buffer solutions -...

March 181 Common Ion Effect

17.1 Common Ion EffectBuffer Solutions

The resistance of pH change

Dr. Fred Omega GarcesChemistry 201Miramar College


Common Ion EffectIonization 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 ? (Pure) HC2H3O2 + H2O D C2H2O2

- + H3O+ Ka =1.8•10-5 M Solving the iCe problem:ka = =1.8•10-5 M = [C2H3O2

-] [H3O+] /0.10 M g [H3O+]=1.34•10-3 M% a = (1.34•10-3 / 0.10 ) * 100 = 1.34 % pH = 2.87

ii) What is % a if 0.100 M HC2H3O2 is mix w/ 0.100M NaC2H3O2 ? (Buffer) HC2H3O2 + H2O D C2H2O2

- + H3O+ Ka =1.8•10-5 M i 0.100 Lots 0.100 1•10-7

C -x -x +x +x[c] 0.100-x Lots 0.100+x 1•10-7+xka = 1.8•10-5 M = [0.100+ x ] [x] /( 0.100 -x) [0.100] [x] /( 0.100 )

x = [H3O+]= 1.8•10 -5 M pH = 4.74% a = (1.8•10-5 / 0.10 ) * 100 = 0.0180 %

Ionization % decrease in presence of common ion !!


Common Ion Effect EquationConsider the previous problem in which a common ion is in the same solution.

HC2H3O2 + H2O D C2H3O2- + H3O+ Ka =1.8•10-5 M

i 0.100 Lots 0.100 1•10-7

C -x -x +x +x[c] 0.100-x Lots 0.100+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


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-]Where HA & A-

are the weak base and its conjugate and Kb is for A-


Henderson-Hasselbach Equation: Example

Consider the common ion effect problem and lets see how the Henderson-Hasselbachequation can be used to simplify this problem.

What is pH if 0.100 M HC2H3O2 is mix w/ 0.100M NaC2H3O2 ?

HC2H3O2 + H2O D C2H2O2- + H3O+ Ka =1.8•10-5 M

i 0.100 Lots 0.100 1•10-7

C -x -x +x +x[c] 0.100-x Lots 0.100+x 1•10-7+x

Using the Henderson-Hasselbach equation:pH = - log (4.3•10-7 ) + log (0.100 / 0.100) pH = 4.74 + log 1 pH = 4.74 + 0 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.


Henderson-Hasselbach Equation and Buffer Problems (sRF)

A buffer 0.100 M acetate and 0.200 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 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.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) g pH = 4.44

ii) C2H3O2- + H3O+ D HC2H2O2 + H2O

s 10mmol 0.1 mmol 20 mmol LotsR -0.1 -0.1 +0.1 -f 9.9 0 20.1 Lots[c] 9.9/101 0 20.1/101 VT = 101 mL

pH = -log (1.8•10-5)+log [(9.9/101) / [20.1/101)] = 4.74 - 0.31 ® pH = 4.43

pH (initial) = 4.44, pH (final) 4.43, DpH (change) = -0.01


...Continue: Henderson-Hasselbach Equation and Buffer Problems

...continueA buffer 0.100 M acetate and 0.200 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 0.100 M HCl is

added to 100.0 mL of water.Note: HCl = 0.100 M • 1.00mL = 0.100 mmol.

iii) HCl + H2O D H3O+ + Cl-s 0.100 mmol 1•10-7M -R -0.100mmol +0.100mmol -f 0 0.100 mmol[c] 0 0.100mmol / 101 mL

[H3O+] = 9.9•10-4 M g pH = 3.00

pH (initial) = 7.00 , pH(final) 3.00, DpH(change) = -4.00


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.


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 - 7.45Buffer system in body -1. Proteins2. Phosphates HPO4

2- / H2PO4- : 1.6 / 1

3. Carbonates H2CO3 / HCO3- : 10 / 1

Reaction:H3O+ + HCO3

- D� H2CO3 + H2OH2CO3 g H2O + CO2 (exhale)


AcidosisBlood pH i 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 AcidosisStarvation or fastingHeavy exercise

Mechanism:1. Respiratory Acidosis

CO2 doesn�t leave lungs which result in the build up of H2CO3in the blood

2. Metabolic AcidosisIf body doesn�t have enough food then Fatty acids (Fat) are used. Fatty Acids g Acidic. Furthermore, exercise leads muscle to produce lactic acid.


AlkalosisBlood pH h 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 AlkalosisMetabolic irregularities or by excess vomiting

Mechanism:1. Respiratory Alkalosis

Excessive loss of CO2lowers H2CO3 and raise HCO3

- level (Can be remedied by breathing in a bag)

2. Metabolic AlkalosisVomiting removes excess acidic material from stomach. (pH of stomach equals one).


Buffer System at WorkBuffer - 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- g� HA

H3O+ Buffer H2O

Remember pH = Conc. of H3O+

Your blood Rxn: HCO3- D H2CO3

Acidosis Excess H3O+ + HCO3- g H2CO3 + H2O

H3O+ CO2 + H2O

Alkalosis Excess OH- + H2CO3 D HCO3- + H2O

OH-


Summary

Equation / Concept Function1 [H+] [OH-] = Kw Permits the calculation of [H+] or [OH-] when the other is known.2 p X = - log X This equation is the basis of the p-scale.3 pH + pOH = 14.00 This equation shows the relationship between the pH and the pOH4 HA ! H++ A-

Ka =[H+ ] [A −]

[HA]

This is the Mass Action Equation for the ionization of a weak acid inwater. This equation yields the ka given the equilibriumconcentration of all specie. The equation also yields the [H3O+]given the initial concentration of the weak base [HA] and the ka.

5 B + H2O ! HB + OH-

Kb =[HB] [OH−]

[B]

This is the Mass Action Equation for the ionization of a weak basein water. This equation yields the kb given the equilibriumconcentration of all specie. The equation also yields the [OH-]given the initial concentration of the weak base [B] and the kb.

6 Percent ionization (α)

α =amount ionizedinitial amount

×100%

The percent ionization can be calculated from the initialconcentration of the acid (or base) and the change in theconcentration of the ions. Given the percent ionization (α) and thepH, the ka (or kb) can be determined.

7 Ka • Kb = Kw This equation relates Ka and Kb for conjugate pairs in aqueoussolution,

8 ID of the solute as :i) only a weak acidii) only a weak baseiii) a mixture of a weak acidand its conjugate base

Identification of the function of the solute leads to the correct MassAction expression and thereby leading to the correct equilibriumlaw. This is a critical first step to solve any acid-base equilibria

9 Identification of acidiccations and basic anions

Identification of function of cation and anion of a salt lead to pH ofthe salt solution. Given the ka or kb of the conjugates of these ionsleads to the calculation of the pH or pOH

10 Assumption whichsimplifies Mass Action

In order to simplify the math calculation of a Mass Actionexpression, assumption can be made base on the ka or kb value.

11 Reactions when H+ or OH-are added to a buffersolution.

Understanding the buffer reaction permits the determination of theeffect of a strong acid or strong base on the pH of the solution.Adding H+ lowers the [A-] and raises [HA], adding OH- lowers[HA] and raises [A-].

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

buffer solutions -...

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