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 HC2H3O2  H+ + C2H3O2 - 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 = 0.00133 pH = 2.88

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 HC2H3O2  H+ + C2H3O2 - 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

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!!!

(Not within a factor of 10) Buffers [NaC2H3O2 ] [HC2H3O2 ] Buffer? 0.10M Yes 1.0M 0.01M 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 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 HC2H3O2  H+ + C2H3O2 - Ka = 1.8 x 10-5 I 0.0833 0.33 C -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

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 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 0.253 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 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

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

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 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 C -x +x E 0.225-x 0.125+x x CH3NH2 + H2O  CH3NH3+ + 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(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

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

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 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

Example (ICE TABLE) 3.60 M [F-] = 3.60 M NaF M = mol/L 3.60 mol NaF x 41.99 g NaF 1 mol NaF 151 g NaF

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