anonymous
  • anonymous
A volume of 26.65 ± 0.04 mL of HNO3 solution was required for complete reaction with 0.8328 ± 0.0007 g of Na2CO3, (FM 105.988 ± 0.001). Find the molarity of the HNO3 and its absolute uncertainty.
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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aaronq
  • aaronq
Are you having problems finding the molarity of the uncertainty?
anonymous
  • anonymous
Well I need both the molarity and the uncertainty and I am not sure how to do any of it.
aaronq
  • aaronq
can you write an equation for the reaction?

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More answers

anonymous
  • anonymous
2H+ + Na2CO3 -> 2Na+ + H2O+ CO2
aaronq
  • aaronq
so first were gonna find the volume, then we'll do the uncertainty after. find the moles of Na2CO3, then using the stoichiometric coefficients find the moles of HNO3
aaronq
  • aaronq
\(n_{Na_2CO_3}=\dfrac{m_{Na_2CO_3}}{M_{Na_2CO_3}}\), btw.
anonymous
  • anonymous
I got 0.007857 moles of Na2CO3. Then I couldn't find the stoichiometric coefficients. The one for Na2CO3 is 1mole but I dont know what the one is for HNO3?
aaronq
  • aaronq
the stoichiometric coefficients are present in the balanced equation, now you wanna find moles of HNO3 (or as it's written in your equation H+). build a ratio, like this: e.g. for a general reaction: \(\color{red}{a}A + \color{blue}{b}B\) \(\rightleftharpoons\) \( \color{green}{c}C\) where upper case are the species (A,B,C), and lower case (a,b,c) are the coefficients (in red), \(\dfrac{n_A}{\color{red}{a}}=\dfrac{n_B}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\)
anonymous
  • anonymous
so is the H+ the same as HNO3?
aaronq
  • aaronq
for the purpose of the reaction, yes, because HNO3 is a strong acid (fully dissociates) and it's monoprotic.
anonymous
  • anonymous
So now I have .015714 moles of HNO3 and ).02665L of HNO3.
anonymous
  • anonymous
Which divid moles and L I get my molarity
aaronq
  • aaronq
great, now find the molarity \(M=\dfrac{n_{solute}}{L_{solution}}\)
anonymous
  • anonymous
now how do I get the absolute uncertainty?
aaronq
  • aaronq
aaronq
  • aaronq
i'll show you the first calculation.
aaronq
  • aaronq
m=0.8328 ± 0.0007 g ; M 105.988 ± 0.001 g/mol a \(\Delta a\) b \(\Delta b\) we divided, so we use \[s_y=n_{Na_2CO_3}*\sqrt{\dfrac{\Delta a}{a}+\dfrac{\Delta b}{b}}\]
aaronq
  • aaronq
damn i forgot to write squared, it should be like this: \[s_y=n_{Na_2CO_3}*\sqrt{(\dfrac{0.0007 }{0.8328})^2+(\dfrac{0.001}{105.988}})^2\]
aaronq
  • aaronq
does it make sense?
anonymous
  • anonymous
I think so, I am trying it right now.
anonymous
  • anonymous
so the n is the moles right?
anonymous
  • anonymous
so where is the +or - 0.04
aaronq
  • aaronq
yeah, thats why you have to do the calculation first, because you need the actual answers to compute the uncertainty. the 0.04 is used later when you're calculating the molarity
aaronq
  • aaronq
These types of calculations can be a pain because you have to do them for every step an uncertainty is given
anonymous
  • anonymous
so I got 6.6*10^-6 is that right?
aaronq
  • aaronq
0.00000660465 yes
anonymous
  • anonymous
It said it was wrong?
anonymous
  • anonymous
the second part.
aaronq
  • aaronq
which is the second part?
anonymous
  • anonymous
the absolute uncertainty
aaronq
  • aaronq
that's not the final answer, you have to do it again when you're finding molarity
anonymous
  • anonymous
ok........
aaronq
  • aaronq
use the uncertainty from: moles of Na2CO3 = \(0.00785 \pm6.6*10^-6 \) moles as the uncertainty in moles of HNO3, then use: volume of 26.65 ± 0.04 mL of HNO3 for the next calculation
anonymous
  • anonymous
Ok I got it as 0.001 thanks!
aaronq
  • aaronq
good stuff !

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