Updates

• Midterms marked; solutions are posted

• Assignment 03 is in the box

• Assignment 04 is up on ACME and is due Mon., Feb. 26 (in class)

Acids and BasesChapter 16

How do we measure pH?

• For less accurate measurements, one can use– Litmus paper

• Turns blue above ~pH = 8

• Turns red below ~pH = 5

– An indicator

How do we measure pH?

For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

Strong acids

• You will recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4.

• These are, by definition, strong electrolytes and exist totally as ions in aqueous solution.

• For the monoprotic strong acids,

[H3O+] = [acid].

Strong bases

• Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+).

• Again, these substances dissociate completely in aqueous solution.

Dissociation constants

• For a generalized acid dissociation,

the equilibrium expression would be

• This equilibrium constant is called the acid-dissociation constant, Ka.

[H3O+] [A−][HA]

Kc =

HA(aq) + H2O(l) A−(aq) + H3O+(aq)

Dissociation constants

The greater the value of Ka, the stronger the acid.

What is the pH of a 0.5 M HF solution (at 250C)?

HF (aq) H+ (aq) + F- (aq) Ka =[H+][F-][HF]

= 7.1 x 10-4

HF (aq) H+ (aq) + F- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.50 0.00

-x +x

0.50 - x

0.00

+x

x x

Ka =x2

0.50 - x= 7.1 x 10-4

Ka x2

0.50= 7.1 x 10-4

0.50 – x 0.50Ka << 1

x2 = 3.55 x 10-4 x = 0.019 M

[H+] = [F-] = 0.019 M pH = -log [H+] = 1.72

[HF] = 0.50 – x = 0.48 M16.5

When can I use the approximation?

0.50 – x 0.50Ka << 1

When x is less than 5% of the value from which it is subtracted.

x = 0.0190.019 M0.50 M

x 100% = 3.8%Less than 5%

Approximation ok.

What is the pH of a 0.05 M HF solution (at 250C)?

Ka x2

0.05= 7.1 x 10-4 x = 0.006 M

0.006 M0.05 M

x 100% = 12%More than 5%

Approximation not ok.

Must solve for x exactly using quadratic equation.

16.5

Solving weak acid ionization problems:

1. Identify the major species that can affect the pH.

• In most cases, you can ignore the autoionization of water.

• Ignore [OH-] because it is determined by [H+].

2. Use ICE to express the equilibrium concentrations in terms of single unknown x.

3. Write Ka in terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly.

4. Calculate concentrations of all species and/or pH of the solution.

16.5

What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 x 10-4?

HA (aq) H+ (aq) + A- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

Ka =x2

0.122 - x= 5.7 x 10-4

Ka x2

0.122= 5.7 x 10-4

0.122 – x 0.122Ka << 1

x2 = 6.95 x 10-5 x = 0.0083 M

0.0083 M0.122 M

x 100% = 6.8%More than 5%

Approximation not ok.

16.5

Ka =x2

0.122 - x= 5.7 x 10-4 x2 + 0.00057x – 6.95 x 10-5 = 0

ax2 + bx + c =0-b ± b2 – 4ac

2ax =

x = 0.0081 x = - 0.0081

HA (aq) H+ (aq) + A- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

[H+] = x = 0.0081 M pH = -log[H+] = 2.09

16.5

percent ionization = Ionized acid concentration at equilibrium

Initial concentration of acidx 100%

For a monoprotic acid HA

Percent ionization = [H+]

[HA]0

x 100% [HA]0 = initial concentration

16.5

Calculating Ka from the pH

• The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature.

• We know that

[H3O+] [COO−][HCOOH]

Ka =

Calculating Ka from the pH

• The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this temperature.

• To calculate Ka, we need the equilibrium concentrations of all three things.

• We can find [H3O+], which is the same as [HCOO−], from the pH.

Calculating Ka from the pH

pH = −log [H3O+]

2.38 = −log [H3O+]

−2.38 = log [H3O+]

10−2.38 = [H3O+]

4.2 10−3 = [H3O+] = [HCOO−]

Calculating Ka from pH

Now we can set up a table…

[HCOOH], M [H3O+], M [HCOO−], M

Initially 0.10 0 0

Change −4.2 10-3 +4.2 10-

3

+4.2 10−3

At Equilibrium

0.10 − 4.2 10−3

= 0.0958 = 0.10

4.2 10−3 4.2 10−3

Calculating Ka from pH

[4.2 10−3] [4.2 10−3][0.10]

Ka =

= 1.8 10−4

Polyprotic Acids

• Have more than one acidic proton.

• If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.

NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)

Weak Bases and Base Ionization Constants

Kb =[NH4

+][OH-][NH3]

Kb is the base ionization constant

Kb

weak basestrength

16.6

Solve weak base problems like weak acids except solve for [OH-] instead of [H+].

16.6

16.7

Ionization Constants of Conjugate Acid-Base Pairs

HA (aq) H+ (aq) + A- (aq)

A- (aq) + H2O (l) OH- (aq) + HA (aq)

Ka

Kb

H2O (l) H+ (aq) + OH- (aq) Kw

KaKb = Kw

Weak Acid and Its Conjugate Base

Ka = Kw

Kb

Kb = Kw

Ka