Solubility and Solubility Product Material

8/12/2019 Solubility and Solubility Product Material

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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.

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Solubility and Solubility Product Material