Answer to Question #327392 in Physical Chemistry for Khoja

Question

Calculate threshold coagulation of 1 L Zn(OH)2 sol by 0.45 L of Al₂(SO4)3 solution with concentration C(1/6 Al2(SO4)3)= 0.02 mol/L.

Answer 1

V(Zn(OH)₂) = 1.00 L;

V(Al₂(SO₄)₃) = 0.45 L;

C(1/6 * Al₂(SO₄)₃) = 0.02 mol/L;

C(Al₂(SO₄)₃) = 0.12 mol/L;

Tc(threshold coagulation) = C_el * V_el/(V_sol + V_el) = C_Al2(SO4)3 * V_Al2(SO4)3/(V_Zn(OH)2 + V_Al2(SO4)3) = 0.12 * 0.45/(1.00 + 0.45) = 0.054/1.45 = 0.0372 mol/L = 37.2 mmol/L.

Answer: 37.2 mmol/L.