Find the molarity of the HNO3 solution and its absolute uncertainty.

Question

A solution of HNO3 is standardized by reaction with pure sodium carbonate.

2H+ +Na2CO3⟶2Na+ +H2O +CO2

A volume of 28.24 ± 0.05 mL of HNO3 solution was required for complete reaction with 0.9373 ± 0.0009 g of Na2CO3 , (FM 105.988 ± 0.001 g/mol ).

Find the molarity of the HNO3 solution and its absolute uncertainty.

Note: Significant figures are graded for this problem. To avoid rounding errors, do not round your answers until the very end of your calculations.

I'm not sure what I'm doing wrong, I keep doing the problem over and over and running through my attempts. Can someone please help me figure out what I'm doing wrong. I keep inputting these numbers below. :(

HNO3 = 0.6263 M

±  0.28 M

Answer 1

Finding molarity of HNO3:

2HNO₃ + Na₂CO₃ ==> 2NaNO₃ + CO₂ + H₂O

0.9373 Na2CO3 x 1 mol Na2CO3/105.988 g = 0.0086000 moles Na2CO3

0.0086000 moles Na2CO3 x 2 mol HNO3/mol Na2CO3 = 0.017200 moles HNO3

0.017200 moles HNO3/28.24 ml x 1000 ml/L = 0.609067 M

Uncertainty: 0.05 + 0.0009 + 0.001 = 0.0519 = ±0.05

HNO3 = 0.61 M ± 0.05

Answer 2

Hi again, Angie G.,

Do your calculation using the nominal values, then with those at each limit of the stated ranges (write your neutralization calculation out and figure out which extreme for each datum will generate the maximum, then the minimum, for the MHNO3). Your absolute uncertainty is a +value, and a -value, around that nominal result. Incidentally, your uncertainty is way too large; I didn't check your nominal result.

Your neutralization equation could take the form of MHNO3 * volumeHNO3 = 2 * gNa2CO3/FMNa2CO3 , then solve for just the MHNO3 term.

-- Cheers, -- Mr. d.