1. 1 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  2NaCl + H 2 CO 3 mmol Na 2 CO 3 = ½ mmol HCl (mg Na 2 CO 3 /FW) = ½ x ( M HCl x V mL (HCl) )

2. 2 (mg Na 2 CO 3 /106) = ½ x 0.1067 x 40.72 mg Na 2 CO 3 = 106 * ½ x 0.1067 x 40.72 = 230 % Na 2 CO 3 = (230 x 10 -3 g/0.4671 g ) x 100 = 49.3 %

3. 3 Example How many mL of 0.25 M NaOH will react with 5.0 mL of 0.10 M H 2 SO 4. H 2 SO 4 + 2 NaOH  Na 2 SO 4 + 2 H 2 O mmol NaOH = 2 mmol H 2 SO 4 M NaOH x V mL(NaOH) = 2 {M (H2SO4) x V mL (H2SO4) } 0.25 x V mL = 2 x 0.10 x 5.0 V mL = 4.0 mL

4. 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 H 2 SO 4 ) x (0.10 mmol H 2 SO 4 / mL H 2 SO 4 ) x 5.0 mL H 2 SO 4 = 4.0 mL

5. 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 Na 2 CO 3 + 2 HCl  2 NaCl + H 2 CO 3

6. 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 Na 2 CO 3 Now let us substitute for mmol HCl by M HCl X V mL, and substitute for mmol carbonate by mg carbonate/FW carbonate. This gives

7. 7 M HCl x 35.86 = 2 * 187.6 mg/ (106 mg/mmol) M HCl = 0.09872 M The same result can be obtained using dimensional analysis in one single step: ? mmol HCl/mL = (187.6 mg Na 2 CO 3 /35.86 mL HCl) x ( mmol Na 2 CO 3 /106 mg Na 2 CO 3 ) x (2 mmol HCl/mmol Na 2 CO 3 ) = 0.09872 M

8. 8 Example An acidified and reduced iron sample required 40.2 mL of 0.0206 M KMnO 4. Find mg Fe (at wt = 55.8) and mg Fe 2 O 3 (FW = 159.7 mg/mmol). Solution The first step is to write the chemical equation MnO 4 - + 5 Fe 2+ + 8 H +  Mn 2+ + 5 Fe 3+ + 4 H 2 O mmol Fe = 5 mmol KMnO 4 (1)

9. 9 Now substitute for mmol Fe by mg Fe/at wt Fe and substitute for mmol KMnO 4 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 KMnO 4 /mL) x 40.2 mL x (5 mmol Fe/mmol KMnO 4 ) x ( 55.8 mg Fe/mmol Fe) = 231 mg

10. 10 To calculate the mg Fe 2 O 3 we set the following 2Fe  Fe 2 O 3 mmol Fe = 2 mmol Fe 2 O 3 Substitute for mmol Fe in equation 1 2* [mg Fe 2 O 3 / (159.7 mg/mmol)] = 5 x (0.0206 mmol/mL) x 40.2 mL mg Fe 2 O 3 = 331 mg The last step can also be done using dimensional analysis as follows: ? mg Fe 2 O 3 = (0.0206 mmol KMnO 4 /mL) x 40.2 mL x (5 mmol Fe/mmol KMnO 4 ) x (mmol Fe 2 O 3 /2 mmol Fe) x ( 159.7 mg Fe 2 O 3 /mmol Fe 2 O 3 ) = 331 mg

11. 11 Example A 1.00 g Al sample required 20.5 mL EDTA. Find the % Al 2 O 3 (FW = 101.96) in the sample if 30.0 mL EDTA required 25.0 mL of 0.100 M CaCl 2. Solution We should know that EDTA reacts in a 1:1 ratio with metal ions Therefore, the equation is Al 3+ + EDTA  Al-EDTA mmol Al = mmol EDTA(1) However, 2Al  Al 2 O 3 mmol Al = 2 mmol Al 2 O 3, substitute in equation (1)

12. 12 2*mmol Al 2 O 3 = 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 CaCl 2 Molarity x V mL (EDTA) = Molarity x V mL (CaCl 2 ) M EDTA x 30.0 = 0.100 x 25.0 M EDTA = 0.0833 M

13. 13 Now we can solve the problem using relation (1) 2*[mg Al 2 O 3 / ( 101.96 mg/mmol)] = 0.0833 x 20.5 ? mg Al 2 O 3 = 87.1 mg % Al 2 O 3 = (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 Al 2 O 3 = (0.0833 mmol EDTA/mL) x 20.5 mL x(mmol Al/mmol EDTA) x (mmol Al 2 O 3 /2mmol Al) x (101.96 mg Al 2 O 3 /mmol Al 2 O 3 ) = 87.1 mg

14. 14 We have seen in previous sections that correct solution of any problem involving reactions between two substances requires setting up two important relations: 1.Writing a balanced chemical equation representing stoichiometric relationships. 2.Formulating a relationship between the number of mmol of substance A and mmol of substance B. 3.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 V mL mmol = mg/FW

15. 15 Example Find the volume of 0.100 M KMnO 4 that will react with 50.0 mL of 0.200 M H 2 O 2 according to the following equation: 5 H 2 O 2 + 2 KMnO 4 + 6 H +  2 Mn 2+ + 5 O 2 + 8 H 2 O 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. 16 mmol KMnO 4 = 2/5 mmol H 2 O 2 It is clear that we should substitute M x V mL for mmol in both substances as this information is given to us. 0.100 x V mL = ( 2/5) x 0.200 x 50.0 V mL = 40.0 mL KMnO 4