Monoprotic Acid-Base Equilibria Monoprotic Weak Acids Monoprotic Weak Bases Fraction of...

Monoprotic Acid-Base Equilibria

• Monoprotic Weak Acids

• Monoprotic Weak Bases

• Fraction of Dissociation-Association

• Salts of Weak Acids

• Buffers

Weak AcidsWeak Acid an acid that is not completely

dissociated in aqueous solution

Weak Acid Equilibrium

HA H+ + A-

weak acid conjugate base

Ka= Acid Dissociation Constant

[H+][A-]

[HA]

(Available in Appendix G)

Weak Bases

Weak Base a base in which the hydrolysis reaction does not go to completion in aqueous solution

Weak Base Equilibrium

B + H2O BH+ + OH-

weak base conjugate acid

Kb = Base Hydrolysis Constant

[BH+][OH-]

[B]

Conjugate Relationships

For conjugate acid-base pairs

KaKb=Kw

pKa +pKb=14 (at 25°C)

This is a very useful relation to determine Kb when you can look up Ka

Conjugate Relationships

The conjugate base of a weak acid is a weak base

• Stronger weak acid Weaker conjugate base

• Weaker weak acid Stronger conjugate base

The conjugate acid of a weak base is a weak acid

• Stronger weak base Weaker conjugate acid

• Weaker weak base Stronger conjugate acid

Monoprotic Weak Acid Problems

Method

Write pertinent equations for all equilibria

Write expression for mass balance

Write expression for the charge balance


Equilibria

HA H+ + A- Ka = [H+][A-]

[HA]

H2O H+ + OH- KW = [H+][OH-]

Mass Balance

F = formal concentration the total amount of a compound dissolved in 1 L of solution

F = [HA] + [A-] [HA] = F - [A-]


Charge Balance

[H+] = [A-] + [OH-]

Approximation (a very good one)

Assume that the weak acid reaction products will dominate over the water dissociation reaction in terms of product concentrations

so [A-] >> [OH-] because the weak acid reaction dominates over the water dissociation reaction

Charge balance equation becomes

[H+] [A-]


Using our equilibrium expression

Ka = [H+][A-] = [H+][H+] = [H+][H+]

[HA] [HA] F - [A-]

so let x = [H+]

and Ka = x2

F - x

Monoprotic Weak Acid Example

Consider placing o-hydroxybenzoic acid in aqueous solution at a formal concentration of F = 0.050 F

Ka = 1.07 x 10-3

What is the [H+] ?

Ka = x2 x2 + Ka x - Ka F = 0

F - x

so using the quadratic equation

[H+] =

= 6.80 x 10-3 M and this is ~ [A-]

2

FK4 K K- x aaa

2

2(1)

)(0.05)10 x (1.074)10 x (1.07)10 x (1.07 3-3-3- x

2

Monoprotic Weak Acid Example

What is the [OH-] ?

[OH-] = 10-14 = 10-14 = 1.47 x 10-12 M

[H+] 6.80 x 10-3

Note our approximation [A-] >> [OH-] is justified!

Another useful approximation is to neglect x in the denominator iff x < 0.01 F

Ka = x2 x2

F - x F

Try this with the o-hydroxybenzoic acid problem-

Does this meet our criterion? 0.00680 < (0.01)(0.050) NO

Monoprotic Weak Base Problems

Weak base - similar to the weak acid approach except we solve for [OH-]

Equilibria

B + H2O BH+ + OH- Kb = [BH+][OH-]

[B]

H2O H+ + OH- KW = [H+][OH-]


Mass Balance

F = [BH+] + [B] [B] = F - [BH+]

Charge Balance

[BH+] + [H+] = [OH-]

In the weak base case we want to solve for [OH-]


Approximation (a very good one)

Assume that the weak base reaction products will dominate over the water dissociation reaction in terms of product concentrations

so [BH+] >> [H+] because the weak base reaction dominates over the water dissociation reaction

Charge balance equation becomes

[BH+] [OH-]


