Determination of Ka, Kb & pH

Solving Acid/Base Problems 1. List the major species in solution eg. HA, A-, B:, HB+, H3O+, OH- Look for reactions that are not equilibrium systems. eg. HCl + H2O  H3O+ + Cl- NaOH  Na+ + OH- NaC2H3O2  Na+ + C2H3O2- 3. For above, determine [H3O+] or [OH-] or [A-].

4. Determine the equilibrium systems and pick the equilibrium that will control the pH. eg. Ka > Kb > Kw or Ka1 > Ka2 Calculate the [H3O+]i or [OH-]i or [A-]I from [HA]i or [B:]i and the values from #3 above. Write the equation for the rxn and the Ka or Kb. Create an I.C.E. table, calculate 100 Rule. Substitute [ ]eq into Ka or Kb. 9. Solve for the pH or [ ]eq or Ka or Kb as required.

1) Determining Ka or Kb From pH eg. Calculate the Ka for formic acid (HCO2H) if a 0.100 M solution has a pH of 2.38 HCO2H + H2O  H3O+ + CO2H- ;   ICE 0.100 - x +x +x [H3O+]eq = x = 10-pH = 10 - 2.38 = 0.0042   = 1.8 x 10-4

From p different determination of [H3O+]eq or [OH-]eq. eg. Calculate the Kb for NH3 if a 0.150 M solution has a p of 7.8 %.   NH3 + H2O  NH4+ + OH- ; ICE 0.150 - x +x +x     [OH-]eq = x = [NH4+]eq = 0.012 ; [NH3]eq = 0.140   = 1.0 x 10-3

2) Determining pH from Ka or Kb Development of the ‘Shortcut’ Given:   HA + H2O  H3O+ + A- ; ICE [HA]i – x +x +x If 100 Rule applies, then:   then     and   also

eg. Calculate the pH of a 0.200 M acetic acid sol’n. Let HA = HC2H3O2 = acetic acid   HA + H2O  H3O+ + A- ; ICE 0.100 – x +x +x = 1.8 x 10-5 100 Rule      = 2.87

3) Polyprotic Acids ionize only 1 H+ at a time. each step has its own Ka, designated Ka1, Ka2, etc. Ka1 is the strongest and sets the pH of the sol’n. eg. Calculate the pH of 2.500 M H2CO3 and the [H2CO3]eq, [HCO3-]eq, [CO32-]eq.

eg. Calculate the pH of 2.500 M H2CO3 and the [H2CO3]eq, [HCO3-]eq, [CO32-]eq. H2CO3 + H2O  H3O+ + HCO3- ; ICE 2.500 – x +x +x   100 Rule  = 4.4 x 10-7     = 1.0 x 10-3 [H2CO3] = 2.500 - 0.0010 = 2.499 pH = -log (1.0 x 10-3) = 3.00

eg. Calculate the pH of 2.500 M H2CO3 and the [H2CO3]eq, [HCO3-]eq, [CO32-]eq. HCO3- + H2O  H3O+ + CO32- ; ICE 1.0 x 10-3 – x 1.0 x 10-3 +x + x 100 Rule    = 4.7 x 10-11 so [HCO3- ] = [H3O+ ], Always if 100 Rule applies   or   = 4.7 x 10-11