[GreenGiant87]

Hi to anyone who has opened this topic. Just joined this site a few minutes ago as I am stuck on a question. I would really appreciate if you could show me some guide lines as to how to do it as there are none like this in the book I am using nor could I find any on line
Ok the question is:

Consider the precipitation titration of 25ml of a solution that is 0.10 M in KI and a solution that is 0.05 M in AgNO3. Calculate the concentration of I^- and thus Ag^+ in solution after the addition of a 10ml, 20ml, 45ml, 50ml, 55ml and 60ml of the AgNO3 solution.

Using this data, construct a titration curve, plotting pAg against volume of AgNO3 solution added.
Ksp for AgI= 8.3*10^-17

[Borek]

http://www.titrations.info/precipitation-titration-curve-calculation

[GreenGiant87]

Thanks for the reply Borek. I checked out the link you sent me. Read it about 4 times but I still can't seem to understand the calculations I need to preform in order to get concentrations for I- and Ag+ after each titration.

[Borek]

Try. Explain where you get lost.

[GreenGiant87]

ok my interpretation is that you get the concentrations of both I- and Ag+ from the concentration of whats being titrated against them. Then you take both away leaving you with the difference in concentration and then you can get the pH. Would that be the correct way of doing this question?

[Borek]

Sorry, but I am not sure I understand your description (only thing I am sure about is that you have not meant pH). Could be you are right. Try to apply this approach to the first point on your titration curve - after 10 mL of titrant were added.

[GreenGiant87]

ok I had a look through my book again and this is the method I had been able to come up with. So if its wrong please tell me

Ok the first thing I do is:

I have an addition of 10ml 0.05 M in AgNO3 to the 25ml of 0.10 M in KI. The total volume of the solution is 35ml. The number of moles of AgNO3 in 10.0 ml is:

10.0mL multiplied by 0.05 mol AgNO3/1L AgNO3 multiplied by 1L/1000mL = 5.0*10^-4



Then to get the number of moles of KI originally present in 25mL of solution is:

25.0mL multiplied by 0.10mol KI/ 1L KI multiplied by 1L/1000mL = 2.5*10^-3

Then the amount of KI left after partial neuralisation is (2.5*10^-3)-(5.0*10^-4)= 2.0*10^-3 mol


Then what I think I do next is:

2.0*10^-3 mol KI/35.0ml multiplied by 1000mL/ 1L =  0.0571 mol KI/L
                                                                    = 0.0571 M KI

Thus to get the pH:

pH = -log 0.0571 = 1.24

If all of this is correct then I have no problem with doing the rest as its the same method for the other titration's.

[Borek]

This is pI, not pH. And this is correct result. But you are asked to calculate concentration of Ag⁺ as well, aren't you?

Note that this approach will work up to the equivalence point only.

[GreenGiant87]

Ah ok sorry i missed the Ag+. Ok to calculate that i need to use the Ksp that was given in the question.

So: Ksp for AgI= 8.3*10^-17

therefore, [Ag+] = Ksp/ [I-]     =>    8.3*10^-17/ 0.05714 = 1.5*10-15 M

therefore pAg+= -log [Ag+]= 14.84

Now my problem is what do I do with the pAg+ and pI-


Also I forgot one very important thing to do, get the Ve for Ag+ since thats the equivalence point I need.

I know that one mole of Ag+ reacts with one mol of I-.
So: (0.0250L)(0.10 mol I-/L) = (v)(0.050 mol Ag+/L)

=>Ve = 0.050L = 50ml

So now I know that the equivalence point is at 50ml.

Therefore, at the equivalence point I know that all the Ag+ added has precipitated out with all the I- in solution. Any Ag+ or I- present is due to the dissolution of Agl(s).

So the Ksp can tell me how  much of each ion is present in solution.

=> [Ag+][I-]=Ksp
      (x)  (x) = 8.3*10^-17

=> [Ag+]= 9.1*10^-9

=>pAg+= -log [Ag+]=8.04

Then to calculate after the equivalence point the next titration is 55.0mL.
So the amount past the equivalence point is 5mL.
 
Am I right thus far?

[Borek]

Looks OK so far.

[GreenGiant87]

ok thats good. So now say for when 20mL is added:

20mL (0.05 mol AgNO3)/(1LAgNO3) multiplied by (1L)/(1000mL)= 1.0*10^-3

I already got the number of moles of KI originally present in 25mL of solution so:

(2.5*10^-3)-(1.0*10^-3)=1.5*10^-3

1.5*10^-3 mol KI/45mL multiplied by 1000mL/ 1L = 0.033 M KI

Concentration of Ag+

8.3*10^-17/0.033 = 2.51*10^-15 M

-log [Ag+] = 14.60

[GreenGiant87]

oh and after the equivalence point: the next titration is 55 mL so the amount past the equivalence point is 5mL.

mol Ag+= 5mL (0.05 ml AgNO3)/(1L AgNO3) multiplied by (1L)/(1000mL) = 2.5*10-4

[Ag+] = (2.5*10^-4)/(0.025+0.055L) = 3.12*10^-3

pAg+= 2.51

[Borek]

Looks OK, but I have not checked exact numbers, just skimmed the logic.

[GreenGiant87]

ok thats brilliant, thank you for the help :-)