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Example

Example. A 0.4671 g sample containing Na 2 CO 3 (FW = 106mg/mmol) was dissolved and titrated with 0.1067 M HCl requiring 40.72 mL. Find the percentage of carbonate in the sample. The equation should be the first thing to formulate Na 2 CO 3 +2 HCl g 2NaCl + H 2 CO 3

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Example

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  1. Example A 0.4671 g sample containing Na2CO3 (FW = 106mg/mmol) was dissolved and titrated with 0.1067 M HCl requiring 40.72 mL. Find the percentage of carbonate in the sample. The equation should be the first thing to formulate Na2CO3 +2 HCl g 2NaCl + H2CO3 mmol Na2CO3 = ½ mmol HCl (mg Na2CO3/FW) = ½ x ( MHCl x VmL (HCl) )

  2. (mg Na2CO3/FW) = ½ x ( MHCl x VmL (HCl) ) (mg Na2CO3/106) = ½ x 0.1067 x 40.72 mg Na2CO3 = 106 * ½ x 0.1067 x 40.72 = 230 % Na2CO3 = (230 x 10-3 g/0.4671 g ) x 100 = 49.3 %

  3. Example How many mL of 0.25 M NaOH will react with 5.0 mL of 0.10 M H2SO4. H2SO4 + 2 NaOH gNa2SO4 + 2 H2O mmol NaOH = 2 mmol H2SO4 MNaOH x VmL(NaOH) = 2 {M(H2SO4) x VmL(H2SO4)} 0.25 x VmL = 2 x 0.10 x 5.0 VmL = 4.0 mL

  4. We can also calculate the volume in one step using dimensional analysis: ? mL NaOH = (mL NaOH/0.25 mmol NaOH) x (2 mmol NaOH/mmol H2SO4) x (0.10 mmol H2SO4 / mL H2SO4) x 5.0 mL H2SO4 = 4.0 mL

  5. Example A 0.1876 g of pure sodium carbonate (FW = 106 mg/mmol) was titrated with approximately 0.1 M HCl requiring 35.86 mL. Find the molarity of HCl. Solution The first thing to do is to write the equation for the reaction. You should remember that carbonate reacts with two protons Na2CO3 + 2 HCl g 2 NaCl + H2CO3

  6. The second step is to relate the number of mmol HCl to mmol carbonate, Where it is clear from the equation that we have 2 mmol HCl and 1 mmol carbonate. This is translated to the following mmol HCl = 2 mmol Na2CO3 Now let us substitute for mmol HCl by MHCl X VmL, and substitute for mmol carbonate by mg carbonate/FW carbonate. This gives

  7. MHCl x 35.86 = 2 * 187.6 mg/ (106 mg/mmol) MHCl = 0.09872 M The same result can be obtained using dimensional analysis in one single step: ? mmol HCl/mL = (187.6 mg Na2CO3 /35.86 mL HCl) x ( mmol Na2CO3/106 mg Na2CO3) x (2 mmol HCl/mmol Na2CO3) = 0.09872 M

  8. Example 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). Solution 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(1)

  9. Now substitute for mmol Fe by mg Fe/at wt Fe and substitute for mmol KMnO4 my molarity of permanganate times volume, we then get [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

  10. To calculate the mg Fe2O3 we set the following 2Fe gFe2O3 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 The last step can also be done using dimensional analysis as follows: ? mg Fe2O3 = (0.0206 mmol KMnO4 /mL) x 40.2 mL x (5 mmol Fe/mmol KMnO4) x (mmol Fe2O3 /2 mmol Fe) x ( 159.7 mg Fe2O3/mmol Fe2O3) = 331 mg

  11. Example A 1.00 g Al sample required 20.5 mL EDTA. Find the % Al2O3 (FW = 101.96) in the sample if 30.0 mL EDTA required 25.0 mL of 0.100 M CaCl2. Solution We should know that EDTA reacts in a 1:1 ratio with metal ions Therefore, the equation is Al3+ + EDTA g Al-EDTA mmol Al = mmol EDTA (1) However, 2Al g Al2O3 mmol Al = 2 mmol Al2O3 , substitute in equation (1)

  12. 2*mmol Al2O3 = mmol EDTA The same procedure above is repeated in the calculation. First we substitute mg aluminum oxide/FW for the mmol of aluminum oxide and substitute mmol EDTA by molarity times volume. But we do not have the molarity of EDTA, therefore, let us calculate the molarity of EDTA. mmol EDTA = mmol CaCl2 Molarity x VmL(EDTA) = Molarity x VmL (CaCl2) MEDTA x 30.0 = 0.100 x 25.0 MEDTA = 0.0833 M

  13. Now we can solve the problem using relation (1) 2*[mg Al2O3/ ( 101.96 mg/mmol)] = 0.0833 x 20.5 ? mg Al2O3 = 87.1 mg % Al2O3 = (87.1 mg/1000 mg) x 100 = 8.71% Another approach: After calculation of the molarity of EDTA, one can do the rest of the calculation in a single step as follows: ? mg Al2O3 = (0.0833 mmol EDTA/mL) x 20.5 mL x(mmol Al/mmol EDTA) x (mmol Al2O3/2mmol Al) x (101.96 mg Al2O3/mmol Al2O3) = 87.1 mg

  14. We have seen in previous sections that correct solution of any problem involving reactions between two substances requires setting up two important relations: • Writing a balanced chemical equation representing stoichiometric relationships. • Formulating a relationship between the number of mmol of substance A and mmol of substance B. • The last step in the calculation is to substitute for the mmol A or B by either one of the following according to given information: mmol = M x VmL mmol = mg/FW

  15. Example 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: 5 H2O2 + 2 KMnO4 + 6 H+g 2 Mn2+ + 5 O2 + 8 H2O 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

  16. mmol KMnO4 = 2/5mmol 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

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