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0,1 mmol 0,1 0,1
konsentrasi Cl⁻ = 0,1/ (1002) ≈ 10⁻⁴ M
Pb(NO₃)₂ ⇒ Pb²⁺ + 2NO₃⁻
0,1 mmol 0,1 0,2
konsentrasi Pb²⁺ = 0,1 / (1002) ≈ 10⁻⁴ M
PbCl₂⇒ Pb²⁺ + 2Cl²⁻
Qsp PbCl₂ = (Pb²⁺) (Cl⁻)²
= (10⁻⁴) (10⁻⁴)²
= 10⁻¹²
Qsp > Ksp, maka terbentuk endapan