Answer is: n<span>o, because the ion product is less than the Ksp of lead iodide. </span>
Chemical dissociation 1: KI(s) → K⁺(aq) + I⁻(aq).
Chemical dissociation 2: Pb(NO₃)₂(s) → Pb²⁺(aq) + 2NO₃⁻(aq).
Chemical reaction: Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s).
Ksp(PbI₂) = 7.1·10⁻⁹.
V = 500 mL ÷ 1000 mL/L = 0.5 L.
c(KI) = c(I⁻) = 0.0025 mol ÷ 0.5 L.
c(I⁻) = 0.005 M.
c(Pb(NO₃)₂) = c(Pb²⁺) = 0.00004 mol ÷ 0.5 L.
c(Pb²⁺) = 0.00008 M.
Q = c(Pb²⁺) · c(I⁻)².
Q = 8·10⁻⁵ M · (5·10⁻³ M)².
Q = 2·10⁻⁹; <span> the ion product.</span>
Mass:20g
Mole=20/[2(23)+16]
=0.32 mil