Using our equilibrium expression

Kb = [BH+][OH-] = [OH-][OH-] = [OH]2

[B] [B] F - [BH+]

so let x = [OH-]

and Kb = x2

F - x


[OH-] =

Again a useful approximation is to neglect x in the denominator iff x < 0.01 F

Kb = x2 x2

F - x F

2

FK4 K K- x bbb

2

Fraction of Dissociation (for an acid)

Fraction of Dissociation the fraction of acid that is present in the form A-

= [A-] = x

[A-] + [HA] x + (F -x)

= x F

Fraction of Association (for a base)

Fraction of Association the fraction of base that has reacted with water

= [BH+] = x

[BH+] + [B] x + (F -x)

= x F

Salt of Weak AcidsSodium acetate dissolved in water

Salts completely dissociate in solution

Na+CH3COO- Na+ + CH3COO-

In solution

CH3COO- + H20 CH3COOH + OH-

Ka=1.75 x 10 -5

A- + H20 HA + OH-

Salt of Weak AcidsWhat if we dissolve F of Na+CH3OO- in solution?

How would we find the pH?

If you dissolve the salt of a weak acid the result is a weak base equilibrium problem.

Equilibria

CH3COO- + H20 CH3COOH + OH-

Kb = [HA][OH-] = KW = 1.01 x 10-14 = 5.77 x 10-10

[A-] Ka 1.75 x 10-5

H2O H+ + OH- KW = [H+][OH-]

Salt of Weak AcidsMass Balance

F = [HA] + [A-] [A-] = F - [HA]

Charge Balance

[H+] = [OH-] + [A-]

Weak base case we want to solve for [OH-]

Kb = [HA][OH-] = KW = 1.01 x 10-14 = 5.77 x 10-10

[A-] Ka 1.75 x 10-5

[HA][OH-] = x2 = 5.77 x 10-10

[A-] F - x

Solve for x = [OH-]

Buffer SolutionsBuffer a solution that resists changes in pH when

acid or base is added or when the solution is diluted

Typical Buffer

• A mixture of an acid and base (salt of the acid) conjugate pair which resists changes in pH

• The weak acid and conjugate base should have similar concentrations (10:1)

• Buffers break down at extremes of pH or dilution

• Buffers are most effective near the pKa of the acid

Buffer SolutionsHow do mixtures of weak acids and their conjugate

bases act as buffers?

Consider a weak acid of F =0.10 M and Ka =1.0 x 10-

4

HA + H2O H3O+ + A- Ka = [H+][A-] = 1.0 x 10-4

0.1-x x x [HA]

Recall Ka = x2

F - x

Solve for x

x = 3.1 x 10-3 M

= x/F = (3.1 x 10-3)/0.1 = 0.31 3.1 % dissociated

Addition of additional A- makes HA dissociate less!

Buffer SolutionsConsider a weak base of F =0.10 M and Kb =1.0 x 10-10

A- + H2O HA + OH- Kb = [HA][OH-] = 1.0 x 10-4

0.1-x x x [A-]

Recall Kb = x2

F - x

Solve for x

x = 3.2 x 10-6 M

= x/F = (3.2 x 10-6)/0.1 = 3.2 x 10-5

0.0032 % associated

Addition of additional HA makes A- associate less!

Buffer SolutionsThe concentration of HA and A- in solution are

similar to the amounts added to solution

A buffer acts through the reaction of added base with excess HA and the reaction of added acid with excess A-

This is how it resists changes in pH

As more A- is formed from the former reaction or as more HA is formed from the latter reaction

the pH changes very little until we have a greater ratio than 1:10 or 10:1 between the

acid and its salt

Henderson-Hasselbach EquationHenderson-Hasselbach Equation

HA + H2O H3O+ + A- Ka = [H+][A-] [HA]

[H+] = Ka [HA]

[A-]

-log [H+] = -log Ka - log [HA] + log [A-]

pH = pKa + log [A-] Use formal concentrations for [HA] all species

pH = pKa + log [B] Use the Ka for the acid form

[BH+]

Henderson-Hasselbalch EquationHenderson-Hasselbalch Equation

If [HA] = [A-] then pH = pKa

Effect of the [A-]/[HA] ratio on pH (Table 10-1)

