Equivalence point - point at which the titrant and the analyte are present in stoichometrically equivalent amounts

throughout the reaction 1) 0.2 M HNO3 [H+] = 0.2 M pH = 0.70 2) = 1.00 x 10-3 mols NaOH 10 mL of 0.1 M NaOH 50 mL of 0.2 M HNO3 = 1.00 x 10-2 mols HNO3 * i . . . . doesn’t change throughout the reaction HNO3 +NaOH → NaNO3 + H2O HNO3 NaOH NaNO3 H2O i 1.00 x 10-2 mols 1.00 x 10-3 mols 0 mols 0 mols c -y -y +y +y e 9.0 x 10-3 mols 0 mols 1.0 x 10-3 mols 1.0 x 10-3 mols

= 0.15 M HNO3 pH = 0.82 HNO3 + NaOH → NaNO3 + H2O 3) 9.0 x 10-3 mols HNO3 pH = 0.82 50 mL + 10 mL = 60 mL 0.06 L 3) = 1.00 x 10-2 mols NaOH 100 mL of 0.1 M NaOH 50 mL of 0.2 M HNO3 = 1.00 x 10-2 mols HNO3 * duh HNO3 + NaOH → NaNO3 + H2O HNO3 NaOH NaNO3 H2O i 1.00 x 10-2 mols 1.00 x 10-2 mols 0 mols 0 mols c -y -y +y +y e 0 mols 0 mols 1.0 x 10-2 mols 1.0 x 10-2 mols

pH = 7.00 HNO3 +NaOH → NaNO3 + H2O = 0.025 M NaOH [OH-] = 0.025 4) [H+] = [OH-] = @ equivalence point = 1.50 x 10-2 mols NaOH 150 mL of 0.1 M NaOH 50 mL of 0.2 M HNO3 * duh = 1.00 x 10-2 mols HNO3 HNO3 +NaOH → NaNO3 + H2O HNO3 NaOH NaNO3 H2O i 1.00 x 10-2 mols 1.50 x 10-2 mols 0 mols 0 mols c -y -y +y +y e = 0.025 M NaOH 0 mols 5.0 x 10-3 mols 1.0 x 10-2 mols 1.0 x 10-2 mols 5.0 x 10-3 mols NaOH [OH-] = 0.025 150 mL + 50 mL = 200 mL pH = 12.4 0.2 L

Titration of a weak acid with a strong base ∙ before equivalence point use Ka ∙ @ equivalence point and beyond use Kb

HC2H3O2 + NaOH → NaC2H3O2 + H2O HC2H3O2 ⇆ H+ + C2H3O2- 1) HC2H3O2 + NaOH → NaC2H3O2 + H2O Neutralization reaction (this does not occur . . . . yet., no NaOH) HC2H3O2 ⇆ H+ + C2H3O2- Equilibrium dissociation 50 mL of 0.10 M HC2H3O2 HC2H3O2 H+ C2H3O2- [i] 0.10 M 0 M 0 M c -y +y +y [e] 0.10 - y y y pH = 2.87 y = 1.34 x 10-3 M [H+] = 1.34 x 10-3 M

throughout the reaction 2) HC2H3O2 + NaOH → NaC2H3O2 + H2O = 1.00 x 10-3 mols NaOH 50 mL of 0.10 M HC2H3O2 = 5.00 x 10-3 mols HC2H3O2 10 mL of 0.10 M NaOH * i . . . . doesn’t change throughout the reaction HC2H3O2 NaOH NaC2H3O2 H2O i 5.00 x 10-3 mols 1.00 x 10-3 mols 0 mols 0 mols c -y -y +y +y e 4.00 x 10-3 mols 0 mols 1.0 x 10-3 mols 1.0 x 10-3 mols

Reaction to completion yields: = 0.067 M HC2H3O2 4.0 x 10-3 mols HC2H3O2 0.06 L 10 mL + 50 mL = 60 mL = 0.0167 M C2H3O2- 1.0 x 10-3 mols C2H3O2- 0.06 L

HC2H3O2 ⇆ H+ + C2H3O2- pH = 4.14 [H+] = 7.22 x 10-5 M HC2H3O2 H+ [i] 0.067 M 0 M 0.0167 M c -y +y +y [e] 0.067 - y y 0.0167 + y y = 7.22 x 10-5 M [H+] = 7.22 x 10-5 M pH = 4.14

@ equivalence point and beyond use Kb 4) HC2H3O2 + NaOH → NaC2H3O2 + H2O = 5.00 x 10-3 mols NaOH 50 mL of 0.10 M HC2H3O2 = 5.00 x 10-3 mols HC2H3O2 50 mL of 0.10 M NaOH HC2H3O2 NaOH NaC2H3O2 H2O i 5.00 x 10-3 mols 5.00 x 10-3 mols 0 mols 0 mols c -y at equivalence point!! -y +y +y e 0 mols 0 mols 5.0 x 10-3 mols 5.0 x 10-3 mols @ equivalence point and beyond use Kb

