Question

1. What is a “back-titration”?

2. Why is a back titration necessary in this experiment?

3. Calculate the number of moles of base equivalents in: a) 675 mg CaCO3 b) 135 mg Mg(OH)2 - Base equivalents = number of moles of acid (H+ ) consumed

4. Calculate the number of moles of acid (H+ ) in 33.6 mL of (a) 0.10 M HCl and (b) 0.10 M H2SO4.

5. Calculate the pH of each of the following solutions: a) 0.10 M HCl b) 2.0 x 10-4 M HNO3 c) 0.015 M NaOH d) 12 M HCl

6. Suppose that the pH meter indicates that the pH of three solutions is 2.00, 7.15, and 10.50, respectively. Calculate the [H3O+ ] of each of these solutions in molarity.

7. What quantities are plotted on the x- and y-axes of an acid-base titration curve?

8. What do the rapid changes or “breaks” in a titration curve represent?

9. What does the equivalence point in a titration represent?

Answer 1

1) A titration in which the concentration of the analyte (substance) is found by reacting it with the known excess reagent is known as a back titration.

2) Back titration is useful determine the concentration of unknown analyte.

3) Number of equivalent of CaCO3 = 0.675 g / 50.045 g/eq = 2.69e-4 Eq [ Eq.wt = 50.045 g/Eq]

Number of equivalent of Mg(OH)2 = 0.135 g/29.16 g/eq = 4.62e-3 eq [Eq. wt = 29.16 g/eq]

4) Number of mols of H+ in HCl = (0.0336 L)(0.10 mol/L) = 0.00336 mol

Number of mols of H+ in H2SO4 = (0.0336 L)(0.10 mol/L)2 = 0.00672 mol