Question

A sample containing 2.00 x 10-2 mmol of Cu(NO3)2 and 3.00 x 10-2 mmol of K2SO4 is passed through an anion-exchange column. Calculate the volume of 0.0100 M HCl needed to titrate the eluate.

Answer 1

Given:

mili moles of Cu(NO3)2 = 2.00 E-2

m moles of K2SO4= 3.00 E-2

[HCl] = 0.0100 M

Solution:

We have to calculate volume of HCl

Lets write reaction between copper nitrate and potassium sulfate.

Cu(NO3)2 (aq) + K2SO4 -- > CuSO4 (s) + 2KNO3 (aq)

From this reaction we can write mole ratio of potassium sulfate and copper nitrate

Mole ratio of copper nitrate to potassium sulfate is 1 : 1

Calculation of limiting reactant:

Moles of potassium sulfate required to react with 2.00E-2 m moles of copper nitrate

= 2.00 E-2 m moles of copper nitrate x 1 moles of potassium sulfate / 1m mol copper nitrate

= 2.00 E-2 m moles of potassium sulfate

Actually there are 3.00 E-2 m moles of potassium sulfate so copper nitrate is limiting reactant.

So the moles of Copper sulfate formation depends on the moles of copper nitrate.

Moles of copper sulfate = 2.00 E-2 m moles of copper nitrate x 1 mol copper sulfate / 1 m mol copper nitrate

= 2.00 E-2 m moles of copper sulfate.

Calculation of m moles of HCl:

Lets write reaction between HCl and copper sulfate

2HCl (aq ) + CuSO4 (s) ---- > H2SO4 (aq) + CuCl2 (s)

Moles of HCl required

= 2.0 E-2 m moles of copper sulfate x 2 m moles of HCl / 1 m moles of copper sulfate

= 4.0 E-2 m moles of HCl

Lets convert m moles of HCl to moles

= 4.0 E-5 m moles x 1 mol / 1000 m moles

= 4.0 E-4 mole HCl

Calculation of volume of HCl

= 4.0 E-5 mol HCl / 0.0100 M HCl

= 0.004 L

conversion f volume into mL

= 0.004 L x 1000 mL / 1 L

= 4.0 mL

So volume of HCl required to react = 4.0 mL