# Homework Help: Acid-Base Titrations - Soda Lime and Hydrochloric Acid

1. Oct 24, 2009

### ajassat

1. The problem statement, all variables and given/known data
"Soda lime is 85.0% NaOH and 15.0% CaO. What volume of 0.500M Hydrochloric Acid is required to completely neutralise 2.50g of the Soda Lime?"
Hint: Consider the reactions separately

2. Relevant equations

Formation of Soda Lime:
CO2 + H2O >> H2CO3
H2CO3 + 2NaOH >> Na2CO3 + heat
Na2CO3 + Ca(OH)2 >> CaCO3 + 2NaOH

number of moles = mass / relative formula mass
volume = number of moles divided by concentration

3. The attempt at a solution

1)Find the number of moles of Soda Lime (from the equations describing its formation)
2)Deduce the number of moles of HCL from the equation describing the titration
3)Work out volume of HCL using number of moles/0.500

4. Main problems encountered

*I don't know the chemical formula for soda lime/I cannot deduce its relative formula mass
*Because of the above I cannot write a balanced equation for the titration

Guidance is greatly appreciated.

Regards,

2. Oct 24, 2009

### Staff: Mentor

Forget about soda lime and its production, it doesn't matter. You have 2.5 g of a mixture of NaOH and CaO. How much NaOH in the mixture? How much acid required for titration of NaOH? What about CaO?

Last edited by a moderator: Aug 13, 2013
3. Oct 24, 2009

### ajassat

Since there 85% of NaOH, the mass of NaOH in the 2.5g mixture is:
(85/100)*2.5 = 2.12500

Therefore the mass of CaO in the mixture is:
2.5 - 2.12500 = 0.375g

The titration of NaOH:
NaOH + HCL >> NaCl + H20
number of moles of NaOH = m/Mr = 2.12500/40 = 0.053125 mol
From equation number of moles of HCL = 0.053125 mol
Therefore volume(HCL) = n/c = (0.053125/0.500)*1000 = 106.25 cm3

Is this first value for volume correct?

4. Oct 24, 2009

### Staff: Mentor

Looks OK.

Last edited by a moderator: Aug 13, 2013
5. Oct 24, 2009

### ajassat

Continuing with the titration of CaO

CaO + 2HCL >> CaCl2 + H2O

number of moles = m/mr = 0.375/56 = 0.00669642857
From equation, number of moles (HCL) = 2*0.00669642857 = 0.013938571
Therefore:
V = n/c = (0.0134/0.500)*1000 = 26.8cm3

Last edited: Oct 24, 2009
6. Oct 24, 2009

### ajassat

Adding the two calculated volumes gives
total volume of HCL required = 26.8 + 106.25 = 133.05cm3 or 133cm3 (to 3 sig fig0

Last edited: Oct 24, 2009
7. Oct 24, 2009

### Staff: Mentor

Approach is correct, but you have eaten one digit, so the final result is off.

Last edited by a moderator: Aug 13, 2013
8. Oct 24, 2009

### ajassat

It should be: 0.0133928571
Which when rounded to 3 sig fig is 0.0134