Solubility Product and Common ion effect Experiment #9
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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
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