Question

The zinc content of a 1.40 g ore sample was determined by dissolving the ore in HCl, which reacts with the zinc, and then neutralizing excess HCl with NaOH. The reaction of HCl with Zn is shown below.

Zn (s) + 2HCl (aq) --> ZnCl2 (aq) + H2 (g)

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 9.97 mL of 0.518 M NaOH for the HCl present to be neutralized. What is the mass percentage (%w/w) of Zn in the ore sample?

Answer 1

You need to calculate how much HCl (acid) there was initially and how much acid was left after reacting with the ore.

initial moles HCl = (Molarity HCl)(L HCl) = (0.600)(0.150) = 0.0900 moles HCl

You took a 20.0-mL aliquot from the 300 mL of dissolved ore to test for HCl content. HCl reacts with NaOH in a 1:1 mole ratio: NaOH + HCl ==> H2O + NaCl.

moles NaOH used in the titration = (Molarity NaOH)(L NaOH) = (0.518)(0.00997) = 0.00516446 moles NaOH

Since NaOH and HCl react in a 1:1 mole ratio, moles HCl = moles NaOH = 0.00516446 in the 20.0 mL aliquot.

moles HCl in the 300 mL of dissolved ore = moles HCl in 20.0 mL x dilution factor = (0.00516446)(300/20) = 0.0774669 moles HCl that did not react with the ore.

moles HCl that reacted with ore = initial moles HCl - moles of unreacted HCl = 0.0900 - 0.0774669 = 0.0125331 moles HCl

The balanced equation tells us that it takes 2 moles of HCl to react with 1 mole of Zn.

0.0125331 moles HCl x (1 mole Zn / 2 moles HCl) = 0.00626655 moles Zn

0.00626655 moles Zn x (65.38 g Zn / 1 mole Zn) = 0.4097 g Zn

%Zn = (grams Zn / grams of ore) x 100 = (0.4097 / 1.40) x 100 = 29.26 % Zn