Posted by Abigail on .
Question:A solution of AgNO3 (45 mL/0.45 M) was mixed with solution of NaCl (85 mL/1.35 x 102 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?

chemistry 
Abigail,
b) Would a precipitate form? The Ksp of AgCl(s) is 1.8 x 1010.
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.