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.