Monoprotic AcidBase Equilibria Review of Fundamentals 1 Acids

Monoprotic Acid-Base Equilibria Review of Fundamentals 1. ) Acids and Bases are essential to virtually every application of chemistry Ø Ø Analytical procedures such as chromatography and electrophoresis Protein purification, chemical reactions, environmental issues Pollutants Contribute to Acid Rain Forest Destruction Urban Stone Decay Yellowstone Air Pollution (same view)

Monoprotic Acid-Base Equilibria Protein activity is p. H dependent Review of Fundamentals 2. ) Knowledge of Acid-Base Equilibrium is Necessary to Understand: Ø Ø Buffer preparation and p. H control Acid-Base Titrations Complexation, etc. Illustration: Protein function and stability depends on p. H, temperature and other conditions. - Blood and milk have high protein concentrations - Common problem in mechanical heart valve is clot formation - Protein precipitation from milk provides a cheap and easy mimic of blood clotting for testing new heart valves avoids expensive and long animal studies Blood clot formation on heart valve Milk precipitation on heart valve

Monoprotic Acid-Base Equilibria Review of Fundamentals 3. ) Strong Acids and Bases Ø Completely dissociates Ø [H 3 O+] or [OH-] equals concentration of strong acid or base - What is the p. H of a 0. 1 M solution of HCl? - What is the p. H of a 0. 1 M solution of KOH?

Monoprotic Acid-Base Equilibria Review of Fundamentals 3. ) Strong Acids and Bases Ø p. H at other concentrations of a strong base [OH-] (M) [H+] (M) p. H 1 x 10 -13 13. 0 1 x 10 -2 1 x 10 -12 12. 0 1 x 10 -3 1 x 10 -11 11 1 x 10 -4 1 x 10 -10 10 1 x 10 -5 1 x 10 -9 9 Ø Acid Ka HCl 103. 9 HBr 105. 8 HI 1010. 4 HNO 3 101. 4 Relationship between p. H and p. OH:

Monoprotic Acid-Base Equilibria Review of Fundamentals 3. ) Strong Acids and Bases Ø Dilemma: What is the p. H of 1. 0 x 10 -8 M KOH? How can a base produce an acidic solution? Ø Wrong Assumption!!

Monoprotic Acid-Base Equilibria Review of Fundamentals 3. ) Strong Acids and Bases Ø Wrong Assumption!! For large concentration of acid or base, [H+] = [acid] or [OH-] = [base] For small concentration, must account for water dissociation In pure water [OH-] = 1. 0 x 10 -7 M, which is greater than [KOH] = 1 x 10 -8 M Must Use Systematic Treatment of Equilibrium

Monoprotic Acid-Base Equilibria Review of Fundamentals 3. ) Strong Acids and Bases Ø Systematic Treatment of Equilibrium Step 1: Pertinent reactions: Kw Completely dissociates, not pertinent Step 2: Charge Balance: Step 3: Mass Balance: All K+ comes from KOH

Monoprotic Acid-Base Equilibria Review of Fundamentals 3. ) Strong Acids and Bases Ø Systematic Treatment of Equilibrium Step 4: Equilibrium constant expression (one for each reaction): Step 5: Count equations and unknowns: Three equations: (1) (2) (3) Three unknowns:

Monoprotic Acid-Base Equilibria Review of Fundamentals 3. ) Strong Acids and Bases Ø Systematic Treatment of Equilibrium Step 6: Solve (Seeking p. H [H+]): Set [H+] =x, and substitute mass balance equation into charge balance equation: From mass balance Substitute OH- equation into equilibrium equation:

Monoprotic Acid-Base Equilibria Review of Fundamentals 3. ) Strong Acids and Bases Ø Systematic Treatment of Equilibrium Step 6: Solve (Seeking p. H [H+]): Solve the quadratic equation: Use quadratic equation Negative number is physically meaningless p. H slightly basic, consistent with low [KOH]

Monoprotic Acid-Base Equilibria Review of Fundamentals 3. ) Strong Acids and Bases Ø Ø Systematic Treatment of Equilibrium Three Regions depending on acid/base concentrations High concentrations (≥ 10 -6 M), p. H considered just from the added H+, OH- intermediate concentrations, (106 -10 -8 M), H O ionization ≈ H+, OH 2 systematic equilibrium calculation necessary low concentrations (≤ 10 -8 M), p. H=7 not enough H+, OH- added to change p. H

Monoprotic Acid-Base Equilibria Review of Fundamentals 4. ) Water Almost Never Produces 10 -7 M H+ and OHØ Ø p. H=7 only true for pure water Any acid or base suppresses water ionization Follows Le Châtelier’s principal In 10 -4 M HBr solution, water dissociation produces only 10 -10 M OH- and H+

