1. Common Ion Effect Buffer Solutions The resistance of pH change

2. Common Ion Effect decreased 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 0.100 M acetic acid ? HC 2 H 3 O 2 + H 2 O  C 2 H 2 O 2 - + H 3 O + K a =1.810 -5 M Solving the i  e problem: k a = =1.810 -5 M = [C 2 H 3 O 2 - ] [H 3 O + ] /0.10 M  [H 3 O + ]=1.3410 -3 M 1.34 %pH = 2.87 %  = (1.3410 -3 / 0.10 ) * 100 = 1.34 %pH = 2.87 ii) What is %  if 0.100 M HC 2 H 3 O 2 is mix w/ 0.100M NaC 2 H 3 O 2 ? HC 2 H 3 O 2 + H 2 O  C 2 H 2 O 2 - + H 3 O + K a =1.810 -5 M i0.100Lots0.100110 -7  -x-x+x+x [c]0.100-xLots0.100+x110 -7 +x k a = 1.810 -5 M = [0.100+ x ] [x] /( 0.100 -x)  [0.100] [x] /( 0.100 ) pH = 4.74 x = [H 3 O + ]= 1.810 -5 M pH = 4.74 0.0180 % %  = (1.810 -5 / 0.10 ) * 100 = 0.0180 % Ionization % decrease in presence of common ion !! 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. HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2 - + H 3 O + K a =1.810 -5 M i0.100Lots0.100110 -7  -x-x+x+x [c]0.100-xLots0.100+x110 -7 +x or [c][HC 2 H 3 O 2 ]Lots [C 2 H 3 O 2 - ][H 3 O + ] k a = [C 2 H 3 O 2 - ] [H 3 O + ] rearrange the equation [H 3 O + ] = k a [HC 2 H 3 O 2 ] [HC 2 H 3 O 2 ] [C 2 H 3 O - ] Taking the - log of both side - - - log [H 3 O + ] = - log ( k a [HC 2 H 3 O 2 ] / [C 2 H 3 O 2 - ] ) orpH = -log k a - log( [ HC 2 H 3 O 2 ] / [ C 2 H 3 O 2 - ] ) let C a = [ HC 2 H 3 O 2 ] and C b = [ C 2 H 3 O 2 - ] ) therefore pH = pK a - log C a / C b orpH = pK a + log C b / C a This is the Henderson Hasselbach Equation: pH = pK a + log C b / C a or pOH = pK b + log C a / C b

4. Henderson-Hasselbach Equation pH of a solution can be calculated using a useful equation: pH = pK a + log [A - ] / [HA] Where HA & A - are the weak acid and its conjugate and K a is for HA Similarly, pOH = pK b + log [HA] / [A - ] Where HA & A - are the weak base and its conjugate and K b 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 0.100 M HC 2 H 3 O 2 is mix w/ 0.100M NaHCO 3 ? HC 2 H 3 O 2 + H 2 O  C 2 H 2 O 2 - + H 3 O + K a =1.810 -5 M i0.100Lots0.100110 -7  -x-x+x+x [c]0.100-xLots0.100+x110 -7 +x Using the Henderson-Hasselbach equation: pH = - log (4.310 -7 ) + log (0.100 / 0.100) pH = 4.74 + log 1pH = 4.74 Buffer Type Note: When a common ion is present in the same solution, then this is called a Buffer Type of Calculation.

6. Henderson-Hasselbach Equation and Buffer Problems A buffer 0.100 M acetate and 0.200 M acetic acid is prepared (K a = 1.8 10 -5 ). Reger 14.19 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 0.100 M HCl is added to 100.0 mL of the buffer. iii) Calculate the initial pH, final pH, and change in pH that result when 1.00 mL of 0.100 M HCl is added to 100.0 mL of water. Note: HCl = 0.100 M 1.00mL = 0.1 mmol. C 2 H 3 O 2 - =0.100 M 100mL = 10 mmol and HC 2 H 3 O 2 =0.200 M 100mL = 20 mmol i) pH = pKa + log Cb/Ca = -log(1.810 -5 ) + log ( 0.10 / 0.20)  pH = 4.44 ii) C 2 H 3 O 2 - +H 3 O +  HC 2 H 2 O 2 +H 2 O 10mmol0.1 mmol20 mmolLots s 10mmol0.1 mmol20 mmolLots -0.1-0.1+0.1- R -0.1-0.1+0.1- 9.9 020.1Lots f 9.9 020.1Lots 9.9/101020.1/101V T = 101 mL [c] 9.9/101020.1/101V T = 101 mL pH = -log (1.810 -5 )+log [(9.9/101)/ [20.1/101)] = 4.74 - 0.31  pH = 4.43 pH (initial) = 4.44, pH(final) 4.43,  pH(change) = -0.01

7. ...Continue: Henderson-Hasselbach Equation and Buffer Problems...continue A buffer 0.100 M acetate and 0.200 M acetic acid is prepared (K a = 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 0.100 M HCl is added to 100.0 mL of water. Note: HCl = 0.100 M 1.00mL = 0.100 mmol. iii) HCl + H 2 O  H 3 O + +Cl- 0.100 mmol110 -7 M- s 0.100 mmol110 -7 M- -0.100mmol+0.100mmol- R -0.100mmol+0.100mmol- 0 0.100 mmol f 0 0.100 mmol 00.100mmol / 101 mL [c] 00.100mmol / 101 mL [H 3 O + ] = 9.910 -4 M  pH = 3.00 pH (initial) = 7.00, pH(final) 3.00,  pH(change) = -4.00

8. 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 7.35 - 7.45 Buffer system in body - 1. Proteins 2. Phosphates HPO 4 2- / H 2 PO 4 - : 1.6 / 1 3. Carbonates H 2 CO 3 / HCO 3 - : 10 / 1 Reaction: H 3 O + + HCO 3 -  H 2 CO 3 + H 2 O H 2 CO 3  H 2 O + CO 2 (exhale)

9. 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 ComaRIP Causes: 1. Respiratory Acidosis Difficulty Breathing (Hypo-ventilation) Pneumonia, Asthma anything which diminish CO 2 from leaving lungs. 2. Metabolic Acidosis Starvation or fasting Heavy exercise Mechanism: 1. Respiratory Acidosis CO 2 doesn’t leave lungs which result in the build up of H 2 CO 3 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.

10. Alkalosis Blood pH  7.45 (ALKALOSIS) Symptoms: Over simulation of the nervous system Muscle cramps ConvulsionDeath Causes: 1. Respiratory Acidosis Heavy rapid breathing (hyperventilation). Results from - fear, hysteria, fever, infection or reaction with drugs. 2. Metabolic Acidosis Metabolic irregularities or by excess vomiting Mechanism: 1. Respiratory Acidosis Excessive loss of CO 2 lowers H 2 CO 3 and raise HCO 3 - level (Can be remedied by breathing in a bag) 2. Metabolic Acidosis Vomiting removes excess acidic material from stomach. (pH of stomach equals one).

11. Buffer System at Work Buffer - System that resists change in pH when H 3 O + or OH - is added. Buffer solution may be prepared by a weak acid and its conjugate base. How it Works: A -  HA H 3 O + Buffer H 2 O Remember pH = Conc. of H 3 O + Your bloodRxn: HCO 3 -  H 2 CO 3 ExcessH 3 O + + HCO 3 -  H 2 CO 3 + H 2 O H 3 O + CO 2 + H 2 O Excess OH - + H 2 CO 3  HCO 3 - + H 2 O OH -

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