Acid base equilibrium part ii ionization constants k a and k b
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Acid-Base Equilibrium Part II: Ionization Constants K a and K b. Jespersen Chap. 17 Sec 3. Dr. C. Yau Spring 2014. 1. Dissociation Constants of Acids. Strong acids: HCl H + + Cl - HCl + H 2 O H 3 O + + Cl - Weak acids:

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Acid-Base Equilibrium Part II: Ionization Constants K a and K b

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Acid base equilibrium part ii ionization constants k a and k b

Acid-Base EquilibriumPart II: Ionization Constants Ka and Kb

Jespersen Chap. 17 Sec 3

Dr. C. Yau

Spring 2014

1


Dissociation constants of acids

Dissociation Constants of Acids

Strong acids:

HCl H+ + Cl-

HCl + H2O H3O+ + Cl-

Weak acids:

CH3CO2H CH3CO2- + H+

CH3CO2H+ H2O CH3CO2- + H3O+

Keq = ?

Ka = ?


Dissociation of weak acids

Dissociation of Weak Acids

Using HA as general formula of an acid, the chemical equation for the dissociation if HA is a weak acid would be....

Its corresponding equilibrium law would be...


Acid base equilibrium part ii ionization constants k a and k b

Nitrous acid, HNO2, is a weak acid that's formed in the stomach when nitrite food preservatives encounter stomach acid. There has been some concern that this acid may form carcinogenic products by reacting with proteins.

Write the chemical equation for the equilibrium ionization of HNO2 in water and the appropriate Ka expression.

4


Acids that are not molecular

Acids that are not molecular

Unlike in Gen Chem I, you will be seeing some acids that are charged:

CH3CH2NH3+ + H2O ?

The general equation would be written as:

BH+ + H2O

Why do you suppose we use the letter B?

What is the equilibrium law?


Acid base equilibrium part ii ionization constants k a and k b

Write the equation and the equilibrium law for the ionization in water of N2H5+.

Do Practice Exer 17.9 & 17.10 p. 781


Ph of weak acids

pH of Weak Acids

In the previous lecture we see that...

If [HCl] = 10-3 M

[H+] = 10-3 M

pH = 3

If [CH3CO2H] = 10-3 M

pH is not 3. Why?

Would you expect it to be larger or smaller than 3?

To determine the pH, we must know its Ka


K a and pk a

Ka and pKa

Remember pH = - log [H+]

Similarly, pKa = - log Ka

e.g. What is the pKa of acetic acid which has a Ka of 1.8x10-5?


Acid base equilibrium part ii ionization constants k a and k b

Note the inverse relationship of Ka and pKa.

The stronger acid has a larger Ka but smaller pKa


Acid strength vs k a and pk a

Acid Strength vs. Ka and pKa

Acid strength

Weaker acid Stronger acids

Ka smaller larger

pKa larger smaller

Remember:

Larger [H3O+] means more acidic.

Smaller pH means more acidic.

Larger Ka means stronger acid.

Smaller pKa means stronger acid.

STRONG ACID has LARGE Ka and SMALL pKa

Ka = 10-15 Ka = 10-3

(pH = 3 is more acidic than pH =4.)


Using the ka and pka table

Using the Ka and pKa Table

Which is the stronger acid?

  • formic acid (in ant stings)

  • barbituric acid (used in production of barbiturate drugs)

    Phenol has a pKa of 9.89

    And butyric acid has a pKa of 4.82.

    Which is the stronger acid?

  • phenol B. butyric acid

    Do Pract. Exer17.11 & 17.12 p. 781


Dissociation constants of weak bases

Dissociation Constants of Weak Bases

You are already familiar with NH3 being a weak base:

NH3 + H2O NH4+ + OH-

Kb =

Hydrazine, N2H4 is a weak base. Write the equation for the reaction of hydrazine with water and write its expression for its Kb.


Bases that are not molecular

Bases that are not molecular

Remember that weak acids make strong conjugate bases?

Acetic acid is a weak acid. Give the formula of its conjugate base.

Now write the equation for its reaction with water:


Acid base equilibrium part ii ionization constants k a and k b

Solutions of chlorine bleach such as Clorox contain the hypochlorite ion, OCl¯ which is a weak base. Write the chemical equation for the reaction of OCl¯ with water and the appropriate expression for Kb for this anion.