Reaction to completion yields: pH = 8.72 = 0.050 M C2H3O2- 5.0 x 10-3 mols C2H3O2- C2H3O2- ⇆ HC2H3O2 + OH- 0.10 L C2H3O2- HC2H3O2 OH- [i] 0.050 M 0 M 0 M c -y +y +y [e] 0.05 - y y y pOH = 5.28 y = 5.3 x 10-6 M [OH-] = 5.3 x 10-6 M

@ equivalence point and beyond use Kb 5) HC2H3O2 + NaOH → NaC2H3O2 + H2O = 6.00 x 10-3 mols NaOH 50 mL of 0.10 M HC2H3O2 = 5.00 x 10-3 mols HC2H3O2 60 mL of 0.10 M NaOH HC2H3O2 NaOH NaC2H3O2 H2O i 5.00 x 10-3 mols 6.00 x 10-3 mols 0 mols 0 mols c -y -y +y +y e 0 mols 1.0 x 10-3 mols 5.0 x 10-3 mols 5.0 x 10-3 mols @ equivalence point and beyond use Kb

Reaction to completion yields: pH = 11.96 = 0.045 M C2H3O2- 5.0 x 10-3 mols C2H3O2- = 0.009 M OH- 0.11 L C2H3O2- ⇆ HC2H3O2 + OH- 1.0 x 10-3 mols OH- 0.11 L C2H3O2- HC2H3O2 OH- [i] 0.045 M 0 M 0.009 M c -y +y +y [e] 0.045 - y y 0.009 + y pOH = 2.04 y = 3.11 x 10-9 M [OH-] = 9.0 x 10-3 M

Titration of a weak base with a strong acid ∙ before equivalence point use Kb ∙ @ equivalence point and beyond use Ka

HCl + NH3 → NH4Cl NH3 ⇆ NH4+ + OH- 100 mL of 0.05 M NH3 pH = 11.0 1) Neutralization reaction (this does not occur . . . . yet., no HCl) NH3 ⇆ NH4+ + OH- Equilibrium dissociation 100 mL of 0.05 M NH3 NH3 NH4+ OH- [i] 0.05 M 0 M 0 M c -y +y +y [e] 0.05 - y y y pH = 11.0 y = 9.5 x 10-4 M [OH-] = 9.5 x 10-4 M

HCl + NH3 → NH4Cl 15 mL of 0.10 M HCl 100 mL of 0.05 M NH3 2) = 1.5 x 10-3 mols HCl 100 mL of 0.05 M NH3 = 5.00 x 10-3 mols NH3 15 mL of 0.10 M HCl HCl NH3 NH4Cl i 1.5 x 10-3 mols 5.00 x 10-3 mols 0 mols c -y -y +y e 0 mols 3.5 x 10-3 mols 1.5 x 10-3 mols

Reaction to completion yields: = 0.013 M NH4+ 1.5 x 10-3 mols NH4+ 0.115 L 15 mL + 100 mL = 115 mL = 0.0304 M NH3 3.5 x 10-3 mols NH3 0.115 L

NH3 ⇆ NH4+ + OH- pH = 9.62 [OH-] = 4.2 x 10-5 M NH3 NH4+ OH- [i] c [e] -y +y +y [e] 0.0304 - y 0.013 + y y y = 4.2 x 10-5 M [OH-] = 4.2 x 10-5 M pH = 9.62

@ equivalence point and beyond use Ka 3) HCl + NH3 → NH4Cl = 5.00 x 10-3 mols HCl 100 mL of 0.05 M NH3 = 5.00 x 10-3 mols NH3 50 mL of 0.10 M HCl HCl NH3 NH4Cl i 5.0 x 10-3 mols 5.00 x 10-3 mols 0 mols c -y -y +y e at equivalence point!! 0 mols 0 mols 5.00 x 10-3 mols @ equivalence point and beyond use Ka

Reaction to completion yields: = 0.033 M NH4+ 5.0 x 10-3 mols NH4+ NH4+ ⇆ NH3 + H+ 0.15 L NH4+ NH3 H+ [i] 0.033 M 0 M 0 M c -y +y +y [e] 0.033 - y y y pH = 5.37 y = 4.3 x 10-6 M [H+] = 4.3 x 10-6 M

@ equivalence point and beyond use Ka 3) HCl + NH3 → NH4Cl = 7.50 x 10-3 mols HCl 100 mL of 0.05 M NH3 = 5.00 x 10-3 mols NH3 75 mL of 0.10 M HCl HCl NH3 NH4Cl i 7.5 x 10-3 mols 5.00 x 10-3 mols 0 mols c -y -y +y e 2.5 x 10-3 mols 0 mols 5.00 x 10-3 mols @ equivalence point and beyond use Ka

Who cares about the equilibrium?? Reaction to completion yields: = 0.0143 M HCl 2.5 x 10-3 mols HCl Who cares about the equilibrium?? 0.175 L Not me This will be 0.0143 + y Which will be 0.0143 [H+] = 0.0143 pH = 1.84