Advanced Acid/Base Theory

pH of Strong Acids

• Strong acids completely dissociate in water.

• Monoprotic strong acids release one mole of hydrogen ions for every mole of acid introduced into the water.

• This makes pH calculations relatively easy.

Sample Question

• What is the pH of a 0.20 M solution of HCl(aq)?

[H+] = [HCl] because of complete dissociation.

pH = log [H+]

-log 0.20 = 0.70

Calculating pH of a strong bases

• Strong bases also completely dissociate in water.

What is the pH of a 0.028 M solution of NaOH?

What the pH of a 0.0011 M solution of Ba(OH)2?

Another way of calculating it.

• pOH = -log [OH-] ions

What would be the sum of pH and pOH of all aqueous solutions?

So what is the pOH of a 0.028 M solution of NaOH?

What must it’s pH then be?

Another note about strong bases

• Alkali oxides and alkaline earth metal oxides form strongly basic solutions in water.

• Na2O + H2O 2 Na+ + 2 OH-

• O2- + H2O 2 OH-

Note: These are not equilibrium reactions.

Weak Acids and pH

Write the equilibrium expression for the generic acid (HA) in water.

Ka – The acid dissociation constant.

• How does Ka relate to acid strength?– Note Ka for weak acids are typically between

10-3 and 10-10.

• Because dissociation is incomplete is weak acids, [acid] cannot be directly used to calculate pH.

Determining Ka from pH

• A 0.10 M solution of methanoic acid (HCOOH) has a pH of 2.38. What is the value of Ka?

HCOOH + H2O HCOO- + H3O+

Is there anything in this expression that we can determine from the information given?

HCOOH

HCOOOHK a

]][[ 3

• pH = -log[H3O]+

• 10-pH = [H3O+]

• [H3O+] = 10-2.38 = 4.2x10-3 mol dm-3

Using this info, let’s build an ICE chart

ICE, ICE Baby

HCOOH H3O+ HCOO-

Initial 0.10 M 0 0

Change - 4.2x10-3 M + 4.2x10-3 M + 4.2x10-3 M

Equilibrium 0.10 (What???) 4.2x10-3 M 4.2x10-3 M

433

108.110.0

)102.4)(102.4(

aK

Percent Ionization

• Another less common measure of acid strength.

• What is the % ionization in the previous problem?

%100ionconcentrat original

species ionizedion concentrat ionizationPercent

Using Ka to calculate pH

• Calculate the pH of a 0.20 M solution of HCN, Ka = 4.9x10-10.

Remember in weak acids, Δ[acid] is negligible.

Solving Equation

HCN H3O+ CN-

Initial 0.20 M 0 0

Change - x M + x M + x M

Equilibrium 0.20 (Why???) + x M + x M

610

102

10

109.9)20.0(109.4

)20.0(109.4

109.420.0

))((

x

x

xxKa

pH = -log [H+]

pH = - log 9.9x10-6 = 5.00

Does concentration affect % ionization?

• Calculate the percent ionization for HF(aq) solutions of 0.10 M and 0.010 M?

Ka = 6.8x10-4.

Polyprotic Acids

• Acids with more than one dissociable H.

• Examples: H2SO4

H3PO4

H2C6H6O6

H2CO3

Writing Ka Expressions

• H3C6H5O7 H+ + H2C6H5O7- Ka1 = 7.4x10-4

• H2C6H5O7- H+ + HC6H5O7

-2 Ka2 = 1.7x10-5

• HC6H5O7-2 H+ + C6H5O7

3- Ka3 = 4.0x10-7

Calculations with polyprotic acids

• If the difference between successive Ka values is greater than 103, the pH of a solution can be estimated to be wholly due to the first Ka.

• If the difference is less than 103, both values must be considered.

Sample Problem• What is the pH of a 0.0037 M solution of H2CO3?

Ka1 = 4.3x10-7, Ka2 = 5.6x10-11

H2CO3 (aq) H+(aq) + HCO3

- (aq)

0.0037 M 0 0

- x + x + x

0.0037 M - x + x +x

40.4

100.4

0037.0

))((103.4

][

]][[

5

71

32

31

pH

Mx

xxK

COH

HCOHK

a

a

Weak Bases

• Weak bases react with water to remove water and form hydroxide ions.

• Write the reaction of ammonia with water in the space below.

Kb

• Kb is the equilibrium expression for bases. Write the Kb expression for ammonia in water.

Sample Problem

• Calculate the [OH-] of a 0.15 M solution of NH3 (aq). What is the pH of the solution? (Kb = 1.8x10-5)

What are the weak bases?

They are…• Molecules containing atoms with lone

pairs of electrons,

e.g. NH3, any amine (N containing substance with only 3 bond pairs, leaving the 4th pair unattached.

• Anions of weak acids

e.g. CH3COO-,

Ka and Kb for conjugate acid-base pairs

• Write the reaction for ammonia going into aqueous solution with Kb expression.

• Write the forward reaction for the conjugate acid of ammonia reacting with Ka expression .

Sum the two reactions.

Simplify the (Ka)(Kb) expression

Important

• The product of Ka and Kb for any conjugate acid-base pair is always equal to Kw.

• If Ka for HF is 6.8x104, what is the Kb for F-?

BUFFER SOLUTIONS

Look at the following reaction

CH3COOH + H2O CH3COO- + H3O+

Ka = 1.8x10-5

• What’s the acid?• What’s the base?• Conjugate acid?• Conjugate base?

• Is the acid weak or strong?

Shifting the equilibrium

• How would adding CH3COONa (sodium ethanoate) change the equilibrium?