[A-]/ [HA] pH100:1 pKa +210:1 pKa +11:1 pKa

1:10 pKa - 11:100 pKa -2

In any mixture of 5 different acids and bases in solution there will only be a single pH and the 5 Henderson-Hasselbalch Equations must be simultaneously met

Buffers Doing Their Job

We add the acid form to a solution as the chloride salt

(BH+Cl-)

Consider the tris buffer system

tris(hydroxymethyl)aminomethane

This salt completely dissociates when placed in solution

BH+Cl- BH+ + Cl-

to form the weak acid


What is the pH of a solution prepared by dissolving 12.43 g of tris (F.W. = 121.136) to 4.67 g of tris hydrochloride (F.W. = 157.597) in 1.00 L of H2O?

Our equilibrium looks like

BH+ B + H+

[BH+] = [4.67 g(1 mol/157.597 g)]/ 1L = 0.0296 M

[B] = [12.43 g(1 mol/121.136 g)]/ 1L = 0.1026 M

pH = pKa + log [B] = 8.075 + log 0.1026 = 8.61

[BH+] 0.0296

NOTICE : Volume does not matter! You can use the number of moles of acid and conjugate base directly


The reactions written above lie completely to the right (i.e. we do not need an arrow written to the left) and only cease when one of the reactants are completely exhausted

Weak Base + Strong Acid react completely

B + H+ BH+ K = [BH+] = 1

[B] [H+] Ka

Weak Acid + Strong Base react completely

HA + OH- A- + H2O K = [A-] = 1

[HA] [OH-] Kb


One of the two reactants is the limiting reagent and the system reacts until that limiting reagent is completely consumed

Here the mol of H+ from HCl = (0.012 L)(1.00 M) = 0.012 mol

Recall we have 0.1026 mol of B so HCl is the limiting reagent

What is the pH resulting from addition of 12.0 mL of 1.00 M HCl to the tris buffer in the previous problem?

Addition of strong acid reduces the amount of tris base B + H+ BH+

weak base strong acid


pH = pKa + log molesB = 8.075 + log 0.0906 = 8.41

molesBH+ 0.0416

NOTICE : Again volume does not matter!

Now we can use Henderson-Hasselbalch

B + H+ BH+

Initial moles 0.1026 0.0120 0.0296

Final moles 0.0906 -------- 0.0416

Buffer Solutions

A buffer is the most effective at a pH = pKa

How do we prepare a buffer in the lab?• Calculate how many grams (mol) of weak acid to add to make final volume (V) and dilute with H2O to approximately 0.75V

• Monitor the pH

• Add concentrated base dropwise to adjust pH to desired value

• Pour solution into a volumetric flask and rinse initial container with H2O several times into volumetric flask

• Dilute with H2O to desired volume and mix well

Buffer Capacity

Buffer Capacity a measure of how well a solution resists changes in pH when adding

strong acid or strong base

Buffer Capacity = (d Cb/d pH) = - (d Ca/d pH)

where Cb and Ca are the number of moles required to change the pH by 1 unit

0.100 F solution of Ha with a pKa = 5.00

The top curve indicates the concentration of strong base necessary to yield the indicated pH

The bottom curve is the derivative of the top curve (the buffer capacity)

Buffer Solution Facts

• When choosing a buffer you should choose the system with a pKa that is closest to the desired pH

• Buffers are “good buffers” over a range of pH’s corresponding to pKa - 1 to pKa + 1

• Outside of this range there is not enough A- to react with strong acid or HA to react with strong base

• Realize that just as with all thermodynamic equilibrium constants the “true” Henderson-Hasselbalch Equation should contain activities rather than concentrations

• This leads to a dependence of the pKa and solution pH on temperature and the ionic strength of the solution

Buffer Solutions Assignment

In addition to doing all of the assigned problems, look over Table 10-2 on pp. 230-231

Make sure that for each buffer listed you can:

identify the acid and base forms of the buffer

write the equilibrium expression that would be used in conjunction with the Henderson-Hasselbalch Equation

You do not need to memorize the structures of individual buffer systems

However, you should familiarize yourself with the buffers listed as indicated above

Monoprotic Acid-Base Equilibria Monoprotic Weak Acids Monoprotic Weak Bases Fraction of...

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