1. Recognize the effect common ions exert on acid ionization and pH, and mechanisms by which solutions are buffered by ions.
2. Examine acid base titration and the Henderson-Hasselbalch equation and calculate the pH of a system at any stage of titration involving both strong and weak acid bases.
3. Apply Le Chatelier's Principle to explain the effects of common ions and pH on solubility.
4. Describe the effect of complex ion formation on solubility.© BrainMass Inc. brainmass.com December 24, 2021, 7:15 pm ad1c9bdddf
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1) Buffer solution: A solution which resists change in pH upon addition of small amount of acid or base to it. They are of two types' acidic buffer and basic buffers.
Acidic buffer is a mixture of weak acid and its salt
Eg : acetic acid buffer
It is a mixture of acetic acid and sodium acetate
If a small amount of acid is added to this system according to Le Chatlier principle
the system will adjust itself to nullify the effect made by us to the equilibrium
If a small amount of HCl is added to this system
HCl dissociate as H + and Cl- , H+ ions are added to this equilibrium (definition of acid = H+ ion donor)
The concentration of H+ ions will increase in the system, the equilibrium ( Ist equation) will shift to the backward direction ie the formation of acetic acid will be preferred there by maintaining the pH constant
Similarly if a small amount of base say NaOH is added (Base is OH- ion donor)
It combines with acetic acid and eliminates a molecule of water and forms acetate ion which is again in the equilibrium (if the any ion in the equilibrium is formed, the system will adjust itself and keep the pH constant)
2) Consider first a buffer solution containing a weak acid HA and its highly dissociated salt NaA. The hydrogen ion concentration of such a solution is given by the equation
[H+] =Ka [Acid]/[Salt]
Taking the logs and reversing the sign, we have
-log [H+] = -logKa + log [Salt]/[Acid]
pH = pKa + log [Salt]/[Acid] ( Hendersons equation)
Consider the titration of 25 ml acetic acid and(0.1M) against 0.1M solution of NaOH and we want to calculate the pH of the titration after the addition of 20ml of NaOH solution
Amount of acetic acid present in the solution = 25 x0.1= 2.5 mmole
Since acetic acid is a weak acid, its H+ ion concentration is given by √cKa, where 'c' is the molar concentration of the acid Ka its dissociation constant
The initial concentration of H+ in 0.1 CH3COOH = √0.1x 1.75x10-5
H +=1.32 x 10-3 mol -1
pH = -log (1.32 x 10-3) = 2.88
During titration, CH3COOH gets converted into CH3COONa. The pH of the titration solution would thus depend upon the concentration of CH3COOH left and of CH3COONa formed during the titration. In a mixture of weak acid and its salt
pH = pKa + log[salt]/[Acid]
with the addition of 20ml of NaOH solution, the amount of NaOH added to the titration solution = 20 x0.1 mmole
Amount of CH3COONa formed = 2.0mmole since the volume of the solution is the same for both CH3COOH and CH3COONa, the ratio of their amounts in millimoles can be taken as the ratio of their molar concentration
Therefore pH = 4.76 + log 2.0/ 0.5 =5.36
( Ka = 1.75x10-5 , therefore pKa = -logKa = 4.76)
So the pH of the system when 20 ml of NaOH is added to HCl = 5.36
3) Effect of common ion
As mentioned earlier when ever we made any change in equilibrium the system will adjust itself to nullify the effect(LeChtlier principle)
Consider the precipitation of NaCl
NaCl (s, sat solution) Na+ aq+ Cl-aq
If we pass HCl gas to this system the [Cl-] will increase appreciably in the equilibrium equation so the system will adjust to decrease the concentration NaCl will be precipitated ie shifting the equilibrium towards the backward dirction.
Effect of pH on solubility
A substance gets precipitated when the ionic product exceeds the solubility product.
A substance precipitates at a particular pH only
Consider the solubility equation for calcium sulphate
Ca 2+ in water = 0.01M
If the volume is doubled by adding sulphuric acid solution (increasing the pH )
Ca2+ in the mixture = 0.01/2 =0.005 M
Con H2SO4 = 0.001 M
There fore [SO4]2-= 0.001 M
Since the volume is doubled when mixed with equal volume of water
[SO4]2-= 0.001 M
If again the volume is doubled by the addition of water
[SO4]2- ions in the mixture = 0.001/2= 0.005 M
The product of concentration of the ions = [Ca 2+ ][SO4]2-= 0.005x0.005= 2.5x10-6M2 which is lower than Ksp so calcium sulphate will be dissolved.
If again sulphuric acid is added of higher concentration say 0.02M
Now [SO4]2-= 0.02/2= 0.01 M
[Ca 2+ ] = 0.01/2= 0.005M
Now the product of con = 0.005x 0.01 = 5x10-5M2
In this case the product of concentration of ions exceeds the solubility product so calcium sulphate will be precipitated. So pH has a crucible role in solubility and precipitation
4) Effect of complex ion formation solubility
AgCl which is only sparingly soluble in water, dissolves completely in aqueous ammonia solution. The increased solubility is due to the formation of a soluble complex [Ag(NH3)2]Cl. The various equilibria involved in the dissolution of AgCl in aqueous ammonia solution are as follows:
The equilibrium constant for the above reaction is given by
K = [Ag (NH3)2]+/[Ag+ ][NH3]2
Adding the above two equations we get
where Ksp = solubility product and Kstab= stability constant
The complex ion formed in the second reaction is highly stable, the stability constant being as high as 1.67x107 the formation of a stable complex ion in the second reaction upset the equilibrium hereby causing it to shift to the right according to Le chatlier principle. Accordingly, AgCl(s) dissolves to a much larger extent
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