Question

The zinc content of a 1.64 g ore sample was determined by dissolving the ore in HCl, which reacts with the zinc, and then neutralizing excess HCl with NaOH.

The ore was dissolved in 150 mL of 0.600 M HCl, and the resulting solution was diluted to a total volume of 300 mL. A 20.0 mL aliquot of the final solution required 8.71 mL of 0.510 M NaOH for the HCl present to be neutralized. What is the mass percentage (%w/w) of Zn in the ore sample?

Answer 1

Zn    +   2 HCl    ---------------> ZnCl2 + H2

HCl   +   NaOH    -------------> NaCl +   H2O

excess moles of HCl in 20 mL dilute solution = moles of NaOH

moles of NaOH = 8.71 x 0.510 / 1000 = 4.44 x 10^-3

excess moles of HCl in 300 mL dilute solution = 300 / 20 x 4.44 x 10^-3

                                                                        = 0.0666 mol

initial moles of HCl = 150 x 0.6 / 1000 = 0.09

moles of HCl reacted = 0.09 - 0.0666 = 0.02337

moles of Zn = 1/2 x 0.02337 = 0.01168

mass of Zn = 0.01168 x 65.38 = 0.7639 g

% of Zn =( 0.7639 / 1.64 ) x 100

(w/w)% of Zn = 46.58 %