Monoprotic Acid-Base Equilibria Review of Fundamentals 5. ) Weak Acids and Bases Ø Weak acid/base do not completely dissociate Dissociation Ka for the acid HA: Ø Base Hydrolysis constant Kb Ø p. K is negative logarithm of equilibrium constant Ø - As Ka or Kb increase p. Ka or p. Kb decrease Smaller p. Ka stronger acid

Monoprotic Acid-Base Equilibria Review of Fundamentals 5. ) Weak Acids and Bases Ø Conjugate acid-base pair – related by the gain or loss of a proton Formic acid (p. Ka 3. 744) stronger acid than benzoic acid (p. Ka=4. 202) Acid-base pair - - Conjugate base of a weak acid is a weak base Conjugate acid of a weak base is a weak acid Conjugate base of a strong acid is a very weak base or salt

Monoprotic Acid-Base Equilibria Weak Acid Equilibria 1. ) Any Effect that Increases the Stability of the Product Drives the Reaction Forward Ø Formation of internal hydrogen bond for an acid/base Can’t form internal hydrogen bond Ortho isomer is stronger acid because product forms internal hydrogen bond

Monoprotic Acid-Base Equilibria Weak Acid Equilibria 2. ) General Systematic Treatment of Equilibrium Ø Ø Unlike concentrated strong acid, need to account for water ionization Find p. H for a solution of a general weak acid HA Step 1: Pertinent reactions: Ka Kw Step 2: Charge Balance: Step 3: Mass Balance: F – formal concentration of acid Step 4: Equilibrium constant expression (one for each reaction):

Monoprotic Acid-Base Equilibria Weak Acid Equilibria 2. ) General Systematic Treatment of Equilibrium Ø Find p. H for a solution of a general weak acid HA Step 5: Count equations and unknowns: Four Equations: (1) (2) (4) Four Unknowns: Step 6: Solve (Not easy to solve cubic equation results!): - Again, need to make assumptions to simplify equations - The goal is to determine [H+], so we can measure p. H (3)

Monoprotic Acid-Base Equilibria Weak Acid Equilibria 2. ) General Systematic Treatment of Equilibrium Ø Find p. H for a solution of a general weak acid HA Step 6: Solve (Not easy to solve cubic equation results!): Make Some Initial Assumptions: For a typical weak acid, [H+] from HA will be much greater than [H+] from H 2 O Ø If dissociation of HA is much greater than H 2 O, [H+] >> [OH-] Ø Set [H+]=x: substitute

Monoprotic Acid-Base Equilibria Weak Acid Equilibria 2. ) General Systematic Treatment of Equilibrium Ø Find p. H for a solution of a general weak acid HA Step 6: Solve (Not easy to solve cubic equation results!): Substitute into Equilibrium Equation: Rearrange: Solve quadratic equation:

Monoprotic Acid-Base Equilibria Weak Acid Equilibria 2. ) General Systematic Treatment of Equilibrium Ø Find p. H for a solution of a general weak acid HA Step 7: Verify Assumption: Was the approximation [H+] ≈ [A-] justified ([H+] >>[OH-])? Setting F = 0. 050 M and Ka = 1. 07 x 10 -3 for o-hydroxybenzoic acid: Determine [OH-] from water dissociation: [H+] >> [OH-] 6. 8 x 10 -3 M >> 1. 5 x 10 -12 M assumption is justified!

Monoprotic Acid-Base Equilibria Weak Acid Equilibria 3. ) Fraction of Dissociation Ø Fraction of acid HA in the form A-(a): Ø Example: What is the percent fraction dissociation for F = 0. 050 M and Ka = 1. 07 x 10 -3 for o-hydroxybenzoic acid? Ø Percent dissociation increases with dilution

Monoprotic Acid-Base Equilibria Weak Base Equilibria 1. ) Treatment of Weak Base is Very Similar to Weak Acid Ø Assume all OH- comes from base and not dissociation of water Step 1: Pertinent reactions: Kb Kw Step 2: Charge Balance: Step 3: Mass Balance: F – formal concentration of base Step 4: Equilibrium constant expression (one for each reaction):

Monoprotic Acid-Base Equilibria Weak Base Equilibria 1. ) Treatment of Weak Base is Very Similar to Weak Acid Ø Assume all OH- comes from base and not dissociation of water Step 6: Solve (Assume [BH+] >> [H+] [BH+] ≈ [OH-]): Set [OH-]=x and substitute into Equilibrium Equation: Rearrange: Solve quadratic equation:

Monoprotic Acid-Base Equilibria Weak Base Equilibria 2. ) Example What is the p. H of cocaine dissolved in water? F = 0. 0372 M and Kb = 2. 6 x 10 -6 for cocaine? Kb=2. 6 x 10 -4 0. 0372 -x x x

