How many mL of 0.0850 M NaOH are needed to titrate each of the following solutions to the equivalence point?

Question

a. 40.0 mL of 0.900 M HNO₃.

b. 35.0 mL of 0.0850 M HC₂H₃O₂.

Answer 1

The equivalence point is reached when moles of acid equals moles of base.

(a). HNO3 + NaOH = NaNO3 + H2O

moles of acid (HNO3) = 40.0 ml x 1 L / 1000 m x 0.900 mol / L = 0.036 mols

volume of base (NaOH) needed = 0.036 mols HNO3 x 1 mol NaOH/mol HNO3 x 1 L / 0.0850 mol = 0.4235 L

volume of base (NaOH) = 424 mls (3 sig. figs.)

(b). HC2H3O2 + NaOH ==> NaC2H3O2 + H2O

moles of acid (HC3H3O2) = 35.0 ml x 1 L / 1000 ml x 0.0850 mol / L = 0.002975 mols

moles of base (NaOH) needed = 0.002975 mol acid x 1 mol NaOH / mol acid = 0.002975 mols NaOH

volume of NaOH = 0.002975 mols x 1 L / 0.0850 mol = 0.0350 L = 35.0 mls