Common Ion Effect Buffers
Jan 21, 2016
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