![10
[mg Fe/(55.8 mg/mmol)] = 5 x (0.0206
mmol/mL) x 40.2 mL
mg Fe = 231 mg
This can also be done in a single step as
follows:
? mg Fe = (0.0206 mmol KMnO4 /mL) x
40.2 mL x (5 mmol Fe/mmol KMnO4) x
(55.8 mg Fe/mmol Fe) = 231 mg](https://image.slidesharecdn.com/stoichiometric-calculationspart2-230417211604-1a7203e9/85/Stoichiometric-calculations-part-2-ppt-10-320.jpg)
![11
To calculate the mg Fe2O3 we set the following:
2Fe g Fe2O3
mmol Fe = 2 mmol Fe2O3
Substitute for mmol Fe in equation 1
2* [mg Fe2O3/ (159.7 mg/mmol)] = 5 x (0.0206 mmol/mL)
x 40.2 mL
mg Fe2O3 = 331 mg](https://image.slidesharecdn.com/stoichiometric-calculationspart2-230417211604-1a7203e9/85/Stoichiometric-calculations-part-2-ppt-11-320.jpg)
Stoichiometric-calculations part 2.ppt
- 1. 1 Stoichiometric Calculations: Volumetric Analysis In this section we look at calculations involved in titration processes as well as general quantitative reactions. In a volumetric titration, an analyte of unknown concentration is titrated with a standard in presence of a suitable indicator. For a reaction to be used in titration the following characteristics should be satisfied:
- 2. 2 1. The stoichiometry of the reaction should be exactly known. This means that we should know the number of moles of A reacting with 1 mole of B. 2. The reaction should be rapid and reaction between A and B should occur immediately and instantly after addition of each drop of titrant (the solution in the burette). 3. There should be no side reactions. A reacts with B only.
- 3. 3 4. The reaction should be quantitative. A reacts completely with B. 5. There should exist a suitable indicator which has distinct color change. 6. There should be very good agreement between the equivalence point (theoretical) and the end point (experimental). This means that Both points should occur at the same volume of titrant or at most a very close volume.
- 4. 4 Standard solutions A standard solution is a solution of known and exactly defined concentration. Usually standards are classified as either primary standards or secondary standards. There are not too many secondary standards available to analysts and standardization of other substances is necessary to prepare secondary standards. A primary standard should have the following properties:
- 5. 5 1. Should have a purity of at least 99.98% 2. Stable to drying, a necessary step to expel adsorbed water molecules before weighing 3. Should have high formula weight as the uncertainty in weight is decreased when weight is increased 4. Should be non hygroscopic 5. Should possess the same properties as that required for a titration
- 6. 6 Remember!! NaOH and HCl are not primary standards and therefore should be standardized using a primary or secondary standard. NaOH absorbs CO2 from air, highly hygroscopic, and usually of low purity. HCl and other acid in solution are not standards as the percentage written on the reagent bottle is a claimed value and should not be taken as guaranteed.
- 7. 7 Examples A 0.4671 g sample containing NaHCO3 (FW = 84.01 mg/mmol) was dissolved and titrated with 0.1067 M HCl requiring 40.72 mL. Find the percentage of bicarbonate in the sample. We should write the equation in order to identify the stoichiometry NaHCO3 + HCl g NaCl + H2CO3
- 8. 8 Now it is clear that the number of mmol of bicarbonate is equal to the number of mmol HCl mmol NaHCO3 = mmol HCl mmol = M x VmL mmol NaHCO3 = (0.1067 mmol/ml ) x 40.72 mL = 4.345 mmol Now get mg bicarbonate by multiplying mmol times FW mg NaHCO3 = 4.345 mmol x (84.01 mg/mmol) = 365.01 % NaHCO3 = (365.01 x 10-3 g/0.4671 g) x 100 = 78.14%
- 9. 9 An acidified and reduced iron sample required 40.2 mL of 0.0206 M KMnO4. Find mg Fe (at wt = 55.8) and mg Fe2O3 (FW = 159.7 mg/mmol). The first step is to write the chemical equation MnO4 - + 5 Fe2+ + 8 H+ g Mn2+ + 5 Fe3+ + 4 H2O mmol Fe = 5 mmol KMnO4 Now substitute for mmol Fe by mg Fe/at wt Fe and substitute for mmol KMnO4 by molarity of permanganate times volume, we then get:
- 10. 10 [mg Fe/(55.8 mg/mmol)] = 5 x (0.0206 mmol/mL) x 40.2 mL mg Fe = 231 mg This can also be done in a single step as follows: ? mg Fe = (0.0206 mmol KMnO4 /mL) x 40.2 mL x (5 mmol Fe/mmol KMnO4) x (55.8 mg Fe/mmol Fe) = 231 mg
- 11. 11 To calculate the mg Fe2O3 we set the following: 2Fe g Fe2O3 mmol Fe = 2 mmol Fe2O3 Substitute for mmol Fe in equation 1 2* [mg Fe2O3/ (159.7 mg/mmol)] = 5 x (0.0206 mmol/mL) x 40.2 mL mg Fe2O3 = 331 mg
- 12. 12 Homework: Find the volume of 0.100 M KMnO4 that will react with 50.0 mL of 0.200 M H2O2 according to the following equation: 5H2O2 + 2KMnO4 + 6H+ g 2Mn2+ + 5O2 + 8H2O Solution We have the equation ready therefore the following step is to formulate the relationship between the number of moles of the two reactants. We always start with the one we want to calculate, that is
- 13. 13 mmol KMnO4 = 2/5 mmol H2O2 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 = (2/5) x 0.200 x 50.0 VmL = 40.0 mL KMnO4
- 14. 14 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.
- 15. 15 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.
- 16. 16 Why Do We Use Back-Titrations? Back-titrations are important especially in some situations like: 1. When the titration reaction is slow. Addition of an excess reagent will force the reaction to proceed faster. 2. When the titration reaction lacks a good indicator. We will see details of this later. 3. When the analyte is not very stable. Addition of excess reagent will finish the analyte instantly thus overcoming stability problems.
- 17. 17 Examples 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
- 18. 18 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:
- 19. 19 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%
- 20. 20 Homework: 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. 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
- 21. 21 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%
- 22. 22 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
- 23. 23 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.
- 24. 24 Examples 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).
- 25. 25 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
- 26. 26 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%
- 27. 27 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
- 28. 28 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
- 29. 29 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
- 30. 30 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.
- 31. 31 Example What is the titer of a 5.442g/L K2Cr2O7 (FW = 294.2 mg/mmol) in terms of mg Fe2O3 (FW = 159.7 mg/mmol). The equation is: 6 Fe2+ + Cr2O7 2- + 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.
- 32. 32 mmol Fe2+ = 6*mmol K2Cr2O7 2Fe g Fe2O3 mmol Fe = 2 mmol Fe2O3 2*mmol Fe2O3 = 6*mmol K2Cr2O7 2*(mg Fe2O3/159.7) = 6*(5.442/294) * 1 mg Fe2O3 = 8.86 mg Therefore, the titer of K2Cr2O7 in terms of Fe2O3 is 8.86 mg Fe2O3 per mL K2Cr2O7