Calculation question*** (1 Viewer)

[coeyz]

The Winkler method may be used to determine the concentration of DO in a sample of water. In the reaction, 1mole of oxygen produces 2 moles of iodine. The iodine is then titrated with Na2S2O3, using starch as an indicator, according to the equation:

I2 + 2Na2S2O3 --> 2NaI + Na2S4O6

A 500 mL sample of river water was analysed using the Winkler method.
4.80 mL of 0.100mol/L Na2S2O3 solution was required to titrate the iodine that was formed.

a) Calculate the number of moles of iodine that was formed.
b) Calculate the concentration of oxygen in the river water in mol/L

I did a) and got the answer 2.4x10^-4 moles, but have no idea about question b? anyone please help? thanks@!!

 

[Pwnage101]

which past trial/HSC paper does this come from? cant remember

 

[coeyz]

which past trial/HSC paper does this come from? cant remember

I dont know.. coz this question is from the hw given by my tutor..

 

[Pwnage101]

Okay here it is:

your answer to (a) is correct.

for Na2S2O3:

n=cv

c=0.100mol/L and v=0.00480L

therefore n=4.8 x 10^-4 moles

ratio of I2 to Na2S2O3 is 1:2

therefore to get the number of moles of I2, divide number of moled of Na2S2O3 by 2

thus n(I2)=2.4x10^-4 moles

(b) we know 1 mole of O2 produces 2 moles of I2

thus n(O2) = 0.5 X n(I2) = 0.5 x 2.4x10^-4 moles = 1.2x10^-4 moles

but we need concentration of O2

c=n/v

here n = 1.2x10^-4 moles as calculated
v=0.5 litres (given)

thus c(O2)= 2.4 x 10^-4 mol/L

hope that helps

 

[DannyT]

haha i got 2.4 x 10^-4 mol/L as well...same answer as part (a) lol
simple Q

 

[annabackwards]

haha i got 2.4 x 10^-4 mol/L as well...same answer as part (a) lol
simple Q

Same here ^^

 

[coeyz]

Thanks heaps

 

[4lettersdown]

Wimkler method isnt in HSC, so it wont be in any past papers.

 

[nosh22]

Thank god. I was having a heart attack.