Back Titration question! Unable to solve like normal question >_<?

Question

A sample of 6.5g of impure calcium oxide is dissolved completely in 250.0 cm3 of 1 mol/dm3 dilute HNO3. The resulting mixture is then titrated with NaOH solution, 0.5 mol/dm3. 20.0 cm3 of the solution requires 14.10 cm3 of NaOH solution for complete neutralisation.

1) Write down the chemical rxn between the dilute HNO3 and NaOH.

2) How many moles of unreacted nitric acid were in the mixture?

3) How many moles of HNO3 were originally present in the solution.

4) CaO rxts with dilute HNO3 to form calcium nitrate and water. From the answers in (2) and (3), how can I derive the number of moles and hence the mass, of CaO present in the sample?

5) Calculate the % purity of the CaO sample.

(Haha, sorry to post the whole question here, but I seriously haven't learnt back titration, and am just learning it before the teacher starts on it).

Answer 1

I will answer the questions in the order taght you have given them:

1. HNO3 + NaOH → NaNO3 + H2O

2. From the above you will see that 1mol NaOH reacts with 1 mol HNO3, whichg makes our calculations much easier.

What is the molarity of the HNO3 remainig?

M1V1 = M2V2

20.0 * M1  =  14.10*0.5

M1 = 14.10*0.5/20

M1 = 0.3525M

Therefore in the 250ml solution you will have 0.3525/4 = 0.088 moles of HNO3 unreacted by the CaO (answer Q2)

3.) The original HNO3 solution was 1.00M Therefore in 250ml of the solution you originally had 0.25moles of HNO3(answer to Q3)

4)Now think about what has happened. You started off with a solution of HNO3 that contained 0.25mol HNO3. You added some CaO, and you ended up with a solution that contained 0.088moles of HNO3. It is clear that the CaO reacted with :

0.25 - 0.088 = 0.162moles of HNO3

Balanced equation:

CaO + 2HNO3 → Ca(NO3)2 + H2O

1mol CaO reacts with 2mol HNO3

You have found above that the CaO reacted with 0.162mol HNO3. How much CaO reacted? It must be 0.162/2 = 0.081moles CaO

Molar mass CaO = 56.077.

0.081mol = 56.077*0.081 = 4.542g CaO

You started with 6.50g impure CaO, which you found contains 4.542g pure Cao

%purity = 4.542/6.50*100 = 69.88% puruty.