1. 17.1 Common Ion Effect Buffer Solutions The resistance of pH change
Dr. Fred Omega Garces
Chemistry 201
Miramar College

2. Common Ion Effect
Ionization of an electrolyte, i.e., salt, acid or base is decreased when a common ion is added to that solution.
i) What is the % ionization for M acetic acid ? (Pure)
HC2H3O H2O  C2H2O H3O+ Ka =1.8•10-5 M
Solving the iCe problem:
ka = =1.8•10-5 M = [C2H3O2-] [H3O+] /0.10 M  [H3O+]=1.34•10-3 M
%  = (1.34•10-3 / 0.10 ) * 100 = 1.34 % pH = 2.87
ii) What is %  if M HC2H3O2 is mix w/ 0.100M NaC2H3O2 ? (Buffer)
i Lots •10-7
C -x -x +x +x
[c] x Lots x 1•10-7+x
ka = 1.8•10-5 M = [ x ] [x] /( x) [0.100] [x] /( )
x = [H3O+]= 1.8•10 -5 M pH = 4.74
%  = (1.8•10-5 / 0.10 ) * 100 = %
Ionization % decrease in presence of common ion !!

3. Common Ion Effect Equation
Consider the previous problem in which a common ion is in the same solution.
HC2H3O H2O  C2H3O H3O+ Ka =1.8•10-5 M
i Lots •10-7
C -x -x +x +x
[c] x Lots x 1•10-7+x
or [c] [HC2H3O2 ] Lots [C2H3O2- ] [H3O+ ]
ka = [C2H3O2- ] [H3O+] rearrange the equation [H3O+] = ka • [HC2H3O2 ]
[HC2H3O2] [C2H3O- ]
Taking the - log of both side -
- log [H3O+] = - log (ka • [HC2H3O2 ] / [C2H3O2-] )
or pH = -log ka - log( [HC2H3O2] / [C2H3O2 - ] )
let Ca = [HC2H3O2] and Cb = [C2H3O2 - ] ) therefore
pH = pKa - log Ca / Cb
or pH = pKa + log Cb / Ca
This is the Henderson Hasselbach Equation:
pH = pKa + log Cb / Ca or pOH = pKb + log Ca / Cb

4. Henderson-Hasselbach Equation
pH of a solution can be calculated using a useful equation:
pH = pKa + log [A-] / [HA]
Where HA & A-
are the weak acid and its conjugate and Ka is for HA
Similarly,
pOH = pKb + log [HA] / [A-]
are the weak base and its conjugate and Kb is for A-

5. Henderson-Hasselbach Equation: Example
Consider the common ion effect problem and lets see how the Henderson-Hasselbach equation can be used to simplify this problem.
What is pH if M HC2H3O2 is mix w/ 0.100M NaC2H3O2 ?
HC2H3O H2O  C2H2O H3O+ Ka =1.8•10-5 M
i Lots •10-7
C -x -x +x +x
[c] x Lots x 1•10-7+x
Using the Henderson-Hasselbach equation:
pH = - log (4.3•10-7 ) + log (0.100 / 0.100)
pH = log pH = pH = 4.74
Note: When a common ion is present in the same solution, the strategy to solve the problem requires a Buffer Type of calculation.

6. Henderson-Hasselbach Equation and Buffer Problems (sRF)
A buffer M acetate and M acetic acid is prepared (Ka = 1.8 •10-5).
i) What is the pH of the buffer?
ii) Calculate the initial pH, final pH, and change in pH that result when 1.00 mL of M HCl is added to mL of the buffer.
iii) Calculate the initial pH, final pH, and change in pH that result when 1.00 mL of M HCl is added to mL of water.
Note: HCl = M • 1.00mL = 0.1 mmol.
C2H3O2- =0.100 M •100mL = 10 mmol and HC2H3O2 =0.200 M •100mL = 20 mmol
i) pH = pKa + log Cb/Ca = -log(1.8•10-5) + log ( 0.10 / 0.20)  pH = 4.44
ii) C2H3O H3O+  HC2H2O H2O
s 10mmol 0.1 mmol 20 mmol Lots R
f Lots
[c] 9.9/ /101 VT = 101 mL
pH = -log (1.8•10-5)+log [(9.9/101) / [20.1/101)] =  pH = 4.43
pH (initial) = 4.44, pH (final) 4.43, pH (change) = -0.01

