Question

A solution of HNO3 is standardized by reaction with pure sodium carbonate. A volume of 28.94 ± 0.05 mL of HNO3 solution was required for complete reaction with 0.9596 ± 0.0007 g of Na2CO3, (FM 105.988 ± 0.001). Find the molarity of the HNO3 and its absolute uncertainty.

I found the correct molarity to be .626 but I'm having trouble with the uncertainty

Answer 1

2 HNO3 + Na2CO3 2 NaNO3 + H2O + CO2

Mass of Na2CO3 = 0.9596 ± 0.0007 g

Molar mass of Na2CO3 = 105.988 ± 0.001 g/mol

Moles of Na2CO3 = Mass / Molar mass

= 0.9596 ± 0.0007 / 105.988 ± 0.001

= 0.009054 ± 0.000007

Now, from equation we have:

2 moles of HNO3 are required for complete reaction with 1 mole Na2CO3.

So,

Moles of HNO3 required = 2 * (0.009054 ± 0.000007)

= 0.018108 ± 0.000007

Volume of HNO3 = 28.94 ± 0.05 mL

= 0.02894 ± 0.05 L

Molarity of HNO3 = 0.018108 ± 0.000007 / 0.02894 ± 0.05

= 0.626 ± 1.08 M