You are correct that a ppt of AgCl will form BUT you didn't arrive at the answer correctly. If I were grading your paper I would give you no credit.
There is no Ka or Kb to worry about. Not with Ksp. Ksp is what we are concerned with.
mol AgNO3 = M x L = 0.45 M 0.045 L = ??
mol NaCl = M x L = 0.0135 M x 0.085 L = ??
(Ag^+) = mols/total volume
total volume = 85 mL + 45 mL = 130 mL = 0.130 L.
(Cl^-) = mols/total volume.
ion produce is (Ag^+) from above *(Cl^-) from above. Then compare with Ksp.
Please judge my answer
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.
b) Would a precipitate form? The Ksp of AgCl(s) is 1.8 x 10-10.
Answer:
For a, doing the ion product is nessecary to get the answer. For part b of this question, comparing the ion product with Ksp is nessecary. Determining the ion product consists of using the formula Kw=Ka/Kb, for Ka I would do 0.0045/0.0085 = 0.529 and for Kb, I would do 0.45/0.0135 = 33.33. I would then use the formula Kw = Ka/Kb to get a value of 1.0 x 10^-2. This is my ion product value. When comparing the ion product to Ksp, I see that 1.0 x 10^-2 is greater than 1.8 x 10^-10, therefore a precipitate will form.
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