Common Ion Effect

Buffers

Common Ion Effect

Sometimes the equilibrium solutions have 2 ions in common

For example if I mixed HF & NaF The main reaction is HF H+ + F-

But some additional F- ions are being added from the NaF

These are worked the same way, you just start with a different initial amount of the ion

Common Ion Effect

Which way will the reaction shift if NaF is added?

To the reactant side What effect will this have on pH?[H+] will go down…so pH will go up

Example

What is the pH of a 0.10M solution of HC2H3O2 (Ka = 1.8 x10-5)

Example

HC2H3O2 H+ + C2H3O2 - Ka = 1.8 x 10-5

I

C

E

Example

I 0.1 0 0

C -x +x +x

E 0.1-x x x

1.8 x 10-5 = x2 / (3-x)

1.8 x 10-5 (0.1-x) = x2

x = 0.00133

pH = 2.88

HC2H3O2 H+ + C2H3O2 - Ka = 1.8 x 10-5

Example

A mixture contains 0.10M HC2H3O2 (Ka = 1.8 x10-5) & 0.10 M NaC2H3O2. Calculate the pH.

Example

HC2H3O2 H+ + C2H3O2 - Ka = 1.8 x 10-5

I

C

E

Example

I 0.1 0 0.10

C -x +x +x

E 0.1-x x 10. + x

1.8 x 10-5 = x(0.1+x) / (3-x)

1.8 x 10-5 (0.1-x) = x(0.1+x) x = 1.8 x 10-5

pH = 4.74

HC2H3O2 H+ + C2H3O2 - Ka = 1.8 x 10-5

What are buffers?

Buffers resist changes in pH They must have 2 parts…

1. Weak acid & a conjugate base OR

2. Weak base & a conjugate acid The concentrations of the 2 MUST be

with in a factor of 10!!!

Buffers

[NaC2H3O2 ] [HC2H3O2 ] Buffer?

0.10M 0.10M Yes

0.10M 1.0M Yes

0.01M 1.0M NO(Not within a factor of 10)

Example

What is the pH of a solution containing 50. mL of 0.50M NaC2H3O2 & 25 mL of 0.25M HC2H3O2. (Ka = 1.8 x10-5).

NaC2H3O2

M = mol/L 0.5 = mol / 50. 25 mmol NaC2H3O2/ 75 mL = 0.33M HC2H3O2

M = mol/L 0.25 = mol / 25 6.25 mmol HC2H3O2/ 75 mL = 0.083M

Example

HC2H3O2 H+ + C2H3O2 - Ka = 1.8 x 10-5

I

C

E

Example

I 0.0833 0 0.33

C -x +x +x

E 0.0833-x x 0.33 + x

1.8 x 10-5 = x(0.33+x) / (0.0833-x)

1.8 x 10-5 (0.0833-x)= x(0.33+x) x = 4.54 x 10-6

pH = 5.34

HC2H3O2 H+ + C2H3O2 - Ka = 1.8 x 10-5

Henderson Hasselbach Equation

Really easy equation to use ONLY with buffer solutions!!!

pH = pKa + log [B]/[A]

The last example using HH

pH = pKa + log [B]/[A] pH = 4.74 + log [25 mmol / 75 mL]

[6.25 mmol / 75 mL] The mL cross out, so on HH you can use

mmol pH = 4.74 + log (25/6.25) pH = 5.34 Same answer as we got with the ICE table Pick the way you like better & use it!

Example (ICE TABLE)

What is the pH of a solution containing 25 mL of 0.150MHClO & 32mL of 0.45M KClO. Ka = 3.5x10-8

HClO M = mol/L 0.15 = mol / 25 3.75 mmol HClO/ 57 mL = 0.0658M KClO M = mol/L 0.45 = mol / 32 14.4 mmol KClO/ 57 mL = 0.253M

Example (ICE TABLE)

HClO H+ + ClO - Ka = 3.5x10-8

I

C

E

Example (ICE TABLE)

I 0.0685 0 0.253

C -x +x +x

E 0.0685 -x x 0.253 + x

3.5 x 10-8= x(0.253 +x) / (0.0685 -x)

3.5 x 10-8 (0.0685 -x)= x(0.253 +x) x = 9.8 x 10-9

pH = 8.02

HClO H+ + ClO - Ka = 3.5x10-8

Example (HH)

pH = pKa + log [B]/[A]pH = 7.45 + log (14.4/3.75)pH = 8.03

Example (ICE TABLE)

What is the pH of a solution containing 25 mL of 0.50M CH3NH3NO3 is mixed with 75 mL of 0.30 M CH3NH2 (Kb CH3NH2 = 4.38 x10-4)

CH3NH3NO3

M = mol/L 0.50 = mol / 25 12.5 mmol / 100 mL = 0.125 M CH3NH3NO3

CH3NH2 M = mol/L 0.30 = mol / 75 22.5 mmol/ 100 mL = 0.225 M CH3NH2

Example (ICE TABLE)

CH3NH2 + H2O CH3NH3+ + OH - Ka = 4.38x10-4

I

C

E

Example (ICE TABLE)

I 0.225 0.125 0

C -x +x +x

E 0.225-x 0.125+x x

4.38x10-4 = x(0.125 +x) / (0.225 -x)

4.38x10-4(0.225 -x)= x(0.125 +x) x = 7.8 x 10-4

pOH = 3.11pH = 10.89

CH3NH2 + H2O CH3NH3+ + OH - Ka = 4.38x10-4

Example (HH)

pH = pKa + log [B]/[A]pH = 10.65 + log (22.5/12/5)pH = 10.91

Example

Calculate the mass of NaF that must be added to 1000.0 ml of 0.50M HF to form a solution with a pH of 4.00. Ka = 7.2x10-4

Example (ICE TABLE)

HF H + + F - Ka = 7.2x10-4

I

C

E

Example (ICE TABLE)

I 0.5 0 ?

C -1x10-4 +1x10-4 +1x10-4

E 0.5 1x10-4 ?+1x10-4

pH = 4.00[H+] = 1x10-4

7.2x10-4 = 1x10-4(?+1x10-4)/0.57.2x10-4(0.5)= 1x10-4(?+1x10-4) x = 3.60

HF H + + F - Ka = 7.2x10-4

Example (ICE TABLE)

3.60 M [F-] = 3.60 M NaF M = mol/L3.60 = mol / 0.100L3.60 mol NaF x 41.99 g NaF 1 mol NaF151 g NaF

Example (HH)

pH = pKa + log [B]/[A]4.00 = 3.14 + log (x/0.5)0.86 = log (x/0.5)Antilog(0.86) = (x/0.5)7.24 = x/0.5X = 3.62 mol 152 g NaF