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The zinc content of a 1.16 g ore sample was determined by dissolving the ore in

ID: 500126 • Letter: T

Question

The zinc content of a 1.16 g ore sample was determined by dissolving the ore in HCI, which reacts with the zinc, and then neutralizing excess HCI with NaOH. The reaction of HCl with Zn is shown below. Zn(s) + 2HCl(aq) rightarrow ZnCI_2(aq) + H_2(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 8.75 mL of 0.514 M NaOH for the HCl present to be neutralized. What is the mass percentage (%w/w) of Zn in the ore sample?

Explanation / Answer

Answer:

1) The Zn ore sample dissolved in 150.0 mL of 0.6 M HCl and then volume made to 300.0 mL.

say M1 = 0.6 M and V1 = 150.0 mL and M2 = ? if V2 = 300.0 mL.

By diliyion law,

M2V2 = M1V1

M2 x 300.0 = 0.6 x 150.0

M2 = 0.6 x 150.0 / 300.0

M2 = 0.3 M.

Molarity of HCl solution in which ore sample dissolved is 0.3 M.

20.0 mL aliquot of 0.3 M HCl before addition of ore expected to have.

milimoles of HCl = molarity x volume in mL = 0.3 x 20.0 = 6 milimoles.

Expected milimoles of HCl = 6 mmol.

But titration of this 20.0 mL of aliquote required 8.75 mL of 0.514 M NaOH.

So Actual mmol HCl = mmol of NaOH = 8.75 x 0.514 = 4.5 mmol

Hence,

mmol of HCl reacted with Zn in sample = expected mmol of HCl - Actual mmol of HCl = 6 - 4.5 = 1.5

mmol of HCl reacted with Zn in sample = 1.5 mmol

Stoichiometric relation between Zn and HCl say that (1 : 2 stoichiometry),

2 mmol HCl = 1 mmol Zn

so, 1.5 mmol HCl = 0.75 mmol Zn

There are 0.75 mmol Zn in 20.0 mL aliquote then how many in 300.0 mL solution?

For this we write,

If, 20.0 mL solution = 0.75 mmol Zn

Then 300.0 mL solution = say 'A' mmol Zn

On cross multiplication,

A x 20 = 0.75 x 300.0

A = 11.25 mmol Zn = 0.01125 mol Zn.

Molar mass of Zn = 65.38 g/mol

So, Mass of 0.01125 mol Zn = 0.01125 x Molar mass of Zn = 0.01125 x 65.38 = 0.74 g

Mass of Zn in 1.6 g ore sample = 0.74.

% Zn (w/w) = (Actual mass of Zn / Mass of Ore) x 100

= (0.74 / 1.16) x 100

= 63.8

Answer : 63.8 % Zn.

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