Post on 01-Jan-2020
In saturated solutions dynamic equilibrium exists between undissolved solids and ionic species in solutions
Solids continue to dissolve and ion-pairs continue to form solids.
The rate of dissolution process is equal to the rate of precipitation.
Barium carbonate, BaCO3, is rather insoluble in water. Careful experiments show that if solid BaCO3 is placed in pure water and vigorously stirred, a small amount of the BaCO3 dissolves in the water.
For many important ionic compounds that are only
slightly soluble in water, equations are written to
represent the equilibrium between the compound and
the ions present in a saturated aqueous solution.
Solubility is an important feature of chemistry,
medicine and biology. Solubility is determined by the
equilibrium between a solid and a solution of the solid
in a liquid.
BaCO3(s) Ba2+ + CO32-
Precipitate
Solution under precipitate
for BaCO3 at 25 С Кsp = 5∙10-9
Consider the dissolution of barium carbonate in water
solid
eqBaCOc
COcBaсK
)(
)()(
3
2
3
2
.
)()( 2
3
2 COcBaсK sp
but …. reactants are in solid form*, so
* Remember, solids are not in equilibrium expressions!
Applying the Low of mass action
Equilibrium between: dissolved
and undissolved
The solubility product constant, Ksp, is the product of the concentrations of the ions involved in a solubility equilibrium, each raised to a power equal to the stoichiometric coefficient of that ion in the chemical equation for the equilibrium.
Ex. Ksp = с(Ba2+)с(CO32-) = 5∙10-9
Ksp has a fixed value for a given system at a given temperature. The solubility product depends only on the ions in solution and is constant for a set of particular ions.
Ksp = с(Mep+)nс(Ang-)m
In general, the dissolution of a slightly soluble compound and its solubility product expression are represented as :
MenAnm nMep+ + mAng-
If quantities of reactants and products present at equilibrium are expressed in terms of activities the calculations are considered to be more precise.
Ex. Ksp = а(Ba2+)а(CO3
2-)
Solubility is a measure of the extent to which a compound will dissolve in a given solvent. In most cases, we are interested in the solubility of a compound in water. Molar Solubility (S) - the molarity of a solute in a saturated aqueous solution.
Remember • Ksp is not solubility. It is Ksp, it is used to calculate molar solubility. • It is related but not the same • If you are asked about solubility, be sure to calculate it. • S - the molar solubility!!!
The greater the solubility, the smaller the amount of precipitation.
The relationship between solubility and Ksp
22
4
2 )()( SSSSOcBacK sp
spKS
BaSO4 Ba2+ + SO42-
S S
only in case when salts produce the ions with the same number of charge
Molar Solubility (S) - the molarity of a solute in a saturated aqueous solution
KtnAnm nKt + mAn
In general, when salts produce the ions with different number of charges :
nS mS
mn
sp AncKtcK )()(
nSKtc )( mSAnc )(
mn
sp mSnSK )()(
nmmn
sp SmnK )(
mn
spnm
mn
KS nm
mn
sp
mn
KS
Calculating solubility equilibrium fall into two categories:
determining a value of Ksp from experimental data
calculating equilibrium concentrations, solubilities when Ksp is known.
3321 421
spsp KKS
5523 10823
spsp KKS
4431 2731
spsp KKS
Ca3(PO4)2 3Ca2+ + 2PO43-
BiCl3 Bi3+ + 3Cl-
EXAMPLES:
PbCl2 Pb2+ + 2Cl-
nmmn
sp
mn
KS
M+
MX
M+ M+
X-
X-
X-
AFTER PRECIPITATION (Ksp = c(M+)c(X-))
BEFORE PRECIPITATION (Qsp = c(M+)c(X-))
MX M+ + X-
X-
M+
Maximum Concentrations BEFORE Precipitation: c(M+) and c(X–) are the maximum concentrations that can “live with each other” (“peaceful coexistence”) without “attacking each other” and creating a ppt. Maximum Concentrations AFTER Precipitation: c(M+) and c(X–) are the maximum concentrations that can “live with each other” after a precipitate of MX(s) is formed, no matter how much ppt is present: Saturated solution.
