The Student Room Group

Titration help!

The concentration of silver ions in the equilibrium mixture can be found bytitration with potassium thiocyanate. Silver thiocyanate precipitates.Ag+(aq) + CNS–(aq) AgCNS(s) When all the silver ions have reacted, a deep red complex ion ofiron(III) thiocyanate forms. In an experiment, 25.0 cm3 of 0.100 mol dm–3 silver nitrate solution was addedto 25.0 cm3 of 0.100 mol dm–3 of iron(II) sulfate solution, mixed thoroughly, andallowed to stand overnight in an air-tight container. 10.0 cm3 samples of the reaction mixture were then titrated with 0.0200 mol dm–3potassium thiocyanate solution. The average titre was 5.60 cm3. (i) The initial concentrations of silver ions and iron(II) ions in the reactionmixture are the same. Calculate this initial concentration in mol dm–3. can some one help me on this question
I have also just come across this, it's quite infuriating as it's only 1 mark and I seem to be missing something so obvious!
Original post by SOREN_TORETTO
The concentration of silver ions in the equilibrium mixture can be found bytitration with potassium thiocyanate. Silver thiocyanate precipitates.Ag+(aq) + CNS–(aq) AgCNS(s) When all the silver ions have reacted, a deep red complex ion ofiron(III) thiocyanate forms. In an experiment, 25.0 cm3 of 0.100 mol dm–3 silver nitrate solution was added to 25.0 cm3 of 0.100 mol dm–3 of iron(II) sulfate solution, mixed thoroughly, and allowed to stand overnight in an air-tight container. 10.0 cm3 samples of the reaction mixture were then titrated with 0.0200 mol dm–3potassium thiocyanate solution. The average titre was 5.60 cm3. (i) The initial concentrations of silver ions and iron(II) ions in the reactionmixture are the same. Calculate this initial concentration in mol dm–3. can some one help me on this question


Presumably the equilibrium that is set up is:

Ag+ + Fe2+ <==> Ag(s) + Fe3+

Your titration gives the moles of silver ions = volume x molarity of thiocyanate ions.

Then this is in a 10 ml sample so divide by 0.01 dm3 to get molarity of silver ions.
Original post by charco
Presumably the equilibrium that is set up is:

Ag+ + Fe2+ <==> Ag(s) + Fe3+

Your titration gives the moles of silver ions = volume x molarity of thiocyanate ions.

Then this is in a 10 ml sample so divide by 0.01 dm3 to get molarity of silver ions.


What would your answer be?
Original post by Hairysweets
What would your answer be?


0.0056 x 0.02/0.01 = 0.0112 mol dm-3
Original post by charco
0.0056 x 0.02/0.01 = 0.0112 mol dm-3


Nope it's 0.05M (In the Mark scheme). Do we have to x5 for the dilution?

Okay, I got 0.112 x10-3 mol of Ag+ reacted which means there was a concentration of 0.00112M (1000n/v) in the second reaction.
So, if we then x5 for the dilution we get 0.0056 for each reactant mole..?
(edited 8 years ago)
Original post by Hairysweets
Nope it's 0.05M (In the Mark scheme). Do we have to x5 for the dilution?

Okay, I got 0.112 x10-3 mol of Ag+ reacted which means there was a concentration of 0.00112M (1000n/v) in the second reaction.
So, if we then x5 for the dilution we get 0.0056 for each reactant mole..?


The concentration of the sample (10ml) has to be the same as the concentration of the original 50ml.

Concentration doesn't change when you take a sample!

Moles of thiocyanate ions = moles of silver ions = 0.0056 x 0.02 = 0.000112

This is from a 10ml sample

Hence the sample has a concentration of 0.000112/0.01 = 0.0112 mol dm-3

Can you provide a link to the original question?
Original post by charco
The concentration of the sample (10ml) has to be the same as the concentration of the original 50ml.

Concentration doesn't change when you take a sample!

Moles of thiocyanate ions = moles of silver ions = 0.0056 x 0.02 = 0.000112

This is from a 10ml sample

Hence the sample has a concentration of 0.000112/0.01 = 0.0112 mol dm-3

Can you provide a link to the original question?


It's an Edexcel Unit 4, R Paper:

http://qualifications.pearson.com/content/dam/pdf/A%20Level/Chemistry/2013/Exam%20materials/6CH04_01R_que_20130612.pdf

Question 11 (b)(i):

Mark scheme:

http://qualifications.pearson.com/content/dam/pdf/A%20Level/Chemistry/2013/Exam%20materials/6CH04_01R_msc_20130815.pdf


I have worked out the concentration of silver ions at equilibrium!

The INITIAL concentration is just the 25 ml of 0.1 mol dm-3 made up to 50ml (by addition of the volume of iron solution)

Double the volume and halve the concentration = 0.1/2 = 0.05 mol dm-3

Quick Reply

Latest