Solubility Productand

Common ion effect

Experiment #9

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 compoundsthat 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 thesolubility product principle

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

zyyz

sp XMK

Solubility product constant Molar solubility of the ions

Solubility product

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

3223 SBiK sp

Solubility product constant Molar solubility of the ions

Remember that we are dealing with molar solubilities andnot 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

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

3223 SBiK sp

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.

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

)(1

1lub

1lub

gmassMolarmol

LgramsinilitySo

LmolesinilitySo

)(96.513

111078.8

lub13

3232 gmol

Lg

SBiSBiofilitysoMolar

ML

molSBi 15

15

32 10708.1110708.1

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

3223 SBiK sp

Solubility product constant

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

If we wanted to figure out the Ksp of Bi2S3, then we need toknow the molar solubilities of Bi3+ and S2-. The molarsolubilities of the ions are usually figured out from the solubility of the parent compound.

Solubility product constant

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

If the molar solubility of Bi2S3 is “s”, the molar solubilityof Bi2+ is “2s” and the molar solubility of S2- is “3s”.

s 2×s 3×s

This is because, there are 2 ions of Bi3+ produced forEach molecule of the parent, Bi2S3 and 3 ions of S2- produced for each molecule of the parent.

3223 SBiK sp

32 32 ssK sp

32 32 ssK sp

Solubility product constant 32 32 ssK sp

sssssK sp 33322

)()33322( sssssK sp

)()274( 5sK sp

)()108( 5sK sp

Since we already know the value of molar solubility for Bi2S3, which is 1.708×10-15 M

515 )10708.1()108( spK 7210569.1 spK

How to find the molar solubility ifwe know is Ksp?

Find the molar solubility of Ca (OH)2, if the Ksp ofCa(OH)2 is 7.9 ×10-6.

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

Let the molar solubility of Ca(OH)2 be “s”. So, the molarSolubility of Ca2+ should be “1s” and the molar solubility of OH- should be “2s”.

s 1×s 2×s

This is because, there are 1 ion of Ca2+ produced foreach 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 ifwe know is Ksp?

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

s 1×s 2×s

212 OHCaK sp

21 21 ssK sp

)2()2()1( sssK sp

)()221( sssK sp 34sK sp

How to find the molar solubility ifwe know is Ksp?

34sK sp

6109.7, spKBut

36 4109.7 s

36

4109.7

s

361097.1 s

Ms 23 6 1025.11097.1

How to find the molar solubility ifwe know is Ksp?

Ms 23 6 1025.11097.1

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 M2.50 ×10-2 M

Is it possible to find the pH of theCa(OH)2 solution at 25C?

Yes

We know that [OH-] = 2.5 ×10-2 M

OHOHKwwaterofproductIonic 3,

14

3 101)25( OHOHCKw

142

3 101105.2 OH

MOH 13

2

14

3 104105.2

101

MOH 13

2

14

3 104105.2

101

Is it possible to find the pH of theCa(OH)2 solution at 25C?

OHLogpH 3

13104 LogpH

39.12pH

The Ca(OH)2 solution is basic.

Experiment- To determine the Ksp of Potassium Hydrogen Tartarate, KHT

KHT is also called cream of tartar

H-C-OH

H-C-OH

COOH

COOK

KHT

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

s 1×s 1×s

11 HTKK sp

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.

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 gravityfiltration.

Now we have a saturated solution of KHT.

H-C-OH

H-C-OH

COOH

COOK

KHT

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

s 1×s 1×s

11 HTKK sp

HT- can act an acid, so if we titrate it with a known concentration of base (NaOH), we can find the [HT-]

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-]

11 HTKK sp

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 effectSuppression of ionization of a weak electrolyte bythe 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 equivalentto adding a stress to the system.

The system responds to the stress byreducing the solubility of one of the ionsand keeping the Ksp constant.

Calculate the molar solubility of lead iodide PbI2, from its Ksp in water at 25C

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

s 1×s 2×s

212 IPbK sp

21 21 ssK sp

)2()2()1( sssK sp

)()221( sssK sp 34sK sp

34sK sp

9109.7, spKBut

39 4109.7 s

39

4109.7

s

391097.1 s

Ms 33 6 103.11097.1

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

212 IPbK sp

9

2 109.7)( PbIK sp

219 1.0)2(1109.7 ss

Calculate the molar solubility of lead iodide PbI2, in 0.1 M NaI solution

219 1.0)2(1109.7 ss

29 1.02)1(109.7 ss

Because the Ksp of PbI2 is really small, the solubility s is goingto be really small. Hence we can make a simplification.

1.01.02 s

29 1.0)1(109.7 s

Ms 79

109.701.0109.7

A comparison of solubility of PbI2 With and without common ion

With common ion Without common ion

s= 1.3 × 10-3 M s= 7.9 × 10-7 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

11 HTKK sp

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

0.1 M 0.1 M

HT- can act an acid, so if we titrate it with a known concentration of base (NaOH), we can find the [HT-]

Common ion

11 11.01 ssK sp