7. ...Continue: Henderson-Hasselbach Equation and Buffer Problems
A buffer M acetate and M acetic acid is prepared (Ka = 1.8 •10-5). Reger 14.19
iii) Calculate the initial pH, final pH, and change in pH that result when 1.00 mL of M HCl is added to mL of water.
Note: HCl = M • 1.00mL = mmol.
iii) HCl + H2O  H3O Cl-
s mmol 1•10-7M -
R mmol mmol -
f mmol
[c] mmol / 101 mL
[H3O+] = 9.9•10-4 M  pH = 3.00
pH (initial) = 7.00 , pH(final) 3.00, pH(change) = -4.00

8. Essential Feature of Buffer Systems
A buffer solution exhibits very small change in pH changes when H3O+ and OH- is added. A buffer solution consists of relatively high concentration of the components of a conjugate weak acid-base pair. The buffer-components concentration ratio determines the pH, and the ratio and pH are related by the Henderson-Hasselbalch equation. A buffer has an effective range of pKa + 1 pH unit.

9. Blood Buffer System
Buffer - A solution whose pH is resistant to change
Your body uses buffers to maintain the pH of your blood
Blood pH
Buffer system in body -
1. Proteins
2. Phosphates HPO42- / H2PO4- : 1.6 / 1
3. Carbonates H2CO3 / HCO3 - : 10 / 1
Reaction:
H3O+ + HCO3-  H2CO3 + H2O
H2CO3  H2O + CO2 (exhale)

10. Acidosis Blood pH  7.35 (ACIDOSIS)
Depression of the acute nervous symptom. Or respiratory center in the medulla of the brain is affected by an accident or by depressive drugs.
Symptoms:
•Depression of the acute nervous system
•Fainting spells
•Coma
•RIP
Causes:
1. Respiratory Acidosis
Difficulty Breathing
(Hypo-ventilation)
Pneumonia, Asthma anything which diminish CO2 from leaving lungs.
2. Metabolic Acidosis
Starvation or fasting
Heavy exercise
Mechanism:
1. Respiratory Acidosis
CO2 doesn’t leave lungs which result in the build up of H2CO3 in the blood
2. Metabolic Acidosis
If body doesn’t have enough food then Fatty acids (Fat) are used.
Fatty Acids  Acidic. Furthermore, exercise leads muscle to produce lactic acid.
Blood pH below 7.35: Acidosis: Depression of the acute nervous system. Or respiratory center in the medulla of the brain is affected by an accident or by depressive drugs.
Respiratory Acidosis:
Symptoms: Failure to ventilate, suppression of breathing, disorientation weakness, fainting spells, coma death.
Causes: Difficulty breathing (Hyperventilation), lung disease blocking gas diffusion (e.g., emphysema, pneumonia, bronchitis and asthma) depression of respiratory center by drugs, cardiopulmonary arrest, stroke, poliomyelitis, or nervous system disorder
Mechanism CO2 doesn't leave lungs which result in the build up of H2CO3 in the blood.
Treatment Correction of disorder, breathing assistance, infusion of bicarbonate, diuretics / digitalis, antibiotics
Metabolic Acidosis :

11. Alkalosis Blood pH  7.45 (ALKALOSIS)
Hyperventilation during extreme fevers or hysteria. Excessive ingestion of basic antacids and severe vomiting
Symptoms:
•Over simulation of the nervous system
•Muscle cramps
•Convulsion
•Death
Causes:
1. Respiratory Alkalosis
Heavy rapid breathing (hyperventilation). Results from - fear, hysteria, fever, infection or reaction with drugs.
2. Metabolic Alkalosis
Metabolic irregularities or by excess vomiting
Mechanism:
1. Respiratory Alkalosis
Excessive loss of CO2 lowers H2CO3 and raise HCO3- level (Can be remedied by breathing in a bag)
2. Metabolic Alkalosis
Vomiting removes excess acidic material from stomach. (pH of stomach equals one).

12. Buffer System at Work H3O+ Buffer H2O Remember pH = Conc. of H3O+
Buffer - System that resists change in pH when H3O+ or OH- is added.
Buffer solution may be prepared by a weak acid and its conjugate base.
How it Works:
A-  HA
H3O+ Buffer H2O
Remember pH = Conc. of H3O+
Your blood Rxn: HCO3-  H2CO3
Acidosis Excess H3O+ + HCO3-  H2CO3 + H2O
H3O CO2 + H2O
Alkalosis Excess OH- + H2CO3  HCO H2O
OH-

13. Summary
The following summary lists the important tools needed to solve problems dealing with acid-base equilibria.