Question

a) A 0.1500 M of AgNO3 solution was employed to titrate a 25.00 mL of 0.1250 M of NaI and 0.2500 M NaCl. Given that Ksp, AgI(s) = 8.3*10-17, and Ksp, AgCl(s) = 1.8*10-10, please calculate the concentration of Ag+ ion after 6.00 mL of AgNO3 was added.

Answer: 1.2 x 10^-15 M

b) please calculate the pAg after 100.00 mL of AgNO3 was added

Answer: 1.35

I just want to see how they got that.

Answer 1

a)

key point : AgI Ksp value is lesser than AgCl . so AgI precipitate first

AgNO3 + NaI ------------------> AgI + NaNO3

moles of AgNO3 = 0.1500 x 6 / 1000 = 9 x 10^-4

moles of NaI = 25.00 x 0.1250 / 1000 = 3.12 x 10^-3

limiting reagent is AgNO3 .

so moles NaI remaining = (3.12 x 10^-3) - ( 9 x 10^-4) = 2.22 x 10^-3

molarity of NaI = 2.22 x 10^-3 / (6+25)x 10^-3 = 0.0718M

AgI -------------------> Ag+ + I-

Ksp = [Ag+] [I-]

8.3 x 10^-17 = [Ag+] x 0.0718

[Ag+] = 1.2 x 10^-15 M

part b)

moles of AgNO3 = 0.1500 x 100 / 1000 = 0.01500

NaI moles = 3.12 x 10^-3

NaI here limiting reagent

AgNO3 remanimg = 0.01500 - (3.12 x 10^-3) = 0.01188

molarity of AgNO3 = 0.01188 / (100 + 25) x10^-3 = 0.095M

[Ag+] = 0.095

pAg = log [Ag+]

= 1.02

note : i am getting 1.02 for part (b)