Question

Question:A solution of AgNO3 (45 mL/0.45 M) was mixed with solution of NaCl (85 mL/1.35 x 10-2 M)
a) Calculate the ion product of the potential precipitate.


My answer:

mol AgNO3 = M x L
= 0.45 M x 0.045 M
= 0.02025

mol NaCl = M x L
= 0.0135 M x 0.085 L
= 0.0011475

(Ag^+) = mols / total volume
= 0.02025 / 0.130 L
= 0.1558


total volume = 85 mL + 45 mL
= 130 mL
= 0.130 L


(Cl^-) = mols / total volume
= 0.0011475 / 0.130 L
= 0.00883

Ion product = [Ag^+][Cl^-]
= [0.1558][0.00883]
= 0.001375

Is my answer correct?

Answer 1

b) Would a precipitate form? The Ksp of AgCl(s) is 1.8 x 10-10.

Also, please check if my answer to be is also correct or not

When comparing the ion product to Ksp, I see that 0.001375 is greater than 1.8 x 10^-10, therefore a precipitate will form.