Predict whether or not a precipitate forms upon mixing 175.0 ml of a 0.0055 mkcl solution with 145.0 ml of a 0.0015 m agno3 solution. identify the precipitate, if any. express your answer as a chemical formula. enter noreaction if no precipitate is formed.

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.