Only a portion of the weak acids and bases will break apart when in H2O

Only a fraction of the molecules will create H+ or OH-

Use the Keq to determine how much of the chemical breaks up

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

(aq)

Keq = [H+][F-] [HF]This Keq = Ka

Ionization constant for acidsIonization constant for acidsHigher Ka value = more breaking apart of the acid

= more H+ in water= STRONGER ACID

2. Calculating Ka from pHIf we know the pH, we can get the Ka1. Use the equilibrium chart 2. Use pH to determine [H+] at equilibrium3. Determine other concentrations at equilibrium4. Use Ka expression to calculate Ka

ExampleIf the pH of a 0.1 M HX is 4.75, what is Ka?

HX H+ + X-

0.1 M 0.0 M 0.0 M

-x X X

1 x 10-4.751.778 x 10-5 1.778 x 10-50.1 - 1.778 x 10-5

The subtracted amount is so small, we can ignore it!!The subtracted amount is so small, we can ignore it!!

Ka = [H+][X-] [HX]

3. Determining the Percent ionized acidIf we need to know what percent of the acid is ionized

% = . [H+] . original concentration

Percent ionized = 1.778 x 10-5 x 100 0.1

Ka = (1.778 x 10-5)2

0.1

Ka = 3.16 x 10Ka = 3.16 x 10-9-9

0.0178%

Example – Calculate the Ka of a 0.50 M solution of formic acid (HCHO2) with a pH of 2.02. Also determine the percent ionization in the acid.

HCHO2 H+ + CHO2-

0.5 M 0.0 M 0.0 M

-x X X

1 x 10-2.029.55 x 10-3 9.55 x 10-30.5 – 9.55 x 10-3

Again, the subtracted amount is so small, we can ignore it!!

Ka = (9.55 x 10-3)2

0.5

Ka = 1.8 x 10-4Ka = [H+][CHO2] [HCHO2]

Percent ionized = [H+] original [HCHO2]

9.55 x 10-2

0.51.91%

4. Finding pH from KaDone much the same way

1. Set up the chart2. Determine final concentration in terms of x3. Us Ka to solve for x and final [ ] of all

products4. Take the –log of the final [H+]

Example – If HY has a Ka of 5.3 x 10-5, what is the pH of a 0.1 M solution? HY H+ Y-0.1 M 0.0 M 0.0 M

-x X X

x x0.1 - x

But, in previous problems, 0.1 – x is essentially 0.1If ionization is less than 5%, we can ignore the amount of If ionization is less than 5%, we can ignore the amount of ionized acid from the original amountionized acid from the original amount

When we finish the problem, we should check to make sure that the ionization is less than 5%. Ka = X2

0.15.3 x 10-5 = X2

0.1X = 2.30 x 10-3

This means that [H+] is 2.30 x 10-3 MpH = -log (2.3 x 10-3)pH = 2.64

Now check to make sure ionization is less than 5%

% ionized = [H+] x 100 [HY]

2.30 x 10-3 x 100 0.1

2.30 %We’re safe!