Question

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.

Answer 1

Are you having problems finding the molarity of the uncertainty?

Answer 2

Well I need both the molarity and the uncertainty and I am not sure how to do any of it.

Answer 3

can you write an equation for the reaction?

Answer 4

2H+ + Na2CO3 -> 2Na+ + H2O+ CO2

Answer 5

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

Answer 6

\(n_{Na_2CO_3}=\dfrac{m_{Na_2CO_3}}{M_{Na_2CO_3}}\), btw.

Answer 7

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?

Answer 8

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),

\(\dfrac{n_A}{\color{red}{a}}=\dfrac{n_B}{\color{blue}{b}}=\dfrac{n_C}{\color{green}{c}}\)

Answer 9

so is the H+ the same as HNO3?

Answer 10

for the purpose of the reaction, yes, because HNO3 is a strong acid (fully dissociates) and it's monoprotic.

Answer 11

So now I have .015714 moles of HNO3 and ).02665L of HNO3.

Answer 12

Which divid moles and L I get my molarity

Answer 13

great, now find the molarity

\(M=\dfrac{n_{solute}}{L_{solution}}\)

Answer 14

now how do I get the absolute uncertainty?

Answer 15

okay, use these equations:

Answer 16

i'll show you the first calculation.

Answer 17

m=0.8328 ± 0.0007 g ; M 105.988 ± 0.001 g/mol
a \(\Delta a\) b \(\Delta b\)

we divided, so we use \[s_y=n_{Na_2CO_3}*\sqrt{\dfrac{\Delta a}{a}+\dfrac{\Delta b}{b}}\]

Answer 18

damn i forgot to write squared, it should be like this:

\[s_y=n_{Na_2CO_3}*\sqrt{(\dfrac{0.0007 }{0.8328})^2+(\dfrac{0.001}{105.988}})^2\]

Answer 19

does it make sense?

Answer 20

I think so, I am trying it right now.

Answer 21

so the n is the moles right?

Answer 22

so where is the +or - 0.04

Answer 23

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

Answer 24

These types of calculations can be a pain because you have to do them for every step an uncertainty is given

Answer 25

so I got 6.6*10^-6 is that right?

Answer 26

0.00000660465

yes

Answer 27

It said it was wrong?

Answer 28

the second part.

Answer 29

which is the second part?

Answer 30

the absolute uncertainty

Answer 31

that's not the final answer, you have to do it again when you're finding molarity

Answer 32

ok........

Answer 33

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

Answer 34

Ok I got it as 0.001 thanks!

Answer 35

good stuff !