Qsp is the ion product reaction quotient and is based on initial conditions of the reaction.
Qsp can then be compared to Ksp.
To predict if a precipitation occurs:
- Precipitation should occur if Qsp > Ksp.
- Precipitation cannot occur if Qsp < Ksp.
- A solution is just saturated if Qsp = Ksp.
If the ion concentration product is greater than the solubility product, precipitation will take place!!!
The concentration of calcium ion in blood plasma is 0.0025 M. If the concentration of oxalate ion is 1.0x10-7 M, do you expect calcium oxalate to precipitate? Ksp = 2.3x10
-9.
Approach: First decide whether any of the ions present will combine to form a slightly soluble salt. Then look up the Ksp for the substance and from the initial concentrations after mixing, calculate Qsp and compare it to Ksp.
PROBLEM
Three steps:
1. Determine the initial concentrations of ions.
2. Evaluate the reaction quotient Qip.
3. Compare Qsp with Ksp.
Qsp = c(Ca2+)(C2O42-) = (0.0025)(1.0 x 10-7)
Qsp = 2.5∙10-10
Qsp < Ksp therefore, no precipitate will form!!!
Comparing Solubilities A salt’s Ksp value gives us information about its solubility. However, we must be careful in using Ksp values to predict solubilities of a given group of salts. There are two possible cases:
1.The salts being compared produce the same number of ions. For example, consider
AgCl Ksp(AgCl) = 1,7∙10-10 AgBr Ksp(AgBr) = 7,7∙10-13 AgI Ksp(AgI) = 1,5∙10-16
each of these solids dissolve to produce two ions. In this case we can compare the solubilities for these solids by comparing the Ksp values
AgCl > AgBr > AgI
2.The salts being compared produce different numbers of ions. For example, consider CuS Ksp(CuS) = 8,5∙10-43
Ag2S Ksp(Ag2S) = 1,6∙10-49
Bi2S3 Ksp(Bi2S3) = 1,1∙10
-73
These salts produce different numbers of ions when they dissolve. The Ksp values cannot be compared directly to determine solubilities. In this case we mast compare the solubility values (S)
Bi2S3 > Ag2S > CuS
Most Soluble = highest molar solubility!!!! Do not use Ksp!!
2SSSKsp
32 4)2( SSSKsp
532 108)3()2( SSSKsp
2345 102,9105,8spKS
173
49
3 104,34
106,1
4
spKS
155
73
5 100,1108
101,1
108
spKS
Factors that Affect Solubility
2. Addition of a Common ion (Common ion effect)
1. Temperature
3. Salt effect
(Solubility generally increases with temperature)
4. Changes in pH
2. Addition of a Common ion (Common ion effect)
The solubility of a slightly soluble ionic compound is lowered when a second solute that furnishes a common ion is added to the solution. Solubility decreases. Equilibrium shifts to left.
PROBLEM
What is the solubility BaSO4 in 0,01 M Na2SO4 and compare it with the solubility in pure water. Ksp(BaSO4) = 1∙10-10
Solution:
BaSO4 Ba2+ + SO42-
The equation for the dissociation of barium sulfate is
So we must find the concentrations of each
ion and then solve for Ksp.
0 +S
lmolKS sp /1010 510
а) in pure water
22
4
2 )()( SSSSOcBacK sp
Substitute into solubility product expression and solve for S, giving the ion concentrations
BaSO4 Ba2+ + SO42-
Barium sulfate solubility product expression is
initial
equilibrium 0 +S
0 0
H2O
б) in a solution of Na2SO4
Note the presence of 0.01 M Na2SO4. A solution of Na2SO4 contains Na+ ions and SO4
2- ions. The SO42- is
a common ion as it appears in the solubility equilibrium for BaSO4. Now setup the equilibrium table and remember that the SO4
2- ion is present initially:
0 +S
BaSO4 Ba2+ + SO42-
initial
equilibrium 0,01 +S
0,01 0
Substituting in the Ksp expression,
)01,0()()( 2
4
2 SSSOcBacK sp
Na2SO4
We might become concerned here as it appears that the problems can only be solved using the quadratic equation. However, it is possible to consider the chemical implications of the mathematical relationship. In pure water S is 1 ∙ 10-5 M, a very small number, certainly smaller than 0.0100 M. So we would expect that 0.01 + S is approximately the same as 0.01.
