Write the dissolution equation

Setup ICE table to calculate molar solubility for

ions

Substitute values into solubility product constant

expression and solve for “S”

“S” is molar solubility and acts the same way as “X”

in all ICE tables!

Practice!

#3 on page 143

Comparing Solubilities As discussed, Ksp values provide valuable

information about a compound’s solubility

Solubility comparisons can be made between

compounds, but you must take into account the

number of ions present in solution!

If the salts being compared have the same

number of total ions, direct comparison of Ksp is

okay!

If the salts have different number of ions, use

ICE table to solve for the molar solubility “S”!

Practice!

#4 on page 143

Factors That Affect

Solubility

The Effect of Temperature on

Solubility

0

100

200

300

0 20 40 60 80 100

SO2 (g)

KCl (s)

Glycine (s)

NaBr (s)

KNO3 (s)

Sucrose (s)

So

lub

ilit

y

(g/1

00

ml

wa

ter)

Temperature (oC)

The Common-Ion Effect and

Solubility Remember from the acid-base equilibria section, a common ion is defined as the ion in a mixture of ionic substances that is common to the formulas of at least two

So, how does the molar solubility of a solid change when the solid is not dissolved in pure water?

In other words, how does the molar solubility of the solid change if it is dissolved in a solution containing a second ionic substance containing a common ion?

To answer this question, you must remember LeChâtelier’s Principle!

The Common-Ion Effect As with other equilibria we’ve discussed, adding a ‘common’ ion

will result in a shift of a solubility equilibrium

CaF2 (s) ↔ Ca2+(aq) + 2F- (aq)

KSP = [Ca2+] [F-]2

Adding either Ca2+or F- (from NaF, say) to our system will result in a

shift away from the increase (i.e. driving it to the left)

CaF2 is formed and precipitation occurs

The solubility of CaF2 DECREASES!

In general, the solubility of one salt is reduced by the presence of another having a common ion

Ksp will not change; however, the concentrations of ions and mass of solid will!

How to Solve Problems Involving

Solubility and the Common Ion

Effect

Plan:

Write the dissolution equation

Setup ICE table to calculate molar solubility

for ions

Substitute values into solubility product

constant expression and solve for “S”

Remember, “S” is molar solubility!

Practice!

#5 on page 143

Explanation of Previous

Example using LeChâtelier’s

Principle Added silver ions and sulfate ions stress

the system and the system responds by

favoring the reverse reaction

Shifts to the left

More solid silver sulfate will be produced

Solubility is decreased

Less solid is dissolved

Effect of pH on Solubility In general, the solubility of a compound containing a basic

anion (i.e. conjugate base of a weak acid) increases as pH decreases (acidity increases)

Let’s look at an example:

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)

The Common Ion Effect and

Predicting Precipitation

Precipitation Conditions

So far, we have considered the dissolution

process of an ionic solid – the forward

reaction

Now, let’s consider the conditions under

which a solid will precipitate out of a

solution - the reverse reaction!

Determining Whether

Precipitation will Occur To answer this, calculate the reaction quotient, Q,

and compare to Ksp

Remember, Q is the same as the equilibrium-constant expression for a reaction, but instead of only equilibrium concentrations, you can use whatever concentrations are being considered

If Q > Ksp, the reaction will proceed to the left, towards the solid

Precipitation occurs until Q = Ksp

This is because solution is already supersaturated with ions

Supersaturated solution means it is an unstable solution in which more solute is dissolved than in a saturated solution

Determining Whether

Precipitation will Occur

If Q = Ksp, equilibrium exists

The solution is saturated which is the highest concentration the solution can have without precipitating

If Q < Ksp, the reaction will proceed to the right, towards the soluble ions

Solid dissolves until Q = Ksp

This is because you haven’t reached saturation yet

Saturation is a solution that contains the maximum amount of dissolved solute at a given temperature in the presence of undissolved solute

Predicting Whether a

Precipitate Forms

Plan:

Write dissolution equation for solid that is formed

from the two solutions

Calculate initial moles of ions that are a part of

dissolution equation using the volumes and

molarities given in the problem

Calculate the concentration of all ions just after the

solutions are mixed

Remember, volumes are additive!

Calculate Q and compare to Ksp

If Q > Ksp, precipitate will form

Practice!

#1 on page ________

Using Solubility for Selective

Precipitation

Supposed you have a mixture of metal ions in

solution

We can selectively remove one of them based

on the solubilities of their salts!

To do so, think about solubility rules!

Separation of ions in an aqueous solution can be

done using a reagent that forms a precipitate with

one or more (but not all) ions

Performing Selective

Precipitation of Ions

Select an anion that selectively precipitates only one metal ion OR…

Choose an anion that selectively precipitates both metal ions but one precipitates at a lower concentration Based on the common ion effect!

The salt with the lower Ksp will precipitate first

Example:

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

Solving Selective Precipitation

Problems

Plan:

Determine what ion is necessary for precipitation of BOTH species

Write dissolution equations and equilibrium expressions for the two solids that will form

Calculate the concentration of each cation (or anion) already present in solution

Calculate the minimum concentration of the added ion that is needed to precipitate the solid using the appropriate equilibrium expression

You will do this step twice

The salt that requires a lower concentration will precipitate first!

Practice!

# 4 on page ________