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Normality volumetric Calculations

# Normality volumetric Calculations

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## Normality volumetric Calculations

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1. 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 milli equivalents (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

2. The next step is to substitute for meq by one of the following relations meq = N x VmL meq = mg/eq wt meq = n x mmol We should remember from previous lectures that: eq wt = FW/n N = n x M Therefore, change from molarity or number of moles to normality or number of equivalents using the above relations.

3. Example A 0.4671 g sample containing sodium bicarbonate was titrated with HCl requiring 40.72 mL. The acid was standardized by titrating 0.1876 g of sodium carbonate (FW = 106 mg/mmol) requiring 37.86 mL of the acid. Find the percentage of NaHCO3 (FW=84.0 mg/mmol) in the sample. This problem was solved previously using molarity. The point here is to use normalities in order to practice and learn how to use the meq concept. First, let us find the normality of the acid from reaction with carbonate (reacts with two protons as you know).

4. Eq wt Na2CO3 = FW/2 = 53.0 and eq wt NaHCO3 = FW/1 = 84.0 meq HCl = meq Na2CO3 Now substitute for meq as mentioned above: Normality x volume (mL) = wt (mg)/ eq wt N x 37.86 = 187.6 mg/ (53 mg/meq) NHCl= 0.0935 eq/L

5. Now we can find meq NaHCO3 where  meq NaHCO3 = meq HCl mg NaHCO3 / eq wt = N x VmL mg NaHCO3 /84.0 = 0.0935 x 40.72 mg NaHCO3 = 84.0 x 0.0935 x 40.72 mg  % NaHCO3 = (84.0 x 0.0935 x 40.72 mg/476.1 mg) x 100 = 67.2%

6. Example Use normalities to calculate how many mL of a 0.10 M H2SO4 will react with 20 mL of 0.25 M NaOH. Solution We can first convert molarities to normalities: N = n x M N (H2SO4) = 2 x 0.10 = 0.20 N (NaOH) = 1 x 0.25 = 0.25

7. meq H2SO4 = meq NaOH Substitute for meq as usual (either NVmL or mg/eq wt) 0.20 x VmL = 0.25 x 20 VmL = 25 mL

8. Example Find the normality of sodium carbonate (FW = 106) in a solution containing 0.212 g carbonate in 100 mL solution if: The carbonate is used as a monobasic base. The carbonate is used as a dibasic base. Solution If carbonate is a monobasic base then eq wt = FW/1 = 106/1 = 106 mg/meq

9. To find the normality of the solution we find the weight per mL and then convert the weight per mL to meq/mL. We have 212 mg/100 mL which means 2.12 mg/mL. Now the point is how many meq per 2.12 mg sodium carbonate. meq = mg/eq wt = 2.12/106 = 0.02 meq Then the normality is 0.02 N If carbonate is to be used as a dibasic salt then the eq wt = FW/2 = 106/2 = 53 mg/meq.

10. To find the normality of the solution we find the weight per mL and then convert the weight per mL to meq/mL. We have 212 mg/100 mL which means 2.12 mg/mL. Now the point is how many meq per 2.12 mg sodium carbonate. meq = mg/eq wt = 2.12/53 = 0.04 meq Then normality will be 0.04 N

11. Example How many mg of I2 (FW = 254 mg/mmol) should you weigh to prepare 250 mL of 0.100 N solution in the reaction: I2 + 2e g 2 I- Solution To find the number of mg I2 to be weighed and dissolved we get the meq required. We have: meq = N x VmL meq = 0.100 x 250 = 25.0 meq mg I2 = meq x eq wt = meq x FW/2 = 25.0 x 254/2 = 3.18x103 mg

12. Example Find the normality of the solution containing 0.25 g/L H2C2O4 (FW = 90.0 mg/mmol). Oxalic acid reacts as a diacidic substance. Solution Since oxalic acid acts as a diacidic substance then its eq wt = FW/2 = 90.0/2 = 45.0 mg/meq

13. We convert the mg acid per mL to meq acid per mL We have 0.25 mg acid/mL meq acid = 0.25 mg/45.0 mg/meq = 0.0056 meq Therefore, the normality is 0.0056 meq/mL

14. The Titer Concept In many situations where routine titrations are carried out and to avoid wasting time in performing calculations, one can calculate the weight of analyte in mg equivalent to 1 mL of titrant. The obtained value is called the titer of the titrant. For example, an EDTA bottle is labeled as having a titer of 2.345 mg CaCO3. This means that each mL of EDTA consumed in a titration of calcium carbonate corresponds to 2.345 mg CaCO3. If the titration required 6.75 mL EDTA then we have in solution 2.345 x 6.75 mg of calcium carbonate.

15. Example What is the titer of a 5.442g/L K2Cr2O7 (FW = 294.2 mg/mmol) in terms of Fe2O3 (FW = 159.7 mg/mmol). The equation is: 6 Fe2+ + Cr2O72- + 14 H+g 6 Fe3+ + 2 Cr3+ + 7 H2O Solution 1 mL of K2Cr2O7 contains 5.442 mg K2Cr2O7 per mL. Therefore let us find how many mg Fe2O3 corresponds to this value of K2Cr2O7.

16. mmol K2Cr2O7 per mL = 5.442 mg/294.2 (mg/mmol) = 0.0185 mmol Fe2O3 = 3 mmol K2Cr2O7 mmol Fe2O3 = 3 x 0.0185 = 0.0555 mmol mg = mmol x FW mg Fe2O3 = 0.0555 mmol x (159.7 mg/mmol) = 8.86 mg Therefore, the titer of K2Cr2O7 in terms of Fe2O3 is 8.86 mg Fe2O3 per mL K2Cr2O7