)01,0()()( 2
4
2 SSSOcBacK sp
01,0)()( 2
4
2 SSOcBacK sp
lmolс
KS
SONa
sp/10
01,0
10 810
42
BaSO4 is much less soluble in the presence of Na2SO4. Adding more SO42- ions
shifts the equilibrium back to the reactants, which is solid BaSO4.
Conclusion:
PROBLEM
Solve the previous problem using activities instead of concentrations.
Solution:
)()()(
)()()()()()(
2
4
2
42
2
4
22
4
22
4
2
SOfBafSONacS
SOfBafSOcBacSOaBaaKsp
Let’s determine ionic strength. It depends only on Na2SO4, because solubility of BaSO4 is very small.
Na2SO4 2Na+ + SO42-
03.0)2101102()( 2222
2122
21
24
zczcISONa
According to Activity Coefficient table: 54.024
2 SOBaff
)()()(
)()()()()()(
2
4
2
42
2
4
22
4
22
4
2
SOfBafSONacS
SOfBafSOcBacSOaBaaKsp
lmolSOfBafс
KS
SONa
sp/104.3
54.010
10
)()(
8
22
10
2
4
2
42
3. Salt effect
Ionic interactions are important even when an ion is not apparently participating in the equilibrium. Uncommon ion tend to increase solubility. For example, solubility of PbSO4 increase in the presence of KNO3.
The salt effect can be explained by applying the following formula to calculate solubility product
Ksp = а(Pb2+)а(SO42-) = с(Pb2+)с(SO4
2-)f(Pb2+)f(SO42-)
)()()()(
2
4
2
2
4
2
SOfPbf
KSOcPbc
sp
In the presence of uncommon ion the activity coefficients f(Pb2+) and f(SO4
2-) usually decrease due to increasing of the ionic strength of the solution. As a result the solubility of PbSO4 increases also.
PROBLEM
What is the solubility BaSO4 in 0,1 M KNO3.
Solution:
)()( 2
4
22 SOfBafSK sp lmolf
KS
sp/1003.3
33.0
10 510
Ksp = а(Ba2+)а(SO42-) = с(Ba2+)с(SO4
2-)f(Ba2+)f(SO42-)
с(Ba2+) = с(SO4
2-) = S
KNO3 K+ + NO3-
1.0)110110()( 2121
2122
21
3
zczcINOK
According to Activity Coefficient table: 33.024
2 SOBaff
Compare the solubility of BaSO4
- - - + + +
+ - - +
Na2SO4 H2O KNO3
- - - + + +
+ - - +
- - - + + +
+ - - +
BaSO4 BaSO4 BaSO4
S = 10-5 mol/l
S = 3,4∙10-8 mol/l S = 3,03∙10-5 mol/l
S S
4. Effect of pH on Solubility
The solubility of an ionic solute may be greatly affected by pH if an acid-base reaction also occurs as the solute dissolves.
In other words, some salts will not dissolve well in pure water, but will dissolve in an acid or a base.
If the anion (A-) of the precipitate is that of a weak acid, the precipitate will dissolve more when in a strong acid (H+ ions will form HA with A-)
However, if the anion of the precipitate is that of a strong acid, adding a strong acid will have no effect on the precipitate dissolving more.
Cl- is an anion of a strong acid.
It will not react with H+ ions, so there is no effect.
How would the addition of HCl affect the solubility of MnS?
