solubility and solubility product material
TRANSCRIPT
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SOLUBILITY ND SOLUBILITY PRODUCT
A.
SOLUBILITYThe solubility of a substance is the
amount of the substance that will dissolve in a
given amount of solvent. Solubility is a
quantutative term.
How is the solubility of a substance
determined? A known amount of the solventfor
example, 100 mLis put in a container. Then the
substance whose solubility is to determined is
added until, even vigorous and prolonged stirring,
some of that substance does not dissolve. Such a
solution is said to be saturated becaise it contains
as much solute as possible at that temperature. In
this saturated solution, the amount of solute is the
solubility of that substance at that temperature in
that solvent. Doing thius experiment with water
as the solvent and sodium chloride as the solute,
we find that, at 20 oC, 35,7 grams of the salt
dissolve in 100 mL of water. The solubility of
sodium chloride is 35,7 g/100 mL water at 20 oC.Sodium chloride is a moderately soluble salt. The
solubility of sodium nitrate is 92,1 g/100 mL
water at 20 oC; sodium nitrate is a very soluble
salt. At the opposite ed of the scale is barium
sulfate, which has a solubility of 2,3 x 10 -4g/100
mL watre at 20 oC. Barium sulfate is an insoluble
salt.
B. MOLAR SOLUBILITY (S)
The molar solubility of a substance is the numberof mole of that substance yhat will dissolve in 1 litre of
solvent. If in 1 litre of water can dissolve 1,31 x 10 -5mol
of AgCl, then the molar solubility of AgCl is 1,31 x 10 -5
mol L-1.
The molar solubility can be calculated by using
formula:
Where:
S = molar solubility
Mr= relative molecular mass
a = the mass of solute (in gram)
S =
Three important definitions:
solubility: quantity of a
substance that dissolves to
form a saturated solution
molar solubility: the
number of moles of the
solute that dissolves to
form a liter of saturated
solution
Ksp(solubility product):
the equilibrium constant
for the equilibrium
between an ionic solidand its saturated solution
Picture 1. Solubility of salt in water
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v= the volume of solution (in mL)
Answer:
C. THE SOLUBILITY PRODUCT (Ksp)
Most substances are soluble in water to
at least some slight extent. If an insoluble or
slightly soluble material is placed in water,
an equilibrium is established when the rate of
dissolution of ions from the solid equals the
rate of precipitation of ions from the saturated
solution. Thus, an aquilibrium exist between
solid silver chloride and a saturated solution of
silver chloride.
AgCl (s) Ag+
(aq) + Cl-
(aq)
The equilibrium constant is
K=[][][]
Since the concmetration of a pure solid is a constant, [AgCl] may be combined with
Kto give,
Ksp= K [AgCl] = [Ag+] [Cl-]
The constant Kspis called a solubility product. The ionic concentration of the
expression are those for a saturated solution at the reference temperature. Since the
solubillityof a salt usually varies widely with temperature, the numerical value for Ksp
for a salt changes with temperature.
For salts that have more than two ions per formula unit, the ion concentration
must be raised to the powers indicated by coefficients of the balanced chemical
equation.
Mg(OH)2(s) Mg2+
+ 2 OH- Ksp= [Mg
2+] [OH-]2
Bi2S3(s) 2 Bi3+
+ 3 S2-
Ksp= [Bi3+
]2
[S2-
]3
Picture 2. AgCl in saturated solution
Example 1
At 25oC, 0.00188 g of AgCl dissolves in 1 litre of water. What is the molar solubility
of AgCl? (Ar Ag = 108, Cl = 35,5)
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Ca3(PO4)2 3 Ca2++ 2 PO4
3- Ksp= [Ca2+]3[PO4
3-]2
For a salt of this type the calculations of the Ksp from the molar solubility
requires that the chemical equation representing the dissociation process be carefully
interpreted.
D. THE RELATIONSHIP BETWEEN MOLAR SOLUBILITY (S) AND KSP
It is possible to relate Ksp to an experimentally determined solubility, as in
Example 2, 3, and 4 or to calculate the solubility from a tabulated value of Ksp, as in
Example 5 and 6.
Answer:
Table 1. the Kspfor some electrolytes
Example 2:
At 25 oC 1.31 x 10-5mol of AgCl dissolves in 1 litre of water. What is the KspofAgCl?
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Answer:
Answer:
Example 3:
At 25 oC 7.8 x 10-5 mol of silver chromate dissolves in 1 litre of water. What is the
Kspof Ag2CrO4?
Example 4:
A 100.0 mL sample is removed from a water solution saturated in MgF2at 18oC.
The water is completely evaporated from the sample and a 7.6 mg deposit of MgF2
is obtained. What is Kspfor MgF2at 18oC?
