Question: A solution of AgNO3 (45 ml/0.45 M) was mixed with a solution of NaCl (85 ml/1.35 x 10-2 M) a) Calculate the ionic product of the potential precipitate.
My answer:
mol AgNO3 = M x L
= 0.45M x 0.045M
= 0.02025
mol NaCl = M x L
= 0.0135M x 0.085L
= 0.0011475
(Ag^+) = moles / total volume
= 0.02025 / 0.130L
= 0.1558
total volume = 85ml + 45ml
= 130mL
= 0.130L
(Cl^-) = moles / total volume
= 0.0011475 / 0.130L
= 0.00883
Ionic product = [Ag^+][Cl^-]
= [0.1558][0.00883]
= 0.001375
Is my answer correct? b) Would a precipitate form? The Ksp of AgCl(s) is 1.8 x 10-10.
Also please check if my answer is also correct or not
Comparing the ion product to Ksp, I see that 0.001375 is greater than 1.8 x 10^-10, so a precipitate will form.