What volume of the solution would produce 0.287 g of AgCl if 26 ml of a HCl solution required 26.25 ml of 0.095 N NaOH in titration?

Answer 1

Reaction 1: HCl + NaOH ==> NaCl + H2O

mols NaOH used = 26.25 ml x 1 L / 1000 ml x 0.095 mol / L = 0.002494 mols NaOH

mols HCl in 26 ml of HCl solution = 0.002494 mols since it is present in a 1:1 mol ratio to NaOH

Reaction 2: HCl(aq) + AgX(aq) ==> AgCl(s) + HX(aq)

0.287 g AgCl x 1 mol AgCl / 143.3 g = 0.00200 mols AgCl

To obtain 0.00200 mols AgCl would require 0.00200 moles HCl. The HCl solution from (1) above has 0.002494 moles HCl in 26 mls. Therefore...

0.002494 mols / 26 mls = 0.00200 mols / x mls

x = 20.9 mls HCl required