Question

0.1200 g of potassium dichromate primary standard is dissolved, quantitatively transferred into a 250-mL volumetric flask and diluted to volume. 0.3000 g an unknown iron ore sample was dissolved in hydrochloric acid, treated with stanous chloride and mercury(II) chloride, quantitatively transferred to a 100-mL volumetric flask and diluted to volume. 25.00 mL of the unknown iron(II) solution was pipetted into 500-mL Erlenmyer flask and titrated with the potassium dichromate solution. 28.00 mL of potassium dichromate is required to reach the end point.

Calculate the moles of iron(II) in the original unknown sample.

I've tried:

0.3000g * (1 mol /55.845g) = 0.005372 mol Fe(2)

but it says it's a wrong answer.

Answer 1

Solution :-

Balanced redox equation for the reaction between the iron (ll) ion and potassium dichromate is as follows

K2Cr2O7 +6Fe^2+(aq)+14H^+(aq)---- > 2Cr^3+(aq) +6Fe^3+(aq)+7H2O(l)+2K^+(aq)

The mole ratio of the potassium dichromate and iron ion is 1 : 6

Step 1) Calculating the moles of potassium dichromate

Moles=mass/molar mass

Moles of K2Cr2O7=0.1200 g/(294.185 g/mol)

                                 =0.000408 mol

Calculating the moles of potassium dichromate in the 28.0 mL solution out of 250 mL total solution.

(0.000408 mol K2Cr2O7*28.0 mL)/(250 mL) =4.57*10^-5 mol K2Cr2O7

Now lets calculate the moles of iron (ll) ions in the 25.0 mL unknown solution.

(4.57*10^-5 mol K2Cr2O7 * 6 mol Fe^2+)/(1 mol K2Cr2O7) = 2.74*10^-4 mol Fe^2+

Now lets calculate the moles of iron (ll) ion in the 100 mL solution.

(2.74*10^-4 mol Fe^2+)*(100 mL)/(25.0 mL) =0.001096 mol Fe^2+ ions

Therefore the number of moles of the iron (ll) ions in the unknown iron sample solution are

0.001096 mol Fe^2+ ions