Step 1: Write down the chemical reaction
KCl(aq) + AgNO3(aq) → AgCl(s) + KNO3(aq)
The precipitate that can be expected is AgCl
Step 2: Calculate the moles of KCl and AgNO3
# moles of KCl = V(KCl) * M(KCl)
= 0.175 L * 0.0055 moles/L = 9.63*10⁻⁴ moles
# moles of AgNO3 = V(AgNO3) * M(AgNO3)
= 0.145 L * 0.0015 moles/L = 2.18*10⁻⁴ moles
Since moles of AgNO3 < KCl, the former is the limiting reagent
Therefore, moles of AgCl formed = 2.18*10⁻⁴ moles
Step 3: Predict if AgCl precipitate will be formed
The solubility product Ksp for AgCl = 1.6 *10⁻¹⁰
i.e.
AgCl(s) ↔ Ag⁺(aq) + Cl⁻(aq)
Ksp = [Ag+][Cl-]
if [Ag+][Cl-] > Ksp then precipitation will occur
Now total volume of the solution = 175 + 145 = 320 ml = 0.320 L
[Ag+] = [Cl-] = 2.18*10⁻⁴ moles/0.320 L = 6.81*10⁻⁴ M
[Ag+][Cl-] = (6.81*10⁻⁴)² = 4.64 *10⁻⁷
Since [Ag+][Cl-] > Ksp, AgCl precipitate will be formed.
