17.1 Common Ion Effect Buffer Solutions The resistance of pH change Dr. Fred Omega Garces Chemistry 201 Miramar College Common Ion Effect 2/28/20 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 0.100 M acetic acid ? (Pure) HC2H3O2 + H2O C2H2O2- + H3O+ Ka =1.810-5 M Solving the iCe problem: ka = =1.810-5 M = [C2H3O2-] [H3O+] /0.10 M [H3O+]=1.3410-3 M % = (1.3410-3 / 0.10 ) * 100 = 1.34 % pH = 2.87 ii) What is % if 0.100 M HC2H3O2 is mix w/ 0.100M NaC2H3O2 ? (Buffer) HC2H3O2 + H2O C2H2O2- + H3O+ i 0.100 Lots 0.100 110-7 C -x [c] 0.100-x -x +x Ka =1.810-5 M
+x Lots 0.100+x 110-7+x ka = 1.810-5 M = [0.100+ x ] [x] /( 0.100 -x) x = [H3O+]= 1.810 -5 [0.100] [x] /( 0.100 ) M pH = 4.74 % = (1.810-5 / 0.10 ) * 100 = 0.0180 % Ionization % decrease in presence of common ion !! Common Ion Effect 2/28/20 Common Ion Effect Equation Consider the previous problem in which a common ion is in the same solution. HC2H3O2 + or H2O C2H3O2- + i 0.100 Lots 0.100 110-7 C -x +x +x
[c] 0.100-x Lots 0.100+x [c] [HC2H3O2 ] Lots ka = [C2H3O2- ] [H3O+] -x [C2H3O2- ] H3O+ Ka =1.810-5 M 110-7+x [H3O+ ] rearrange the equation [H3O+] = ka [HC2H3O2 ] [HC2H3O2] [C2H3O- ] Taking the - log of both side - - log [H3O+] or pH let Ca = [HC2H3O2] or = - log (ka [HC2H3O2 ] / [C2H3O2-] ) = -log ka and Cb =
log( [HC2H3O2] / [C2H3O2 - ] ) [C2H3O2 - ] ) therefore pH = pKa - log Ca / Cb pH = pKa + log Cb / Ca This is the Henderson Hasselbach Equation: pH = pKa + log Cb / Ca or Common Ion Effect pOH = pKb + log Ca / 2/28/20 Henderson-Hasselbach Equation pH of a solution can be calculated using a useful equation: pH = pKa + log [A-] / [HA] Where HA & Aare the weak acid and its conjugate and Ka is for HA Similarly, pOH = pKb + log [HA] / [A-] Where HA & Aare the weak base and its conjugate and Kb is for A- Common Ion Effect 2/28/20 Henderson-Hasselbach Equation: Example Consider the common ion effect problem and lets see how the HendersonHasselbach equation can be used to simplify this problem. What is pH if 0.100 M HC2H3O2 is mix w/ 0.100M NaC2H3O2 ? HC2H3O2 + H2O C2H2O2- + H3O+ Ka =1.810-5 M i 0.100 Lots 0.100
C -x +x [c] 0.100-x -x 110-7 +x Lots 0.100+x 110-7+x Using the Henderson-Hasselbach equation: pH = - log (4.310-7 ) + 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. Common Ion Effect 2/28/20 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(1.810 -5) + log ( 0.10 / 0.20) pH = 4.44 C2H3O2-
and Buffer Problems ...continue A 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. HCl + iii) H2O H3O+ + s 0.100 mmol 110-7M R -0.100mmol +0.100mmol f 0 [c] 0 Cl- - 0.100 mmol 0.100mmol / 101 mL [H3O+] = 9.910-4 M
pH = 3.00 pH (initial) = 7.00 , pH(final) 3.00, pH(change) = -4.00 Common Ion Effect 2/28/20 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 HendersonHasselbalch equation. A buffer has an effective range of pKa + 1 pH unit. Common Ion Effect 2/28/20 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 HPO42- / H2PO4- : 1.6 / 1 3. Carbonates H2CO3 / HCO3 - : 10 / 1 Reaction: H3O+ + HCO3H2CO3 H2CO3 + H2O H2O + CO2 (exhale)
Common Ion Effect 2/28/20 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: Mechanism: 1. Respiratory 1. Respiratory Acidosis Acidosis CO2 doesnt leave lungs Difficulty Breathing which result in the build (Hypo-ventilation) up of H2CO3 in the blood Pneumonia, Asthma 2. Metabolic Acidosis anything which diminish CO2 from If body doesnt have enough food then Fatty leaving lungs. acids (Fat) are used. 2. Metabolic Acidosis Fatty Acids Acidic. Starvation or fasting Furthermore, exercise Heavy exercise leads muscle to produce lactic acid. Common Ion Effect 2/28/20
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 Common Ion Effect 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). 2/28/20
Buffer System at Work 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 Acidosis Excess Rxn: HCO3- H3O+ + HCO3- H3O+ Alkalosis Excess OH- H2CO3 H2CO3 + H2O CO2 + H2O OH- + H2CO3 HCO3- + H2O Common Ion Effect 2/28/20 Equation / Concept Summar y The following summary lists the important tools needed to solve problems dealing with
acid-base equilibria. Function 1 [H+] [OH-] = Kw Permits the calculation of [H +] or [OH-] when the other is known. 2 3 p X = - log X pH + pOH = 14.00 HA D H++ A[H + ] [A - ] K a = [H A ] This equation is the basis of the p-scale. This equation shows the relationship between the pH and the pOH This is the Mass Action Equation for the ionization of a weak acid in water. This equation yields the ka given the equilibrium concentration of all specie. The equa tion also yields the [H3O+] given the initial concentration of the weak base [HA] and the k a. This is the Mass Action Equation for the ionization of a weak base in water. This equation yields the k b given the equilibrium concentration of all specie. The equa tion also yields the [OH-] given the initial concentration of the weak base [B] and the k b. The percent ionization can be calculated from the initial concentration of the acid (or base) and th e change in the concentration of the ions. Given the percent ionization ( ) and the pH, the ka (or kb) can be determined. This equation relates Ka and Kb for conjugate pairs in aqueous solution, Identification of the function of the solute leads to the correct Mass Action expression and thereby leading to the correct equ ilibrium law. This is a critical first step to solve any acid-base equilibria 4 5 B + H 2O D HB + OH[H B ] [O H - ] K b = [B ] 6
Percent ionization () a m o u n t io n iz e d = 100% in itia l a m o u n t 7 Ka K b = Kw 8 ID of the solute as : i) only a weak acid ii) only a weak base iii) a mixture of a weak acid and its conjugate base Identification of acidic cations and basic anions 9 10 Assumption which simplifies Mass Action 11 Reactions when H+ or OHare added to a buffer solution. Common Ion Effect Identification of function of cation and anion of a salt lead to pH of the salt solution. Given the k a or kb of the conjugates of these ions leads to the calculation of the pH or pOH In order to simplify the math calculation of a Mass Ac tion expression, assumption can be made base on the ka or kb value. Understanding the buffer reaction permits the determination of the effect 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-]. 2/28/20
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