1. Common Ion Effect
Buffers

2. 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

3. 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

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

5. Example
HC2H3O2  H+ + C2H3O Ka = 1.8 x 10-5 I C E

6. Example
HC2H3O2  H+ + C2H3O Ka = 1.8 x 10-5 I
0.1 C -x +x E
0.1-x x
1.8 x 10-5 = x2 / (3-x)
1.8 x 10-5 (0.1-x) = x2
x =
pH = 2.88

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

8. Example
HC2H3O2  H+ + C2H3O Ka = 1.8 x 10-5 I C E

9. Example
HC2H3O2  H+ + C2H3O Ka = 1.8 x 10-5 I
0.1
0.10 C -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

10. What are buffers? Buffers resist changes in pH They must have 2 parts…
Weak acid & a conjugate base OR
Weak base & a conjugate acid
The concentrations of the 2 MUST be with in a factor of 10!!!

11. (Not within a factor of 10)
Buffers
[NaC2H3O2 ]
[HC2H3O2 ]
Buffer?
0.10M
Yes
1.0M
0.01M NO
(Not within a factor of 10)

12. 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
0.25 = mol / 25
6.25 mmol HC2H3O2/ 75 mL = 0.083M

13. Example
HC2H3O2  H+ + C2H3O Ka = 1.8 x 10-5 I C E

14. Example
HC2H3O2  H+ + C2H3O Ka = 1.8 x 10-5 I
0.0833
0.33 C -x +x E x x x
1.8 x 10-5 = x(0.33+x) / ( x)
1.8 x 10-5 ( x)= x(0.33+x)
x = 4.54 x 10-6
pH = 5.34

15. Henderson Hasselbach Equation
Really easy equation to use ONLY with buffer solutions!!!
pH = pKa + log [B]/[A]

16. The last example using HH
pH = pKa + log [B]/[A]
pH = log [25 mmol / 75 mL] [6.25 mmol / 75 mL]
The mL cross out, so on HH you can use mmol
pH = log (25/6.25)
pH = 5.34
Same answer as we got with the ICE table
Pick the way you like better & use it!

17. 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 = M
KClO
0.45 = mol / 32
14.4 mmol KClO/ 57 mL = 0.253M

18. Example (ICE TABLE)
HClO H ClO Ka = 3.5x10-8 I C E

19. Example (ICE TABLE) I 0.0685 0.253 C -x +x E 0.0685 -x x 0.253 + x
HClO H ClO Ka = 3.5x10-8 I
0.0685
0.253 C -x +x E x x x
3.5 x 10-8= x( x) / ( x)
3.5 x 10-8 ( x)= x( x)
x = 9.8 x 10-9
pH = 8.02

20. Example (HH) pH = pKa + log [B]/[A] pH = 7.45 + log (14.4/3.75)

21. 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 = M CH3NH3NO3
CH3NH2
0.30 = mol / 75
22.5 mmol/ 100 mL = M CH3NH2

22. Example (ICE TABLE)
CH3NH2 + H2O  CH3NH OH Ka = 4.38x10-4 I C E

23. Example (ICE TABLE) I 0.225 0.125 C -x +x E 0.225-x 0.125+x x
CH3NH2 + H2O  CH3NH OH Ka = 4.38x10-4 I
0.225
0.125 C -x +x E
0.225-x
0.125+x x
4.38x10-4 = x( x) / ( x)
4.38x10-4( x)= x( x)
x = 7.8 x 10-4
pOH = 3.11
pH = 10.89

24. Example (HH) pH = pKa + log [B]/[A] pH = 10.65 + log (22.5/12/5)

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

26. Example (ICE TABLE)
HF  H F Ka = 7.2x10-4 I C E

27. Example (ICE TABLE) I C E HF  H + + F - Ka = 7.2x10-4 0.5 ? -1x10-4? C
-1x10-4
+1x10-4 E
1x10-4
?+1x10-4
pH = 4.00
[H+] = 1x10-4
7.2x10-4 = 1x10-4(?+1x10-4)/0.5
7.2x10-4(0.5)= 1x10-4(?+1x10-4)
x = 3.60

28. Example (ICE TABLE) 3.60 M [F-] = 3.60 M NaF M = mol/L
3.60 mol NaF x g NaF
1 mol NaF
151 g NaF

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