Example   Find the volume of 0.100 M KMnO4 that will react with 50.0 mL of 0.200 M MnSO4 according to the following equation: 3 MnSO4 + 2 KMnO4 + 4 OH- g 5 MnO2 + 2 H2O + 2 K+ + 3 SO42- Solution Again, we have the equation ready therefore we turn to formulating a relationship for the number of mmol for both substances. mmol KMnO4 = 2/3 mmol MnSO4

mmol KMnO4 = 2/3 mmol MnSO4   It is clear that we should substitute M x VmL for mmol in both substances as this information is given to us. 0.100 x VmL = 2/3 x 0.200 x 50.0 VmL = 66.7 mL KMnO4

5 H2C2O4 + 2 KMnO4 + 6 H+ g 10 CO2 + 2 Mn2++ 8 H2O + 2 K+ Example   Find the volume of 0.100 M KMnO4 that will react with 500 mg of H2C2O4 (FW = 90.0mg/mmol) according to the following equation: 5 H2C2O4 + 2 KMnO4 + 6 H+ g 10 CO2 + 2 Mn2++ 8 H2O + 2 K+ Solution We have the equation ready therefore we turn to formulating a relationship for the number of mmol for both substances. mmol KMnO4 = 2/5 mmol H2C2O4

mmol KMnO4 = 2/5 mmol H2C2O4   It is clear that we should substitute M x VmL for mmol permanganate while substitute for mmol oxalic acid by mg/FW. This gives: 0.100 x VmL = 2/5 x 500 mg / (90.0 mg/mmol) VmL = 22.2 mL KMnO4

Back-Titrations In this technique, an accurately known amount of a reagent is added to analyte in such a way that some excess of the added reagent is left. This excess is then titrated to determine its amount and thus:   mmol reagent taken = mmol reagent reacted with analyte + mmol reagent titrated  Therefore, the analyte can be determined since we know mmol reagent added and mmol reagent titrated.

mmol reagent reacted = mmol reagent taken – mmol reagent titrated   Finally, the number of mmol reagent reacted can be related to the number of mmol analyte from the stoichiometry of the reaction between the two substances.

Why Do We Use Back-Titrations?   Back-titrations are important especially in some situations like: When the titration reaction is slow. Addition of an excess reagent will force the reaction to proceed faster. When the titration reaction lacks a good indicator. We will see details of this later. When the analyte is not very stable. Addition of excess reagent will finish the analyte instantly thus overcoming stability problems.

mmol EDTA reacted = mmol EDTA taken – mmol EDTA titrated Example   A 2.63 g Cr(III) sample was dissolved and analyzed by addition of 5.00 mL of 0.0103 M EDTA. The excess EDTA required 1.32 mL of 0.0112 M Zn(II). Calculate % CrCl3 (FW=158.4 mg/mmol) in the sample. Solution We should remember that EDTA reacts in a 1:1 ratio. The first step in the calculation is to find the mmol EDTA reacted from the relation: mmol EDTA reacted = mmol EDTA taken – mmol EDTA titrated

mmol EDTA reacted = mmol CrCl3. mmol EDTA titrated = mmol Zn(II). We substitute for both mmol EDTA taken and mmol EDTA titrated by M x VmL for each. Also since EDTA reacts in a 1:1 ratio, we can state the following:   mmol EDTA reacted = mmol CrCl3. mmol EDTA titrated = mmol Zn(II). Therefore, we now have the following reformulated relation: mmol CrCl3 = mmol EDTA taken – mmol Zn(II)  Substitution gives: mmol CrCl3 = 0.0103 x 5.00 – 0.0112 x 1.32 mmol CrCl3 = 0.0367 mmol

We can then find the number of mg CrCl3 by multiplying mmol times FW.  Mg CrCl3 = 0.0367 mmol x 158.4 mg/mmol = 5.81 mg  % CrCl3 = (5.81 mg/2.63x103 mg) x 100 = 0.221%  

Example   A 0.500 g sample containing sodium carbonate (FW=106 mg/mmol) was dissolved and analyzed by addition of 50.0 mL of 0.100 M HCl solution. The excess HCl required 5.6 mL of 0.05 M NaOH solution. Find the percentage of Na2CO3 n the sample. Solution First, write the chemical equations involved Na2CO3 + 2 HCl g 2 NaCl + H2CO3 HCl + NaOH g NaCl + H2O mmol HCl reacted = mmol HCl taken – mmol HCl titrated

mmol HCl reacted = 2 mmol Na2CO3 mmol HCl titrated = mmol NaOH   Now we can reformulate the above relation to read 2 mmol Na2CO3 = mmol HCl taken – mmol NaOH mmol Na2CO3 = 1/2 ( 0.100 x 50.0 – 0.050 x 5.6) = 2.36 mmol ? mg Na2CO3 = 2.36 mmol x 106 mg/mmol = 250 mg % Na2CO3 = (250 mg/0.500 x103 mg) x 100 = 50.0%

2 Fe2+ + MnO2 + 4 H+ g 2 Fe3+ + Mn2+ + 2 H2O Example   A 0.200 g sample containing MnO2 was dissolved and analyzed by addition of 50.0 mL of 0.100 M Fe2+ to drive the reaction 2 Fe2+ + MnO2 + 4 H+ g 2 Fe3+ + Mn2+ + 2 H2O The excess Fe2+ required 15.0 mL of 0.0200 M KMnO4. Find % Mn3O4 (FW= 228.8 mg/mmol) in the sample. 5 Fe2+ + MnO4- + 8 H+ g 5 Fe3+ + Mn2+ + 4 H2O Solution First, calculate mmol MnO2 then change it to mmol Mn3O4 . We have the relation

mmol Fe2+ reacted = mmol Fe2+ taken – mmol Fe2+ titrated Now we should perform the following substitutions:   mmol Fe2+ reacted = 2 mmol MnO2 mmol Fe2+ titrated = 5 mmol KMnO4 Now reformulate the above relation: 2 mmol MnO2 = mmol Fe2+ taken – 5 mmol KMnO4

Substitution gives:   mmol MnO2 = 1/2( 0.100 x 50.0 – 5 x 0.0200 x 15.0) = 1.75 mmol 3 MnO2 = Mn3O4 + O2 mmol Mn3O4 = 1/3 mmol MnO2 mmol Mn3O4 = 1/3 x 1.75 mmol = 0.5833 ?mg Mn3O4 = 0.5833 mmol x 228.8 mg/mmol = 133.5 mg % Mn3O4 = (133.5 mg/200 mg) x 100 = 66.7%

Normality volumetric Calculations We have seen previously that solving volumetric problems required setting up a relation between the number of mmoles or reacting species. In case of normality, calculation is easier as we always have the number of milliequivalents (meq) of substance A is equal to meq of substance B, regardless of the stoichiometry in the chemical equation. Of course this is because the number of moles involved in the reaction is accounted for in the calculation of meqs. Therefore, the first step in a calculation using normalities is to write down the relation   meq A = meq B