(Ar Mg = 24, F = 19)
MgF2(s) Mg2+(aq) + 2 F-(aq)
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ARE YOU WONDERING .........
When comparing molar solubilities, is a solute with a larger value of Ksp
always more soluble than one with smaller value?
If the solutes being compares are of the same type (all of the type MX, MX 2,
M2X, ...), their molar solubilities will be related in the same way as their Kspvalues.
That is, the solute with the largest Kspvalue will have the greatest molar solubility.
Thus, AgCl (Ksp= 1.8 x 10-10) is more soluble than AgBr (Ksp= 5.0 x 10
-13). For these
particular solutes, the molar solubility is s = .If the solutes are not of same type, youll have to calculate, or at least estimate,
each molar solubility and compare the results. Thus, even though it solubility product
constant is smaller, Ag2CrO4(Ksp= 1.1 x 10-12) is more soluble than AgCl (Ksp= 1.8
x 10-10). For Ag2CrO4, the molar solunility is s =
= 6.5 x 10-5 M, whereas for
AgCl, it is s = = 1.3 x 10-5 M.E. THE COMMON ION EFFECT IN SOLUBILITY EQUILIBRIA
The common ion effect is responsible for the reduction in solubility of an ionic
precipitate when a soluble compound combining one of the ions of the precipitate is
added to the solution in equilibrium with the precipitate.It states that if the
concentration of any one of
the ions is increased, then,
according to Le Chatelier's
principle, the ions in excess
should combine with the
oppositely charged ions.
Some of the salt will be
precipitated until the ionic
product is equal to the
solubility of the product. In
simple words, common ion
effect is defined as the
suppression of the degree of
dissociation of a weak
electrolyte containing a
common ion.
As an example, consider the system
PbCrO4 (s) Pb2+(aq) + CrO42-(aq)
Picture 3. The effect of a common ion on solubility
http://en.wikipedia.org/wiki/Le_Chatelier%27s_principlehttp://en.wikipedia.org/wiki/Le_Chatelier%27s_principlehttp://en.wikipedia.org/wiki/Le_Chatelier%27s_principlehttp://en.wikipedia.org/wiki/Le_Chatelier%27s_principlehttp://en.wikipedia.org/wiki/Le_Chatelier%27s_principle -
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The addition of chromate ion to a saturated solution of Lead (II) chromate will cause
the equilibrium to shift to the left. The concebtration of Pb2+ will decrease, and
PbCrO4 will precipitate. Since the product [Pb2+] [CrO4
2-] is a constant, increasing
[CrO42-] will cause [Pb2+] to decrease (see Picture 3).
The solubility of a
sparingly soluble salt is
reduced in a solution that
contains an ion in common
with that salt. For instance,
the solubility of silver
chloride in water is
reduced if a solution of
sodium chloride is added
to a suspension of silver
chloride in water.
A practical example used very widely in areas drawing drinking water from
chalk or limestone aquifers is the addition of sodium carbonate to the raw water to
reduce the hardness of the water. In the water treatment process, highly soluble
sodium carbonate salt is added to precipitate out sparingly solublecalcium carbonate.
The very pure and finely divided precipitate of calcium carbonate that is generated is a
valuable by-product used in the manufacture oftoothpaste.The salting out
process used in the
manufacture of soaps
benefits from the common
ion effect. Soaps are sodium
salts of fatty acids.Addition
of sodium chloride reduces
the solubility of the soap
salts. The soaps precipitatedue to a combination of
common ion effect and
increasedionic strength.
Sea, brackish and other waters that contain appreciable amount of Na+
interfere with the normal behavior of soap because of common ion effect. In the
presence of excess Sodium ions the solubility of soap salts is reduced, making the
soap less effective.
