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 reaction of HCl with Zn is shown below.

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

First, calculate:

initial amount of HCl used:

mmol = Macid * Vacid = 150mL*0.6 = 90 mmol of HCl

then, dilution to

V = 300 mL

[HCl] = mmol/V = 90/300 = 0.3 M of HCl

only V = 20 mL are used

mmol of HCl actually used = MV = 0.3*20 = 6 mmol of HCl used

mmol of NaOH = MV = 8.71 mL * 0.51 = 4.4421 mmol of NaOH used

mmol used in neutralization of NaOH = 4.4421 mmol of HCl are neutralized

therefore, this is the excess

the (6-4.4421) = 1.5579 mmol of H+ was ued in the zinc ore

Zn(s) + 2H+(aq) = H2(g) + Zn+2(aq)

then

1 mmol of Zn = 2 mmol of H+

1.5579 mmol of H+ = 1/2*1.5579 = 0.77895 mmol of Zn present

mass of Zn = mmol*MW = 0.77895*65.38 = 50.927751 mg of Zn present = 50.927751*10^-3 g

now,

% mass of zinc = mass of zinc / total sample * !005 = 50.927751*10^-3)/(1.64) * 100

% mass of Zinc in ore = 3.1053 % approx