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

Monoprotic Acid-Base Equilibria Review of Fundamentals 2.)Knowledge of Acid-Base Equilibrium is Necessary to Understand:  Buffer preparation and pH control  Acid-Base Titrations  Complexation, etc.  Illustration: Blood clot formation on heart valve Milk precipitation on heart valve Protein function and stability depends on pH, 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 Protein activity is pH dependent

Monoprotic Acid-Base Equilibria Review of Fundamentals 3.)Strong Acids and Bases  Completely dissociates  [H 3 O + ] or [OH - ] equals concentration of strong acid or base - What is the pH of a 0.1M solution of HCl? - What is the pH of a 0.1M solution of KOH?

Monoprotic Acid-Base Equilibria Review of Fundamentals 3.)Strong Acids and Bases  pH at other concentrations of a strong base  Relationship between pH and pOH: [OH - ] (M)[H + ] (M)pH 1x x x x x x x x x x AcidKaKa HCl HBr HI HNO

Monoprotic Acid-Base Equilibria Review of Fundamentals 3.)Strong Acids and Bases  Dilemma: What is the pH of 1.0x10 -8 M KOH? How can a base produce an acidic solution?  Wrong Assumption!!

Monoprotic Acid-Base Equilibria Review of Fundamentals 3.)Strong Acids and Bases  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 - ] = 1.0x10 -7 M, which is greater than [KOH] = 1x10 -8 M Must Use Systematic Treatment of Equilibrium

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

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

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

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

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

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

Monoprotic Acid-Base Equilibria Review of Fundamentals 5.)Weak Acids and Bases  Conjugate acid-base pair – related by the gain or loss of a proton -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 Acid-base pair Formic acid (pK a 3.744) stronger acid than benzoic acid (pK a =4.202)

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

Monoprotic Acid-Base Equilibria Weak Acid Equilibria 2.)General Systematic Treatment of Equilibrium  Unlike concentrated strong acid, need to account for water ionization  Find pH for a solution of a general weak acid HA Step 1: Pertinent reactions: KwKw KaKa 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  Find pH for a solution of a general weak acid HA Step 5: Count equations and unknowns: Four Equations: Four Unknowns: (1) (4) (2) (3) 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 pH

Monoprotic Acid-Base Equilibria Weak Acid Equilibria 2.)General Systematic Treatment of Equilibrium  Find pH 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  If dissociation of HA is much greater than H 2 O, [H + ] >> [OH - ] Set [H + ]=x: substitute

Monoprotic Acid-Base Equilibria Weak Acid Equilibria 2.)General Systematic Treatment of Equilibrium  Find pH 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  Find pH for a solution of a general weak acid HA Step 7: Verify Assumption: Was the approximation [H + ] ≈ [A - ] justified ([H + ] >>[OH - ])? Setting F = M and K a = 1.07x10 -3 for o-hydroxybenzoic acid: Determine [OH - ] from water dissociation: [H + ] >> [OH - ] 6.8x10 -3 M >> 1.5x M assumption is justified!

Monoprotic Acid-Base Equilibria Weak Acid Equilibria 3.)Fraction of Dissociation  Fraction of acid HA in the form A - (  ):  Example:  Percent dissociation increases with dilution What is the percent fraction dissociation for F = M and K a = 1.07x10 -3 for o-hydroxybenzoic acid?

Monoprotic Acid-Base Equilibria Weak Base Equilibria 1.)Treatment of Weak Base is Very Similar to Weak Acid  Assume all OH - comes from base and not dissociation of water Step 1: Pertinent reactions: KwKw KbKb 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  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 pH of cocaine dissolved in water? F = M and K b = 2.6x10 -6 for cocaine? x x x K b =2.6x10 -4

Monoprotic Acid-Base Equilibria Weak Base Equilibria 2.)Example What is the pH of cocaine dissolved in water? F = M and K b = 2.6x10 -6 for cocaine? Because x=[OH - ], we need to solve for [H + ]

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

Monoprotic Acid-Base Equilibria Weak Base Equilibria 4.)Fraction of Association  Fraction of Base B in BH + form (  ):  Example: What is the percent fraction dissociation of cocaine reacted with water? F = M and K b = 2.6x10 -6 for cocaine?

Monoprotic Acid-Base Equilibria Weak Acid Base Equilibria 5.)Example A M solution of benzoic acid has a pH of Calculate pK a for this acid. What is the percent fraction dissociation?

Monoprotic Acid-Base Equilibria Buffers 1.)A buffered solution resists changes in pH when acids or bases are added  Buffer: is a mixture of a weak acid and its conjugate base - Must be comparable amounts of acid & base  For an organism to survive, it must control the pH of each subcellular compartments - Enzyme-catalyzed reactions are pH dependent Thermophilic archaea Picrophilus oshimae and Picrophilus torridus grow at pH =0.7 (Stomach acid 1-3 pH) Nature (London) (1995), 375(6534), Bacteria growing in hot springs (acidic pH) Bacteria growing in lung tissues (neutral pH)

Monoprotic Acid-Base Equilibria Buffers 2.)Mixing a Weak Acid and Its Conjugated Base  Very little reaction occurs  Very little change in concentrations  Example: Consider a 0.10 M of acid with pK a of x x x

Monoprotic Acid-Base Equilibria Buffers 2.)Mixing a Weak Acid and Its Conjugated Base  Example: Consider adding 0.10 M of conjugate base with pK b of x x x HA dissociates very little and A - reacts very little with water

Monoprotic Acid-Base Equilibria Buffers 3.)Henderson-Hasselbalch Equation  Rearranged form of K a equilibrium equation: Take log of both sides: rearrange: pH pKa

Monoprotic Acid-Base Equilibria Buffers 3.)Henderson-Hasselbalch Equation  Determines pH of buffered solution - Need to know ratio of conjugate [acid] and [base] - If [A - ] = [HA], pH = pK a - 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 pKa is for this acid [A - ]/[HA]pH 100:1pK a :1pK a + 1 1:1pKapKa 1:10pK a - 1 1:100pK a - 2

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

Monoprotic Acid-Base Equilibria Buffers 3.)Henderson-Hasselbalch Equation  Example: Calculate how many milliters of M KOH should be added to 5.00 g of MOBS to give a pH of 7.40? What is the pH if an additional 5 mL of the KOH solution is added? FW = pK a = 7.48

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

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

Wide number of buffers available that cover an essential complete range of pHs. Choose a buffer with a pK a as close as possible to the desired pH.