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Experiment #9. Solubility Product and Common ion effect. What are we doing in this experiment?. Determine the molar solubility and solubility product constant (Ksp) of potassium hydrogen tartarate (KHT). Study the effect of common ion on the Ksp

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Experiment #9

Solubility Product

and

Common ion effect


What are we doing in

this experiment?

Determine the molar solubility and solubility

product constant (Ksp) of potassium hydrogen

tartarate (KHT).

Study the effect of common ion on the Ksp

of KHT and the molar solubilites of its ions.


Remember!!

In this experiment, we are dealing with compounds

that are very slightly soluble that they are called

“Insoluble compounds”.

Our bones and teeth are mostly calcium phosphate,Ca3(PO4)2, a very slightly soluble compound.


Solubility product

In general, the solubility product expression for a compound

is the product of the concentration (molar solubility) of its

constituent ions, each raised to the power that corresponds to

the number of ions in one formula unit of the compound. The

quantity is constant at constant temperature for a saturated

solution of the compound.This statement is called the

solubility product principle

MyXz (s) yMZ+ (aq) + zXY-(aq)

Molar solubility of the ions

Solubility product constant


Solubility product

Bi2S3 (s) 2Bi3+ (aq) + 3S2-(aq)

Molar solubility of the ions

Solubility product constant

Remember that we are dealing with molar solubilities and

not concentration.

For a saturated solution, molar solubility is equal to molar concentration.


Different types of solution

Unsaturated solution: More solute can be dissolved

in it.

Saturated solution: No more solute can be dissolved

in it. Any more of solute you add

will not dissolve. It will precipitate

out.

Super saturated solution: Has more solute than can

be dissolved in it. The

solute precipitates out.


Molar solubility

Let us say, we try to dissolve 1 g of Bi2S3 in 1 L of water. If

only 8.78×10-13 g out of the 1.0 g dissolves, we can make the

following conclusions:

1. The solution is saturated with Bi2S3.

2. If we filter out the undissolved Bi2S3, the amount of solute

that dissolved (soluble) in 1. 0 L of water is 0.0025 g.

Bi2S3 (s) 2Bi3+ (aq) + 3S2-(aq)

So we can say, the solubility of Bi2S3 is 8.78 ×10-13 g per liter


Molar solubility

Molar solubility is solubility in moles per liter


Solubility product constant

Bi2S3 (s) 2Bi3+ (aq) + 3S2-(aq)

Bi2S3 (s) 2Bi3+ (aq) + 3S2-(aq)

If we wanted to figure out the Ksp of Bi2S3, then we need to

know the molar solubilities of Bi3+ and S2-. The molar

solubilities of the ions are usually figured out from the

solubility of the parent compound.


Solubility product constant

If the molar solubility of Bi2S3 is “s”, the molar solubility

of Bi2+ is “2s” and the molar solubility of S2- is “3s”.

Bi2S3 (s) 2Bi3+ (aq) + 3S2-(aq)

s 2×s 3×s

This is because, there are 2 ions of Bi3+ produced for

Each molecule of the parent, Bi2S3 and 3 ions of S2-

produced for each molecule of the parent.


Solubility product constant

Since we already know the value of molar solubility for

Bi2S3, which is 1.708×10-15 M


How to find the molar solubility if

we know is Ksp?

Find the molar solubility of Ca (OH)2, if the Ksp of

Ca(OH)2 is 7.9 ×10-6.

Ca(OH)2 (s) Ca2+ (aq) + 2OH-(aq)

s 1×s 2×s

Let the molar solubility of Ca(OH)2 be “s”. So, the molar

Solubility of Ca2+ should be “1s” and the molar solubility

of OH- should be “2s”.

This is because, there are 1 ion of Ca2+ produced for

each molecule of the parent, Ca(OH)2 and 2 ions of OH-

produced for each molecule of the parent.


How to find the molar solubility if

we know is Ksp?

Ca(OH)2 (s) Ca2+ (aq) + 2OH-(aq)

s 1×s 2×s



How to find the molar solubility if

we know is Ksp?

So the molar solubility of Ca(OH)2 = s = 1.25 ×10-2 M

The molar solubility of Ca2+ = [Ca2+] =1s = 1.25 ×10-2 M

The molar solubility of OH- = [OH-] = 2s = 2×1.25 ×10-2 M

2.50 ×10-2 M


Is it possible to find the pH of the

Ca(OH)2 solution at 25C?

Yes

We know that [OH-] = 2.5 ×10-2 M


Is it possible to find the pH of the

Ca(OH)2 solution at 25C?

The Ca(OH)2 solution is basic.


COOK

H-C-OH

H-C-OH

COOH

Experiment- To determine the Ksp of

Potassium Hydrogen Tartarate, KHT

KHT is also called cream of tartar

KHT (s) K+ (aq) + HT-(aq)

s 1×s 1×s

If we want to determine the Ksp of KHT,

we need to know the molar solubilities of

K+ and HT-. Also remember that Ksp is

measured for a saturated solution.

KHT


COOK

H-C-OH

H-C-OH

COOH

How do we determine [K+] and [HT-]?

Firstly prepare a saturated solution of KHT.

3.0 g of KHT in 200 ml of water.

Filter out the undissolved KHT using gravity

filtration.

Now we have a saturated solution of KHT.

KHT (s) K+ (aq) + HT-(aq)

s 1×s 1×s

HT- can act an acid, so if we titrate it with a known

concentration of base (NaOH), we can find the [HT-]

KHT


How do we determine [K+] and [HT-]?

Once we know the concentration of HT-, based on

1 to 1 molar relationship between K+ and HT-,

[K+]= [HT-]

NaOH is hygroscopic, so the NaOH solution needs to

be standardized by using KHP


LeChatelier’s Principle

If a stress (change of condition) is applied to a

system at dynamic equilibrium,the system shifts

in the direction that reduces the stress.

Common ion effect

Suppression of ionization of a weak electrolyte by

the presence in the same solution of a strong

electrolyte containing one of the same ions as

the weak electrolyte.


About Common ion effect

Common ion effect is a special case of

LeChatelier principle

Addition of a common ion is equivalent

to adding a stress to the system.

The system responds to the stress by

reducing the solubility of one of the ions

and keeping the Ksp constant.


Calculate the molar solubility of lead iodide PbI2,

from its Ksp in water at 25C

PbI2 (s) Pb2+ (aq) + 2I-(aq)

s 1×s 2×s


Calculate the molar solubility of lead iodide PbI2,

in 0.1 M NaI solution

PbI2 (s) Pb2+ (aq) + 2I-(aq)

s 1×s 2×s

NaI (s) Na+ (aq) + I-(aq)

0.1M 0.1M

Common

ion


Calculate the molar solubility of lead iodide PbI2,

in 0.1 M NaI solution

Because the Ksp of PbI2 is really small, the solubility s is going

to be really small. Hence we can make a simplification.


A comparison of solubility of PbI2

With and without common ion

Without common ion

With common ion

s= 7.9 × 10-7 M

s= 1.3 × 10-3 M

Solubility decreases because of the presence of

common ion


To study the effect of common ion on the solubility of KHT

KHT (s) K+ (aq) + HT-(aq)

s 1×s 1×s

KCl (s) K+ (aq) + Cl-(aq)

0.1 M 0.1 M

Common ion

HT- can act an acid, so if we titrate it with a known

concentration of base (NaOH), we can find the [HT-]


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