Acid-Base Titration Q

[luken93]

25. (a) 1.461 g of dry sodium carbonate, Na2CO3, is dissolved in water in a 250 mL volumetric flask.
(i) Why can this be called a standard solution?
(ii) Calculate the concentration of the solution.

      (b) 20 mL aliquots of this solution were titrated with nitric acid, HNO3, three times. The average concordant titre was found to be 22.17 mL.
(i) Write the equation for the reaction of sodium carbonate with nitric acid.
(ii) Find the concentration of the nitric acid.

Thanks

[andy456]

a) i) You can accurately determine the concentration of the solution
ii)figure out n(Na2CO3)= mass/molar mass
then C=number of moles/ volume

b) i) Na2CO3+ 2HNO3 > 2NaNO3 + H2O + CO2
ii) number of moles of HNO3 = 2 x number of moles of Na2CO3
C=n/v

[Russ]

a)
i) There are various criteria for a standard solution, I can't remember them all off the top of my head, but it's stuff like "known formula" and "can be stored without reacting" etc.

ii) C = n/v
n = m/M = 1.461/106 = .01378 mol
v = .250 L
C = .01378 / .250 = .0551M

b)
i) Na2CO3 + 2HNO3 ---> CO2 + H2O + 2NaNO3

ii)

[luken93]

a) i) You can accurately determine the concentration of the solution
ii)figure out n(Na2CO3)= mass/molar mass
then C=number of moles/ volume

b) i) Na2CO3+ 2HNO3 > 2NaNO3 + H2O + CO2
ii) number of moles of HNO3 = 2 x number of moles of Na2CO3
C=n/v

yeh i got a and b) i), but i'm not sure what volume you use to find the concentration. is it the concordant or the aliquots?

[andy456]

a) i) You can accurately determine the concentration of the solution
ii)figure out n(Na2CO3)= mass/molar mass
then C=number of moles/ volume

b) i) Na2CO3+ 2HNO3 > 2NaNO3 + H2O + CO2
ii) number of moles of HNO3 = 2 x number of moles of Na2CO3
C=n/v

yeh i got a and b) i), but i'm not sure what volume you use to find the concentration. is it the concordant or the aliquots?

yes.... that is the only volume they give in regards to the nitric acid

[luken93]

a) i) You can accurately determine the concentration of the solution
ii)figure out n(Na2CO3)= mass/molar mass
then C=number of moles/ volume

b) i) Na2CO3+ 2HNO3 > 2NaNO3 + H2O + CO2
ii) number of moles of HNO3 = 2 x number of moles of Na2CO3
C=n/v

yeh i got a and b) i), but i'm not sure what volume you use to find the concentration. is it the concordant or the aliquots?

yes.... that is the only volume they give in regards to the nitric acid

ummm yes to which one??? haha
soz if im a noob haha

[andy456]

sorry i only read to concordant then zoned out.... my bad,
its the concordant titres because that is the volume of nitric acid used
the aliquots are of the Na2CO3 solution

[luken93]

sorry i only read to concordant then zoned out.... my bad,
its the concordant titres because that is the volume of nitric acid used
the aliquots are of the Na2CO3 solution

ok thanks
so it should be:
Mol of Nitric Acid =
Concentration = ? or is it just 1 titre of .02217?

Anyways, the answer in the back is 0.099 mol L^-1?

Thanks again

[luken93]

sorry i only read to concordant then zoned out.... my bad,
its the concordant titres because that is the volume of nitric acid used
the aliquots are of the Na2CO3 solution

ok thanks
so it should be:
Mol of Nitric Acid =
Concentration = ? or is it just 1 titre of .02217?

Anyways, the answer in the back is 0.099 mol L^-1?

Thanks again

dw i've worked it out, its:
Concentration =

[andy456]

Ok I've got it.
Such an easy question too...

First:
n(Sodium Carbonate in 250ml) = 1.461g/106gmol^-1
                             = 0.0138mol
C (Sodium Carbonate in 250ml) = 0.0138mol/0.25L
                              = 0.0551mol/L

Na2CO3 + 2HNO3 > 2NaNO3 + H20+ CO2

C(Sodum Carbonate in 250ml) = 0.0551mol/L
n (Sodium Carbonate in 20ml) = 0.0551mol/L x 0.02L
                                        = 0.001102mol
n (Nitric Acid) = 0.001102 x 2
                   = 0.002204mol
C (Nitric Acid) = 0.002204mol/0.02217L
                    = 0.099mol/L

Sorry it took so long for me to explain

[luken93]

Ok I've got it.
Such an easy question too...

First:
n(Sodium Carbonate in 250ml) = 1.461g/106gmol^-1
                             = 0.0138mol
C (Sodium Carbonate in 250ml) = 0.0138mol/0.25L
                              = 0.0551mol/L

Na2CO3 + 2HNO3 > 2NaNO3 + H20+ CO2

C(Sodum Carbonate in 250ml) = 0.0551mol/L
n (Sodium Carbonate in 20ml) = 0.0551mol/L x 0.02L
                                        = 0.001102mol
n (Nitric Acid) = 0.001102 x 2
                   = 0.002204mol
C (Nitric Acid) = 0.002204mol/0.02217L
                    = 0.099mol/L

Sorry it took so long for me to explain

cheers, maybe i need to look into titration abit more haa

[andy456]

They're pretty easy and become mechanical after a while.... I would say the most difficult part is determining when to include a dilution factor in the calculations...

[luken93]

They're pretty easy and become mechanical after a while.... I would say the most difficult part is determining when to include a dilution factor in the calculations...

i'm guessing this occurs in year 12?
just finished the stoich chapter for year 11, veerrrrrryyy mechanical, and 62 chapter review q's........

[Russ]

Oh wow, I just realised I didn't do bii yesterday. Oh well, you got an answer for it eventually, sorry bout that.

Yeah dilutions come up in year 12, year 11 stoich is (from memory) mostly just getting you used to the concept of a mole ratio

[luken93]

Oh wow, I just realised I didn't do bii yesterday. Oh well, you got an answer for it eventually, sorry bout that.

Yeah dilutions come up in year 12, year 11 stoich is (from memory) mostly just getting you used to the concept of a mole ratio

yep pretty much. and 60 review questions at the end of stoich chapter