1.44g of sodium ethanedioate was dissolved in water and the volume made up to 200cm3. 20.00cm3 of this solution, after acidification with dilute sulphuric acid, reduced 17.90cm3 of a solution of potassium manganate(VII).
a) Calculate the conc. of the sodium ethanedioate solution in mold dm-3
1.44/134 = 0.0107.../0.2 = 0.054 mol dm-3
I don't know how to do b) please explain
b) Calculate the conc. of the potassium manganate(VII) solution, in mol dm-3 and g dm-3.
Okay, so you know 5 moles of sodium ethanedioate reacts with 2 moles of potassium permanganate. You know how much ethanedioate you used (1.44g in 200cm^3, you used 20cm^3 so 0.144g) so you can work out that there's going to be 2/5 of that many moles of permanganate, which you can also get in grams, and divide these by the volume of permanganate you used.
We assume that part (a) is correct (assumpition is the mother of all confusion)
Concentration of ester: 0.054 mol dm-3
1000 cm3 contains 0.054 mol
1000 cm3 : 0.054 mol
20.00 cm3 : (0.054 X 20.00/1000) mol
(we simply multiply both sides of ratio by 20/1000)
(0.054 X 20.00/1000) mol ester reacts
now if 1 mol ester reacts, then 2/5 manganate mol reacts (see above)
so if (0.054 X 20.00/1000) mol ester reacts
Then (2/5 X 0.054 X 20.00/1000) mol manganate reacts (but 17.90 cm3 reacts.)
17.90cm3 contains (2/5 X 0.054 X 20.00/1000) mol manganate
17.90cm3 : (2/5 X 0.054 X 20.00/1000) mol manganate
1000cm3 : 1000/17.90 X (2/5 X 0.054 X 20.00/1000) mol manganate
(we simply multiply both sides of the ratio by 1000/17.90)
BUT 1000cm3 = 1 dm3
So concentration of manganate IS:
1000/17.90 X (2/5 X 0.054 X 20.00/1000) mol dm-3
This is how I work out these type of questions. It may be a bit long and tedious, but I find I understand everything this way. If you don't understand, I will try to explain. Other's may have a much better way at working out the problem. PM me anytime.