Verified answer

answer is [HNO3] = 0.696M ± 0.002M

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I'm not sure which class you're in but I'm going to use formal propagation of errors to solve this.

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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)

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now let's setup a conversion..

(0.8935±0.0008g Na2CO3 / 24.24±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

get it?

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now to the calculations..

(0.8935±0.0008 / 24.24±0.06) x (1 / 105.988±0.001) x (2 / 1) x (1000 / 1) = what?

well.. this is all multiplication and division.. and if you notice, the terms in the √ sign for the error propogation equations for multiplication and division are the "relative errors".. (ΔA/A for example).. and for EXACT numbers that ΔA=0.. so (ΔA/A)² = 0 and we can ignore the error terms for exact numbers..

the numbers 1, 2, 1, 1000, and 1 in that equatlion are all exact so we don't have to include those in the calculation of error. and.....

(0.8935±0.0008 / 24.24±0.06) x (1 / 105.988±0.001) x (2 / 1) x (1000 / 1) = (0.8935 / 24.24 / 105.988 x 2 x 1000) ± (0.8935 / 24.24 / 105.988 x 2 x 1000) x √ ((0.0008/0.8935)² + (0.06/24.24)² + (0.001/105.988)² )

and that of course = 0.695561 ± 0.695561 x 0.002632223

and that = 0.695561 ± 0.00183

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now.. what decimal place should we report that to? Let's report the ± with 1 digit and therefore we round 0.00183 to 0.002 and therefore report the value to the 0.001's decimal point.

[HNO3] = 0.696M ± 0.002M

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follow any of that?

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by the way.. if you stopped to think about what the answer would be if we used "sig figs"..

(0.894 / 24.2) x (1 / 105.988) x (2 / 1) x (1000 / 1) = 0.696M

and would be assumed to be ± 1 in the rightmost digit so we would report this as 0.696M± 0.001M which is why we say sig figs is a fairly decent SIMPLE ESTIMATE of error.

You are told that you are completing the reaction of sodium carbonate with nitric acid. So first you need to calculate how many moles of sodium carbonate you have:

0.8935gm Na2CO3 * (mole Na2CO3 / 105.988gm Na2CO3) = 0.00843mole Na2CO3

You can see from the reaction given that 2 moles of nitric acid will be required per mole of sodium carbonate since two hydrogen ions are required to react with each sodium carbonate. So:

0.00843mole Na2CO3 * (2mole HNO3 / mole Na2CO3) = 0.01686mole HNO3

This is the number of moles of nitric acid required to complete the reaction. You were given that 24.24mL of nitric acid solution was required to supply this many moles of nitric acid, so we know that there were 0.01686moles of HNO3 in 24.24mL of solution. So all we need to do now is get the correct units for molarity (moles/liter) and you have your answer!

(0.01686mole HNO3 / 24.24mL solution) * (1000mL / liter) = 0.6956 M HNO3

So this is the "average" answer to your question. To find the absolute uncertainty, you just need to redo the problem using the upper or lower limit numbers for each measured item to determine the absolute highest and lowest values possible. Note that to get the highest limit you will need to use the upper limit of each number in the numerator, but the lower limit for each number in the denominator. And vice versa to get the lowest limit. Good luck!

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