Chemistry help!

ImRebecca

Could someone please show me how to do 2013 question 1 (f)? I'm so confused about how to start it. Thanks in advance.

After the three additions, the thoroughly mixed contents of the sample bottle were titrated in 200 cm3 portions with a 0.02 M solution of sodium thiosulfate (Na2S2O3). The average titre was 9.4 cm3.
The balanced equation for the titration reaction is:
2SO2– +I  SO2– + 2I– 23246
(d) Describe how the burette was rinsed and filled for use in the titrations. (15)
(e) Name the indicator used in the titrations. (3)
(f) Calculate the concentration of iodine (I2) in the sample bottle in moles per litre.
For every one mole of dissolved oxygen (O2) in the water sample, two moles of iodine (I2) are liberated in this experiment. Calculate the concentration of dissolved oxygen in the river water sample
(i) in moles per litre,
(ii) in grams per litre,
(iii) in ppm.

chocksaway

ImRebecca wrote: »

Could someone please show me how to do 2013 question 1 (f)? I'm so confused about how to start it. Thanks in advance.

Want to actuallystick up the question so people can help?

rainbowtrout

OK, the equation you have put up there is wrong but ignoring that for the moment


You need to use the formula

(V1 * M1 )/N1 = (V2 * M2)/N2


Volume thiosulphate V1 = 9.4
Molarity thiosulphate M1 = 0.02
No. moles thiosulphate N1 = 2 (taken from left hand side of equation)

Volume V2 = 200
Molarity M2 = X
No. moles iodine N2 = 1 (taken from left hand side of equation)


Slot these values into equation above

(9.4 * 0.02)/2 = (200 * x)/1

Rearrange to solve for X, x= 0.00047 moles of iodine

You were told for every two moles of iodine there is one mole of O2. Therefore there is twice as much iodine as oxygen so to get moles of oxygen you need to divide 0.00047 by 2

0.00047/2 = 0.000235 moles of O2



Second part : grams of oxgen

Get relative molecular mass of O2 = 16*2 = 32g

multiply grams by moles

32 * 0.000235 = 0.000752g of Oxygen


Third part : ppm of Oxygen

PPM is the number of milligrams of a substance there are. There are 1000 milligrams in a gram

so

0.000752 * 1000 = 7.52 mg = 7.52 ppm of oxygen

ImRebecca

rainbowtrout wrote: »

OK, the equation you have put up there is wrong but ignoring that for the moment


You need to use the formula

(V1 * M1 )/N1 = (V2 * M2)/N2


Volume thiosulphate V1 = 9.4
Molarity thiosulphate M1 = 0.02
No. moles thiosulphate N1 = 2 (taken from left hand side of equation)

Volume V2 = 200
Molarity M2 = X
No. moles iodine N2 = 1 (taken from left hand side of equation)


Slot these values into equation above

(9.4 * 0.02)/2 = (200 * x)/1

Rearrange to solve for X, x= 0.00047 moles of iodine

You were told for every two moles of iodine there is one mole of O2. Therefore there is twice as much iodine as oxygen so to get moles of oxygen you need to divide 0.00047 by 2

0.00047/2 = 0.000235 moles of O2



Second part : grams of oxgen

Get relative molecular mass of O2 = 16*2 = 32g

multiply grams by moles

32 * 0.000235 = 0.000752g of Oxygen


Third part : ppm of Oxygen

PPM is the number of milligrams of a substance there are. There are 1000 milligrams in a gram

so

0.000752 * 1000 = 7.52 mp = 7.52 ppm of oxygen

Thank you so much!