S2- is an anion of a weak acid
Thus, it will react with H+ ions to form H2S,
This will shift the equilibrium to make more MnS dissolve!
How would the addition of HCl affect the solubility of PbCl2?
Consider the following equilibrium:
Mg(OH)2 ⇌ Mg2+ + 2 OH-;
Increasing the pH means increasing c(OH-) and equilibrium will shift to the left, causing some of Mg(OH)2 to precipitate out.
If the pH is lowered, c(OH-) decreases and equilibrium shifts to the right, causing solid Mg(OH)2 to dissolve.
The solubility of compounds of the type M(OH)n decreases as pH is increased, and increases as pH is decreased.
What pH is required to just begin precipitating of Mg(OH)2 from 0,01 M solution of Mg2+. At what conditions the precipitation is complete?
PROBLEM TO COUNT
Solution:
1022 106)()( OHcMgcK sp
Let’s calculate с(ОН-) that is required for Mg(OH)2 precipitation
lmolMgc
KOHc
sp/105.2
101
106
)()( 4
2
10
2
6.3105.2lg 4pOH
4.106.31414 pOHpH
If 99,99% or more of a particular ion has precipitated leaving less than 0,1% (10-5 – 10-6 mol/l) of the ion in
solution – precipitation is complete!
lmolMgc
KOHc
sp/105.2
101
106
)()( 2
6
10
2
58.1105.2lg 2pOH
42.12pH
if с(Mg2+) = 10-6 mol/l
At рН = 10.4 magnesium hydroxide begins to precipitate. At рН > 12.42, the precipitation will be complete.
Conclusion:
FRACTIONAL (or SELECTIVE) PRECIPITATION
A technique in which two or more ions in solution are separated by the proper use of one reagent that causes precipitation of both ions.
Fractional Precipitation - “precipitating” one ion at a time.
Compounds with different solubility can be selectively precipitated by adjusting the concentration of the precipitating reagents.
PROBLEM
Solution:
An aqueous solution of AgNO3 is slowly added from a buret to an aqueous solution that is 0.0100 M in Cl- and also 0.0100 M in CrO4
2-
Which ion. Cl- or CrO42-, is the first to precipitate from solution?
)(2
)(
42
2
4 brownredCrOAgCrOAg
whiteAgClClAg
101056.1)()( ClcAgcK sp
122
4
2 100.9)()( CrOcAgcK sp
lmolClc
KAgc
sp
Cl/1056.1
1.0
1056.1
)()( 9
10
lmolCrOc
KAgc
sp
CrO/105.9
1.0
100.9
)()( 6
12
2
4
24
? Which will ppt. first? Cl-
ELECTROLYTE PbCl2 PbSO4 PbS
Ksp 2.4∙10-4 2.2∙10-8 1.1∙10-29
S 3.9∙10-2 1.5∙10-4 3.3∙10-15
1. Type of precipitating reagent
Factors that Affect Precipitation
2. Quantity of a precipitating reagent
Usually excess of reagent is added (1,2-1,5)
3. Dissociation degree of a precipitating reagent
4. Hydrolysis of a precipitating reagent. Some sparingly soluble salts are more soluble in water than normally expected. Why??? Because the resultant anion can undergo hydrolysis as a base. For example: CO3
2-
5. Reaction conditions (pH, temperature, redox potential, etc.)
Two strategies for dissolving a water–insoluble ionic solid.
If the anion of the solid is a good base, the solubility is greatly increased by acidifying the solution.
In cases where the anion is not sufficiently basic, the ionic solid often can be dissolved in a solution containing a ligand that forms stable complex ions with its cation.
AgCl Ag+ + Cl- +
2NH3
[Ag(NH3)2]+
2. Formation of a complex ion
In formation of complex ion: Removed Ag+ from the equilibrium Equilibrium shifts to right Favors dissolving AgCl
3. Transformation of unsoluble precipitates in more soluble
Transformation of BaSO4 to BaCO3
BaSO4 + CO32- BaCO3 + SO4
2-