Question

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

2H+ + Na2Co3 arrowto 2Na+ + H2O + CO2

A volume of 25.36+/- 0.06mL of HNO3 solution was required for complete reaction with 0.9572 +/- 0.0008 g of Na2CO3 (FM 105.988 +/- 0.001). Find the molarity of the HNO3 and its absolute uncertainty.

            HNO3=                  M          +/-                       M      (sig figs count for this)             If you could show me step-by-step how to do this problem, that would be helpful.

Answer 1

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

2H⁺ + Na₂CO₃ ---> 2Na⁺ +H₂O + CO₂

A volume of 25.36 ± 0.06 mL of HNO₃ solution was required for complete reaction with 0.9572 ± 0.0008 g of Na₂CO₃, (FM 105.988 ± 0.001). Find the molarity of the HNO₃ and its absolute uncertainty.

by formal propogation of errors...

addition and subtraction..
... given the values.. A ± ΔA and B ± ΔB
... .and the operation.. .Z = A + B
.... therefore... Z ± ΔZ = Z ± √ (ΔA² + ΔB²)

multiplication and division...
... given the values.. A ± ΔA and B ± ΔB
... .and the operation.. .Z = A x B
.... therefore... Z ± ΔZ = Z ± Z x √ ((ΔA/A)² + (ΔB/B)²)

*******
now let's setup a conversion..

(0.9572 ± 0.0008 g Na2CO3 / 25.36 ± 0.06 mL HNO3) x (1 mol Na2CO3 / 105.988±0.001g Na2CO3) x (2 moles HNO3 / 1 mol Na2CO3) x (1000mL / 1L) = __ mol / L HNO3

right? if you have trouble with that equation.. look at just the units and let's let A = Na2CO3 and B = HNO3..

.... .... ...... .. .. . ...coefficients of the balanced equation.
... ... ... ... ... ... ..... ...... ..... ..... ... .. ..↓
(g A / mL B) x (mol A / g A) x (mol B / mol A) x (mL / L) = mol B / L
... ... .. .. ... ... .. .... ...↑... ... ... .... ... ... .... ... ... .. .. ..↑
.. .... ... ... .. ...molar mass A.. . ..... .... .converts mL to L

(0.9572 ± 0.0008 / 25.36 ± 0.06) x (1 / 105.988±0.001) x (2 / 1) x (1000 / 1) = (0.9572 / 25.36 / 105.988 x 2 x 1000) ± (0.9572 / 25.36 / 105.988 x 2 x 1000) x √ ((0.0008/0.9572)² + (0.06/25.36)² + (0.001/105.988)² )

and that = 0.71224062149 ± 0.00178717475

now.. what decimal place should we report that to? Let's report the ± with 1 digit and therefore we round 0.00178717475 to 0.002 and therefore report the value to the 0.001's decimal point.

[HNO₃] = 0.712M ± 0.002M

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