Do Pract Exer 17.13 & 17.14 p. 782

14


General equations

General Equations

ACIDS

HA + H2O H3O+ + A−

BH+ + H2O H3O+ + B

BASES

B + H2O BH+ + OH−

B− + H2O BH + OH−

Note the conservation of charges in these equations.


Acid base equilibrium part ii ionization constants k a and k b

pKb = - log Kb

Again, we see the inverse relationship of Kb and pKb.

The stronger base has the larger Kb and smaller pKb.

B + HOH BH+ + OH-

(Strong bases grab protons more greedily.)


Relationship of k a and k b

Relationship of Ka and Kb

Write the general chemical equation and Ka expression for HA.

Write the general chemical equation and Kb expression for the conjugate base of HA.

Using the Ka and Kb expressions you just wrote, write the expression for Ka x Kb.


Summary

Summary

HA + H2O H3O+ + A-

pKa = - log Ka

B + H2O BH+ + OH-

pKb = - log Kb

If B = conjugate base of A =A- ,

A-+H2O HA + OH-

Ka x Kb = 1.0x10-14 pKa + pKb = 14.00


Acid base equilibrium part ii ionization constants k a and k b

What is the Kb value for CN¯?

Examine Table 17.2 for Kb values.

Note that Table 17.1 and 17.2 are for neutral molecules.

You will not find Ka or Kb values for species with a charge.

What are you going to do in order to determine Kb for CN¯?

Look up Ka of HCN.

Calculate Kb from Ka.


Acid base equilibrium part ii ionization constants k a and k b

With the use of the tables for Ka and Kb, calculate the Kb of OCN¯.

Calculate the Ka of the ammonium ion.

Ans. Kb = 5x10-11 Ka = 5.6x10-10

Do Pract Exer 17.15 & 7.16 p. 783


Acid base equilibrium part ii ionization constants k a and k b

Stronger acids give weaker conjugate bases.

CONJUGATE BASES

ACIDS

Fig. 17.3 p. 784


K a x k b 1 0x10 14

Ka x Kb = 1.0x10-14

If Ka is large, Kb must be small.

If Ka is small, Kb must be large.

Ka and Kb are INVERSELY PROPORTIONAL.

When comparing 2 acids,

the weaker acid produces

the stronger conjugate base.


Henderson hasselbalch eqn

Henderson-Hasselbalch Eqn

If we can get [A-] = [HA], how would it simplify this equation?

This allows us to experimentally determine pKa easily.


How k a can be determined experimentally

How Ka can be determined experimentally:

  • Prepare a solution of the weak acid and divide it into two halves by volume.

  • Neutralize one half with a strong base (such as NaOH) and HA become A¯.

  • Add the un-neutralized half to the neutralized solution and we have "half-neutralized" solution. [HA] = [A-]

  • Measure the pH.

    (pH = pKa, calculate Ka from pKa.)


Acid base equilibrium part ii ionization constants k a and k b

For example, to determine the ionization constant of barbituric acid:

1) Prepare 1000 mL of 0.10 M barbituric acid.

2) Pour 500 mL of this solution into a flask and titrate it to the end point with an aqueous solution of NaOH.

3) Add the other 500 mL (un-neutralized) barbituric acid solution and measure the pH.

If the pH is 4.01, what is the Ka of barbituric acid?

HA + NaOH  H2O + Na+A–

500 mL

Combine and get

0.050 mol A–/1000mL = 0.050M A–0.050 mol HA/1000mL= 0.050M HA

[A–] = [HA]

0.050

mol HA

  • 0.050

  • molA–

1000mL

500 mL

0.10 M HA

1.00Lx0.10M M=0.10mol HA

0.050 mol HA


Determination of ka

Determination of Ka

Another method of getting pKa = pH (and then calculating Ka from pKa) is from a titration curve:

  • Record the pH while adding a base to the acid and plot pH vs. Volume of Base Added.

  • Determine on the graph what the volume of base is at the equivalence point (the exact point when all acid has reacted).

  • Divide this volume by two (the point when exactly half of acid has reacted).

  • Locate the pH on the graph at this volume of base. This is the point when [HA] = [A-]! Think about this! This is, therefore, the point when

    pH = pKa.


Acid base equilibrium part ii ionization constants k a and k b

HA + NaOH Na+A- + HOH

Half of 25.00 mL (12.50 mL): Exactly half of acid has reacted

Fig. 17.8 p. 815 Titration of Weak Acid with Strong Base

This is when phenolphthalein would turn pink.

pH= 4.8

Ka = ?


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