Common ion effect

• When a weak electrolyte and a strong electrolyte contain a common ion, the weak electrolyte ionizes less than it would if it were in solution alone.

• What is the pH of a solution made by adding 0.30 mol of acetic acid to 0.30 mol of sodium acetate in enough water to make a 1.0 L solution?

Solving common ion problems

• Identify strong and weak electrolytes.

• Determine the source of H3O+ ions.

• Determine the concentration of ions involved in the equilibrium.

• Use equilibrium constant to make calculations

CH3COOH + H2O CH3COO- + H3O+

I 0.30 0.30 0

C -x +x +x

E (0.30-x) (0.30+x) +x

Solve for hydronium ion concentration

Compare this to the pH of a solution of 0.30 M acetic acid

• Calculate the fluoride ion concentration and pH of a solution that is 0.20 M HF and 0.10 M HCl. HF Ka = 6.8x10-4

Buffered Solutions

• Solutions that resist drastic changes in pH upon the addition of a strong acid or base.

• The solutions contain common ions as discussed previously.

Imagining a Buffered System

• Must contain a weak acid (HX) and the ion that is its conjugate base (X-).

• What would the equation and equilibrium expression for this reaction look like?

What are the major determinants of pH in this system?

Adding a stress

• How would adding NaOH to this solution affect the solution?

• How would adding HCl to this solution affect the solution?

Determing the pH of a buffer system.

• For the hypothetical acid HX, write the equilibrium expression and rearrange to solve for [H+]

• Take the –log of both sides of the equation.

Henderson-Hasselbach Equation

• Determines the pH of a buffer solution

][

]_[log

acid

baseconjugatepKapH

Sample problem

• What is the pH of a buffer solution that is 0.12 M lactic acid (HC3H5O3) and 0.10 M sodium lactate (NaC3H5O3)? Ka for lactic acid is 1.4x10-4.

• Solve this problem two different ways.– ICE chart– Henderson-Hasselbach

Calculating pH change when a strong acid or base is added to a

buffered solution.• A 2.00 L solution containing 0.300 mol of acetic

acid and 0.300 mol of sodium acetate has a pH of 4.74.

• Calculate the pH of the solution after 0.020 mol of NaOH is added.

• What would the be the pH after 0.020 mol of NaOH be if added to pure water?

Write out the reaction and set up an ICE chart

CH3COOH + OH- H2O + CH3COO-

NOTE:

Strong bases react completely with weak acids

Strong acids react completely with weak bases.

Acid-Base Titrations

• A base of known concentration is added to an acid of unknown concentration

OR

• An acid of known concentration is added to a base of unknown concentration

What can we learn from acid-base titrations?

• Concentration of an acid/base

• Ka of a weak acid

• Kb of a weak base

Strong Acid Strong Base Titrations

Curve shows how pH of a 50.00 mL sample of 0.100 M solution of HCl changes as 0.100 M NaOH is added to the solution

Phases of a Titration (SA-SB)

• Initial pH of the acid (or base)

• The time between the initial pH and the equivalence point. pH first increases slowly and then very rapidly as the solution approaches the equivalence point.

Phases of a Titration (SA-SB)

• The equivalence point. Equal numbers of moles of NaOH and HCl have reacted. pH is 7.

– For strong acid-strong base titrations the equivalence point is at pH = 7.00 because Na+ and Cl- have no effect on pH.

– How would the equivalence point change when titrating weak acids and bases?

Phases of Titrations

• After the equivalence point. The pH of the solution is determine by the concentration of the excess NaOH.

Sample problem

• Calculate the pH when 49.0 mL of 0.100 M NaOH are added to 50.0 mL of 0.100 M HCl.

• Calculate the pH when 51.0 mL of 0.100 M NaOH are added to 50.0 mL of 0.100 M HCl.

Weak Acid-Strong Base Titrations

• mL of 0.100 M NaOH added to 50.00 mL of 0.100 M CH3COOH

Phases of Titration (WA-SB)

• Initial pH of the acid. Ka can be used to determine this.

• Between the initial pH and equivalence point. A buffer system is established at this point.

Phases of Titration (WA-SB)

• Equivalence point- Equal number of moles of acid and base are present in the solution.

– pH is not equal to 7.00. Why?• Does Na+ have an effect on pH?• Does CH3COO- have an effect on pH?

Phases of Titration (SB-WA)

• After the equivalence point: The OH- from NaOH has a dramatic effect the [OH-] of the solution.

• The OH- from the reaction of acetate with water in negligable compared to the OH-

from NaOH.

Sample Problem

• Calculate the pH of the solution formed

Calculating pH at Equivalence Point

• Calculate pH at the equivalence point of a titration of 50.0 mL of 0.100 M acetic acid and 0.100 M NaOH.

CH3COOH + OH- CH3COO- + H2O

• What is the original number of moles of acetic acid?

• How many moles of OH- must be added?

• How many mole acetate will form at equivalence point?

CH3COOH + OH- CH3COO- + H2O

• What is the concentration of acetate at the equivalence point?

• Remember acetate is a weak base and the conjugate base of acetic acid.

• For a conjugate acid/base pair Ka + Kb = Kw

• What is the value of Kb?

CH3COOH + OH- CH3COO- + H2O

• What is the equilibrium expression for this Kb value.

• Calculate [OH-] from this data.

• Calculate pOH

• Calculate pH

Determining pKa from a titration

• Write out Henderson-Hasselbach equation.

• How could you make the

log [conj. base]/[acid] = 0?

• At which point during a titration does this happen?

Determining pKa from a titration

• Mathematically, how does pH relate to pKa at the half equivalence point?

• How would you find Ka from pKa?