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AP Chemistry
Solubility and Precipitation
Name: ___________
For double replacement reactions to occur, there needs to be a driving force. In other words, according
to the laws of thermodynamics, the products must be more stable than the reactants.
1. formation of
2. formation of
3. formation of
Precipitation- the process in which ions leave a solution and regenerate an ionic solid. LEARN YOUR
SOLUBILITY RULES!
Always Soluble
·
Alkali metals ions (Li+, Na+,
Generally Soluble
·
K+, Rb+, Cs+), NH4+, NO3-,
ClO3-, ClO4-, C2H3O2-
·
Cl-, Br-, I- Soluble except:
·
O2-, OH- insoluble except:
Ag+, Pb2+, Hg22+ (AP/H).
with alkali metals, and NH4+,
F- Soluble except: Ca2+,
Ca2+, Sr2+, Ba2+ (CBS)
Sr2+, Ba2+, Pb2+, Mg2+
somewhat soluble.
(CBS-PM).
·
Generally Insoluble
·
CO32-, PO43-, S2-, SO32-,
SO42- Soluble except: Ca2+,
C2O42-, CrO42- insoluble
Sr2+, Ba2+, Pb2+,
except: with alkali metals or
(CBS/PBS).
NH4+.
Examples:
1. A solution of silver nitrate is added to a solution of calcium chloride
2. Solid calcium carbonate is reacted with sulfuric acid.
3. Potassium sulfide is reacted with hydrobromic acid.
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Net Ionic Equations
Complete ionic equation – an equation that shows all soluble ionic substances as ions
A solution of copper (II) sulfate is added to a solution of potassium hydroxide.
Complete Ionic Equation:
Notice that the total charge on each side of the equation is zero.
Also, notice only the copper (II) ions and the hydroxide ions undergo a change in the chemical reaction.
The sodium ions and the nitrate ions are unchanged in the reaction.
Spectator Ions- ions that do not take part in a chemical reaction
Net Ionic Equation = Complete Ionic Equation – Spectator Ions:
A net ionic equation includes only those compounds and ions that undergo a chemical change in a reaction.
Dissolution- the process in which an ionic solid dissolves in a polar liquid is called dissolution.
Solid sodium chloride is placed in water
Note: When writing dissolution equations, be certain to balance the reactants and products according to
number of atoms and charge. Both sides must be electrically neutral.
Remember oxidation numbers.
CaCl2 (s) à
(forms a total of __ ions)
Bi2S3 (s) à
(forms a total of ___ ions)
Saturated Solution- a solution that contains as much dissolved solute as it can under a given set of
conditions. No more solid material can dissolve in the given amount of solution.
Solubility Equilibrium- the rate at which an ionic solid dissolves is equal to the rate at which the aqueous
ions in solution precipitate.
AgCl(s) « Ag+ (aq) + Cl- (aq)
Strong electrolytes- dissolve completely in water- will have large Ksp values
Weak electrolytes- do not dissolve completely, instead, form an equilibrium between the solution phase
(as ions) and the solid phase.
CaSO4 (s) « Ca2+ + SO42-
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The Solubility Product
The solubility product constant (Ksp) is a special type of equilibrium constant that measures the
concentration of ionic compounds in water at specific temperatures. A compound with a small Ksp value is
only slightly soluble in water.
For example,
CaSO4 (Ksp = 9.1 X 10-6) at 25oC is only very slightly soluble in water
BaSO4 (Ksp = 9.1 X 10-10) at 25oC is even less soluble
In fact, solubility of BaSO4 is so low that we could classify it as nearly insoluble in water.
Development of Ksp
AB(s) « A+ (aq) + B- (aq)
Keq = [A+ (aq)][ B- (aq)]
[AB (s)]
Since AB is present as a solid, it is absorbed into the equilibrium expression:
Keq[AB(s)] = Ksp = [A+ (aq)][ B- (aq)]
Question:
Write the solubility equation and Ksp expression for BaSO4.
Write the solubility equation and Ksp expression for PbI2.
Finding and Using Solubility Products
The value of Ksp is calculated from concentrations at equilibrium. No units are assigned to the value of
Ksp.
