Chapter 15Part I: Common Ion Effect

When the salt with the anion of a weak acid is added to that acid,

It reverses the dissociation of the acid. Lowers the percent dissociation of the acid.

The same principle applies to salts with the cation of a weak base.

The calculations are the same as last chapter.

The Common Ion Effect

A solution that resists a change in pH. Either a weak acid and its salt or a weak

base and its salt. We can make a buffer of any pH by varying

the concentrations of these solutions. Same calculations as before.

Calculate the pH of a solution that is .50 M HC2H3O2 and .25 M NaC2H3O2

(Ka = 1.8 x 10-5)

Buffered solutions

Na+ is a spectator and the reaction we are worried about is R HC2H3O2 H+ + C2H3O2

-

x

xx-x

0.50-x 0.25+x

I 0.50 M 0 0.25 M

EC

Do the math Ka = 1.8 x 10-5

1.8 x 10-5 =x (0.25+x)

(0.50-x) =

x (0.25)

(0.50)

x = 3.6 x 10-5

pH = -log (3.6 x 10-5) = 4.44

HC2H3O2 H+ + C2H3O2-

General equation

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

so [H+] = Ka [HA]

[A-]

The [H+] depends on the ratio [HA]/[A-] taking the negative log of both sides pH = -log(Ka [HA]/[A-])

pH = -log(Ka)-log([HA]/[A-])

pH = pKa + log([A-]/[HA])

This is called the Henderson-Hasselbach equation pH = pKa + log([A-]/[HA]) pH = pKa + log(base/acid) Works for an acid and its salt Like HNO2 and NaNO2

Or a base and its salt Like NH3 and NH4Cl But remember to change Kb to Ka

pH = pKa + log [base]/[acid]

pH = -log (1.8 x 10-5) + log (.25/.50)

pH = 4.44

Or the easy way….

Let’s take the solution and add 0.01mol of NaOH to it.

1. Write the net ionic equation for the reaction of HC2H3O2 + NaOH

HC2H3O2 + OH- C2H3O2- + H2O

2. Do the stoichiometry using moles (Use a BRA Chart)

With a twist now

HC2H3O2 + OH- C2H3O2- + H2O

B .50mol .01mol .25mol

R -.01 -.01 +.01

A .49 mol 0 .26mol

*Assume 1L of solution and we now have our starting concentrations

I .49 0 .26 C -x +x +x E .49 –x x .26+x

Ka = 1.8 x10-5 = x(.26+x) / (.49-x)

5% rule 1.8 x10-5 = x(.26) / (.49) x = 3.4 x 10-5 pH = 4.47

R HC2H3O2 H+ + C2H3O2-

pH = pKa + log [base]/[acid]

pH = -log (1.8 x 10-5) + log (.26/.49)

pH = 4.47

Again, the easy way

Notice If we had added 0.010 mol of NaOH to 1 L of

water, what would the pH have been. 0.010 M OH-

NaOH Na+ + OH-

.010 .010 .010 pOH = -log(0.010) = 2.00 pH = 12.00 But with a mixture of an acid and its

conjugate base the pH doesn’t change much

Called a buffer.

Calculate the pH 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)

pH = pKa + log(base/acid)pH = -log(1.4 x 10-4) + log (.25/.75)

pH = 3.38

Calculate the pH 0.25 M NH3 and 0.40 M NH4Cl

(Kb = 1.8 x 10-5) Ka = 5.6 x 10-10

remember its the ratio base over acid

pH = -log (5.6 x 10-10) + log(.25/.40)

pH = 9.05

Prove they’re buffers What would the pH be if .020 mol of HCl is

added to the lactic acid solution?

0.75 M lactic acid (HC3H5O3) and 0.25 M

sodium lactate (Ka = 1.4 x 10-4)HC3H5O3 H

+ + C3H5O3-

Before 0.75 mol +.02 0.25 molReactions +.02mol -.02 -.02After 0.77 mol 0 0.23 mol

pH = pKa + log (base/acid)

pH = -log(1.4 x 10-4) + log (.23/.77)

pH = 3.33, compared to 3.38 before adding the HCl

What would the pH be if .020 mol of HCl is added to the lactic acid solution?