Monoprotic Acid-Base Equilibria Weak Base Equilibria 2. ) Example What is the p. H of cocaine dissolved in water? F = 0. 0372 M and Kb = 2. 6 x 10 -6 for cocaine? Because x=[OH-], we need to solve for [H+]

Monoprotic Acid-Base Equilibria Weak Base Equilibria Consider an Illegal Drug Lab (pictured) 3. ) Illustration Criminal Investigation of the illegal lab will require identifying: What drugs are present? Ø How much of each drug is present? Ø What is the purity of the drugs? Ø p. H change or titration indicator may help identify/quantify drug. (positive field test for cocaine) http: //www. stopmethaddiction. com/meth-lab-photos. htm

Monoprotic Acid-Base Equilibria Weak Base Equilibria 4. ) Fraction of Association Ø Fraction of Base B in BH+ form (a): Ø Example: What is the percent fraction dissociation of cocaine reacted with water? F = 0. 0372 M and Kb = 2. 6 x 10 -6 for cocaine?

Monoprotic Acid-Base Equilibria Weak Acid Base Equilibria 5. ) Example A 0. 0450 M solution of benzoic acid has a p. H of 2. 78. Calculate p. Ka for this acid. What is the percent fraction dissociation?

Monoprotic Acid-Base Equilibria Buffers 1. ) A buffered solution resists changes in p. H when acids or bases are added Ø Buffer: is a mixture of a weak acid and its conjugate base - Ø Must be comparable amounts of acid & base For an organism to survive, it must control the p. H of each subcellular compartments - Enzyme-catalyzed reactions are p. H dependent Bacteria growing in hot springs (acidic p. H) Thermophilic archaea Picrophilus oshimae and Picrophilus torridus grow at p. H =0. 7 (Stomach acid 1 -3 p. H) Nature (London) (1995), 375(6534), 741 -2. Bacteria growing in lung tissues (neutral p. H)

Monoprotic Acid-Base Equilibria Buffers 2. ) Mixing a Weak Acid and Its Conjugated Base Ø Ø Ø Very little reaction occurs Very little change in concentrations Example: Consider a 0. 10 M of acid with p. Ka of 4. 00 0. 10 -x x x

Monoprotic Acid-Base Equilibria Buffers 2. ) Mixing a Weak Acid and Its Conjugated Base Ø Example: Consider adding 0. 10 M of conjugate base with p. Kb of 10. 00 0. 10 -x x x HA dissociates very little and A- reacts very little with water

Monoprotic Acid-Base Equilibria Buffers 3. ) Henderson-Hasselbalch Equation Ø Rearranged form of Ka equilibrium equation: Take log of both sides: rearrange: p. H p. Ka

Monoprotic Acid-Base Equilibria Buffers [A-]/[HA] 3. ) Henderson-Hasselbalch Equation Ø Determines p. H of buffered solution - - Need to know ratio of conjugate [acid] and [base] Ø 100: 1 p. Ka + 2 10: 1 p. Ka + 1 1: 1 p. Ka 1: 10 p. Ka - 1 1: 100 p. Ka - 2 If [A-] = [HA], p. H = p. Ka - p. H All equilibria must be satisfied simultaneously in any solution at equilibrium Only one concentration of H + in a solution Similar equation for weak base and conjugate acid p. Ka is for this acid

Monoprotic Acid-Base Equilibria Buffers 3. ) Henderson-Hasselbalch Equation Ø A strong acid and a weak base react “completely” to give the conjugate acid: Weak base Ø Strong acid conjugate acid Also, a strong base and a weak acid react “completely” to give the conjugate base: Weak acid Strong base conjugate base

Monoprotic Acid-Base Equilibria Buffers 3. ) Henderson-Hasselbalch Equation Ø Example: Calculate how many milliters of 0. 626 M KOH should be added to 5. 00 g of MOBS to give a p. H of 7. 40? p. Ka = 7. 48 FW = 223. 29 What is the p. H if an additional 5 m. L of the KOH solution is added?

Monoprotic Acid-Base Equilibria Buffers 4. ) Why Does a Buffer Resist Changes in p. H? Ø Ø Ø Strong acid or base is consumed by B or BH+ Maximum capacity to resist p. H change occurs at p. H=p. Ka Buffer Capacity (b): measure of a solutions resistance to p. H change where Ca and Cb are the number of moles of strong acid and strong base per liter needed to produce a unit change in p. H 5. ) Choosing a Buffer Ø Ø Ø Choose a buffer with p. Ka as close as possible to desired p. H Useful buffer range is p. Ka ± 1 p. H units Buffer p. H depends on temperature and ionic strength activity coefficients

When preparing a buffer, you need to monitor the p. H. Can not assume the added HA and A- will yield the desired p. H dependent on: - activity - temperature - ionic strength

Wide number of buffers available that cover an essential complete range of p. Hs. Choose a buffer with a p. Ka as close as possible to the desired p. H.