Common Ion Effect

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+ + C2H3O2- Ka = 1.8 x 10-5

6. Example HC2H3O2 H+ + C2H3O2- Ka = 1.8 x 10-5 1.8 x 10-5 = x2 / (3-x) 1.8 x 10-5 (0.1-x) = x2 x = 0.00133 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+ + C2H3O2- Ka = 1.8 x 10-5

9. Example HC2H3O2 H+ + C2H3O2- Ka = 1.8 x 10-5 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. Buffers

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

13. Example HC2H3O2 H+ + C2H3O2- Ka = 1.8 x 10-5

14. Example HC2H3O2 H+ + C2H3O2- Ka = 1.8 x 10-5 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

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 = 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!

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 = 0.0658M • KClO • M = mol/L • 0.45 = mol / 32 • 14.4 mmol KClO/ 57 mL = 0.253M

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

19. Example (ICE TABLE) HClO H+ + ClO- Ka = 3.5x10-8 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

20. Example (HH) • pH = pKa + log [B]/[A] • pH = 7.45 + log (14.4/3.75) • pH = 8.03

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 = 0.125 M CH3NH3NO3 • CH3NH2 • M = mol/L • 0.30 = mol / 75 • 22.5 mmol/ 100 mL = 0.225 M CH3NH2

22. Example (ICE TABLE) CH3NH2 + H2O  CH3NH3+ + OH- Ka = 4.38x10-4

23. Example (ICE TABLE) CH3NH2 + H2O  CH3NH3+ + OH- Ka = 4.38x10-4 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 = 3.11 pH = 10.89

24. Example (HH) • pH = pKa + log [B]/[A] • pH = 10.65 + log (22.5/12/5) • pH = 10.91

25. 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

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

27. Example (ICE TABLE) HF  H + + F- Ka = 7.2x10-4 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 / 0.100L • 3.60 mol NaF x 41.99 g NaF • 1 mol NaF • 151 g NaF

29. 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/0.5 • X = 3.62 mol  152 g NaF