Question

Calculate the concentrations of ions in equilibrium in a solution obtained by mixing 25.00 mL of 0.30 M NaCl with 50.00 mL of 0.40 M NaI and 50.00 mL of 0.50 M AgNO₃. K_psAgCl = 1,8·10^-10; K_psAgI = 8.5·10^-17

Answer 1

Moles of NaCl = 0.3*0.025 (M *V in litre gives mole of solute)

= 0.0075

Moles of NaI = 0.40 * 0.050 = 0.02

Moles of AgNO3 = 0.5*0.05 = 0.025

Now Reaction that takes place: Ag + Cl = AgCl (white ppt), Ag + I (yellow ppt) and Na+ and NO3- are spectator ions (remains dissolved in solution)

Since, Ksp of AgI is very high: and moles of Ag+ available is 0.025 from AgNO3 which is higher than that of I- (from NaI) so total AgI formed is 0.02 moles and AgCl is 0.005.

So total ions remaining in concentration is

[Cl-]= Moles in Solution is 0.0025 (since all the Ag is used up in AgCl and AgI, this much amount will remain free in solution)

= 0.0025 / total volume of solution (i.e.0.125L)

= 0.02 M

[Na+] =0.0075+0.02 =0.0275 moles / 0.125 L = 0.22 M

{NO3-] = 0.025/0.125=0.2 M

[Ag+] = negligible, the only source of it is dissociation of salt and its very low