Prove they’re buffers What would the pH be if 0.050 mol of solid

NaOH is added to 1.0 L of the solutions.

0.75 M lactic acid (HC3H5O3) and 0.25 M

sodium lactate (Ka = 1.4 x 10-4)HC3H5O3 + OH-

H2O + C3H5O3-

B 0.75 mol .05 mol 0.25 molR -.05 -.05 +.05A 0.70 mol 0 mol 0.30 mol

pH = pKa + log (base/acid)

pH = -log(1.4 x 10-4) + log (.30/.70)

pH = 3.49, compared to 3.38 before adding the NaOH

What would the pH be if 0.050 mol of solid NaOH is added to 1.0 L of the lactic acid solution.

Prove they’re buffers What would the pH be if .020 mol of HCl is

added to 1.0 L of ammonia solution.

0.25 M NH3 and 0.40 M NH4Cl

Kb = 1.8 x 10-5 so Ka = 5.6 x 10-10

NH3 + H+ NH4

+

B 0.25 mol .02 mol 0.40 molR -.02 -.02 +.02A 0.23 mol 0 mol 0.42 mol

Net Ionic

pH = pKa + log (base/acid)

pH = -log(5.6 x 10-10) + log (.23/.42)

pH = 8.99, compared to 9.05 before adding the HCl

What would the pH be if .020 mol of HCl is added to 1.0 L of ammonia solution.

Prove they’re buffers What would the pH be if 0.050 mol of solid

NaOH is added to 1.0 L of ammonia solution.

0.25 M NH3 and 0.40 M NH4Cl

Kb = 1.8 x 10-5 so Ka = 5.6 x 10-10

NH3 + H2O NH4+ + OH-

B 0.40 mol 0.25 mol .050molR +.05 -.05 -.05A 0.45 mol 0.20 mol 0 mol

pH = pKa + log (base/acid)

pH = -log(5.6 x 10-10) + log (.45/.20)

pH = 9.60, compared to 9.05 before adding the NaOH

What would the pH be if 0.050 mol of solid NaOH is added to 1.0 L of ammonia solution.

Buffer capacity The pH of a buffered solution is determined

by the ratio [A-]/[HA]. As long as this doesn’t change much the pH

won’t change much. The more concentrated these two are the

more H+ and OH- the solution will be able to absorb.

Larger concentrations = bigger buffer capacity.

Buffer Capacity Calculate the change in pH that occurs when

0.020 mol of HCl(g) is added to 1.0 L of each of the following:

5.00 M HAc and 5.00 M NaAc 0.050 M HAc and 0.050 M NaAc Ka= 1.8x10-5

pH = pKa

-5 5.00 MpH = -log(1.8 x 10 ) + log 4.74

5.00 M

Buffer Capacity Calculate the change in pH that occurs when

0.040 mol of HCl(g) is added to 1.0 L of 5.00 M HAc and 5.00 M NaAc

Ka= 1.8x10-5

HAc H+ + Ac-

B 5.00 mol .04mol 5.00 molR +.04 -.04 -/04A 5.04 mol 4.96 molpH = 4.74 Compared to 4.74 before acid was added

Buffer Capacity Calculate the change in pH that occurs when

0.040 mol of HCl(g) is added to 1.0 L of 0.050 M HAc and 0.050 M NaAc

Ka= 1.8x10-5

HAc H+ + Ac-

B 0.050 mol .04mol 0.050 molR +.04 -.04 -.04A 0.090 mol 0 0.010 mol

pH = 3.79Compared to 4.74 before acid was added

Buffer capacity

The best buffers have a ratio [A-]/[HA] = 1 This is most resistant to change True when [A-] = [HA] Makes pH = pKa (since log 1 = 0)