Picture 5. Soap
Picture 4. Common ion effect in equilibrium
http://en.wikipedia.org/wiki/Silver_chloridehttp://en.wikipedia.org/wiki/Silver_chloridehttp://en.wikipedia.org/wiki/Sodium_chloridehttp://en.wikipedia.org/wiki/Drinking_waterhttp://en.wikipedia.org/wiki/Chalkhttp://en.wikipedia.org/wiki/Limestonehttp://en.wikipedia.org/wiki/Aquifershttp://en.wikipedia.org/wiki/Hard_waterhttp://en.wikipedia.org/wiki/Water_treatmenthttp://en.wikipedia.org/wiki/Sodium_carbonatehttp://en.wikipedia.org/wiki/Calcium_carbonatehttp://en.wikipedia.org/wiki/Toothpastehttp://en.wikipedia.org/wiki/Salting_outhttp://en.wikipedia.org/wiki/Soaphttp://en.wikipedia.org/wiki/Fatty_acidhttp://en.wikipedia.org/wiki/Ionic_strengthhttp://en.wikipedia.org/wiki/Sodiumhttp://en.wikipedia.org/wiki/Sodiumhttp://en.wikipedia.org/wiki/Ionic_strengthhttp://en.wikipedia.org/wiki/Fatty_acidhttp://en.wikipedia.org/wiki/Soaphttp://en.wikipedia.org/wiki/Salting_outhttp://en.wikipedia.org/wiki/Toothpastehttp://en.wikipedia.org/wiki/Calcium_carbonatehttp://en.wikipedia.org/wiki/Sodium_carbonatehttp://en.wikipedia.org/wiki/Water_treatmenthttp://en.wikipedia.org/wiki/Hard_waterhttp://en.wikipedia.org/wiki/Aquifershttp://en.wikipedia.org/wiki/Limestonehttp://en.wikipedia.org/wiki/Chalkhttp://en.wikipedia.org/wiki/Drinking_waterhttp://en.wikipedia.org/wiki/Sodium_chloridehttp://en.wikipedia.org/wiki/Silver_chloridehttp://en.wikipedia.org/wiki/Silver_chloride -
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Answer:
At times, the common ion effect is used to prevent the formation of precipitate.
Consider the precipitation of magnesium hydroxide from a solution that contains
Mg2+ions.
Mg(OH)2(s) Mg2+(aq) + 2 OH-(aq)
The precipitation can be prevented by holding the concentration of OH - to a low
value. If the hydroxide ion is supplied by the weak base ammonia
NH3+ H2O NH4+ + OH-
the concentration of OH- can be controlled by the addition of NH4+. When the
common ion NH4+is added, the ammonia equilibrium shifts to the left, which reduces
the concentration of OH-. In this way, the concentration of OH-can be held to a level
that will not cause Mg(OH)2to precipitate.
Answer:
Example 7:
At 25 oC a saturated solution of BaSO4is 3.9 x 10-5M. The Kspof BaSO4is
1.5 x 10-9. What is the solubility of BaSO4in 0.050 M Na2SO4?
Example 8:
What is the concentration of NH4+, derive from NH4Cl, is necessary to prevent the
formation of an Mg(OH)2precipitate in a solution that is 0.050 M in NH3?
Kspof Mg(OH)2is 8.9 x 10-12
.
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F. PRECIPITATION AND THE SOLUBILITY PRODUCT
We have used the term solubility equilibrium to describe the phenomenaencountered to this point. But, as we have seen on previous occasions, an equilibrium
condition can be approached from either direction. If the equilibria of the preceding
sections are approached by starting with ions in solution and producing pure,
undissolved solute, then the process involved is precipitation reaction. There is a
great deal that we can say about precipitation reactions from the standpoint of
solubility product constants.
Criterion for Precipitation from Solution.
The most fundamental question we can ask about a precipitation reaction
whether it will in fact occur for a given set of conditions. Suppose that a solution is
made simultaneously 0.10 M in Ag+and 0.10 M in Cl-. Shoild a precipitate of AgCl(s)
form? To answer this question we bwgin with a chemical equation to represent
equilibrium between the slightly soluble solute and its ions, together with the value of
Kspfor this equilibrium.
AgCl (s) Ag+(aq) + Cl-(aq)
Ksp= [Ag+] [Cl-] = 1.6 x 10-10
Now recall how we dealt with
the question of the direction of netchange in section. We formulated the
quantity called the reaction quotient,
Q, and compared its value with that of
the equilibrium constant, K. In this case,
the reaction quotient is just the product,
[Ag+] [Cl-], based on the initial
concentration of these ions. For
precipitation reaction, Q is sometimes
called the ion product.Q = (0.10) (0.10) = 1 x 10-2> Ksp= 1.6 x 10
-10
We conclude that reaction should occur to the left or in reverse direction in
equationprecipitation should occur.
The general conclusions about precipitation
from solution are
Q > Ksp precipitation should occur
Q < Ksp precipitation does not occur
Q = Ksp a solution is just saturated
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Answer:
Answer:
Example 9:
Should a precipitate form if 10.0 mL 0.0010 M AgNO3(aq) is added to 500 mL0.0020 M K2CrO4(aq)?
Ag2CrO4 2 Ag+(aq) + CrO4
2-(aq) Ksp= 2.4 x 10-12
Example 10:
Will a precipitate form if 10 mL of 0.010 M AgNO3and 10 mL of 0.00010 M NaCl
are mixed? Assume that the final volume of the solution is 20 mL. For AgCl, Ksp=
1.7 x 10-10.