Silver chloride (s) is added to pure water and allowed to come to equilibrium
[Ag+] = 1.3 X 10-5 M
[Cl-] = 1.3 X 10-5 M
Ksp = ?
Small values of Ksp à
Large values of Ksp à
Questions:
At 25oC, the concentration of lead ions in a saturated solution of lead (II) fluoride is 1.9 X 10-3 M. What
is the value of Ksp for lead (II) fluoride?
A sample of cadmium (II) hydroxide (s) is added to pure water and allowed to come to equilibrium at
25oC. The concentration of cadmium ions is 1.7 X 10-5 M at equilibrium. What is the value of Ksp for
cadmium (II) hydroxide?
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A sample of Ce(OH)3 (s) is added to pure water and allowed to come to equilibrium at 25oC. The
concentration of Ce+3 (aq) is 5.2 X 10-6 M at equilibrium. What is the value of Ksp for Ce(OH)3?
Finding Concentrations
Use algebraic techniques to solve for the concentration of ions if given Ksp values
It is also possible to use Ksp to predict the equilibrium concentration of ions in a saturated solution.
Questions:
A sample of calcium sulfate (s) is placed in pure water at 25oC and stirred. What will be the
concentrations of calcium ions and sulfate ions in the solution at equilibrium?
( Ksp = 2.4 X 10-5 @25oC)
What will be the equilibrium concentrations of the dissolved ions in a saturated solution of magnesium
hydroxide at 25oC? (Ksp = 8.9 X 10-12)
What are the equilibrium concentrations of calcium ions and fluoride ions in a saturated solution of CaF2 ?
(Ksp = 3.9 X 10-11)
Predicting the Formation of Precipitates
Saturated solution = a solution that contains as much dissolved solute as it can under a given set of
conditions.
If a solution has more ions dissolved than it can hold, a precipitate will form.
Q (ion product) is used to determine if a precipitate will form. The ion product can be compared with
the solubility product (Ksp) to determine if an aqueous solution of ions is supersaturated and will form a
precipitate.
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Q > Ksp à solution is _____________________________ lowering the concentration of the dissolved
ions to their equilibrium values.
Q < Ksp à solution is _____________________________, more solid can be added to solution and be
dissolved.
Q = Ksp à solution is _____________________________
Question:
Suppose 0.01 mol of lead (II) chloride (s) is dissolved in 150 ml of hot water, and the solution is cooled
slowly to 25oC. The volume of the final solution is 150 ml. What is the relationship between temperature
and solubility in the formation of a precipitate? Is the solution supersaturated? (Ksp = 1.6 X 10-5).
Question:
Will the following double replacement reaction occur?
A 1.0 M solution of copper (II) nitrate is placed in a 1.0 M solution of sodium hydroxide.
Even if a precipitation reaction can proceed, it still does not always proceed. Whether or not a
precipitate actually forms in a double replacement reaction depends upon the concentrations of the
dissolved ions after the solutions are mixed. If the final solution is too dilute, no precipitate can form
and the double replacement does not occur. Only if the ion product exceeds the solubility product will a
precipitate form. And precipitation will continue until the ion concentrations decrease to the equilibrium
levels.
Question:
Will a precipitate form if 20 ml of 0.010 M silver nitrate and 20 ml of 3.0 X 10-4 M potassium bromide are
mixed?
(Ksp silver bromide = 5.0 X 10-13 at 25oC)
Common Ion Effect
Consider the following equilibrium
CaSO4 (s) « Ca2+ + SO42Ksp = 9.1 X10-6 @ 25oC
According to Le Chatelier‛s Principle, what happens to the position of the equilibrium if magnesium sulfate
is added?
What is common to these two substances?
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Addition of magnesium sulfate (a strong electrolyte) shifts the equilibrium to the right, decreasing the
solubility of the calcium sulfate. This phenomenon is called the common ion effect.
Common ion equilibrium problems are solved following the same pattern as other equilibrium problems.
However, the initial concentration of the common ion (the strong electrolyte) must be considered.
Common Ion and Ksp
Silver acetate, AgC2H3O2, has Ksp = 4.0 X 10-3. Would a precipitate of silver acetate form if 18.0 ml of
0.10 M AgNO3 were added to 40.0 ml of 0.024 M NaC2H3O2?