Acid-Base Titrations - Soda Lime and Hydrochloric Acid

[ajassat]

Homework Statement

"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

Homework 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

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,
Adam

 

[Borek]

Hint: Consider the reactions separately

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?

 

[ajassat]

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?

--
chemical calculators - buffer calculator, concentration calculator
www.titrations.info - all about titration methods

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?

 

[Borek]

Looks OK.

 

[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

 

[ajassat]

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

Borek: Please confirm :)

 

[Borek]

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

 

[ajassat]

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

NB: See adjusted Calculation and answer