Question

Consider the reaction: 2K3PO4(aq)+3NiCl2(aq)→ Ni3(PO4)2(s)+6KCl(aq)

What volume of 0.205 M K3PO4 solution is necessary to completely react with 146 mL of 0.0122 M NiCl2?

Answer 1

2K₃ PO₄(aq)+3NiCl₂(aq)→ Ni₃(PO₄)₂(s)+6KCl(aq)

V₁ x M₁=V₂ x M₂

V₁= volume of K₃ PO₄

M₁= Molarity of K₃ PO₄

V₂= volume of NiCl₂

M₂= Molarity of NiCl₂

V₁ x M₁=V₂ x M₂ (This equation is applicable only when the stoichiometric coefficient of both reactants are the same. It is not applicable when the stoichiometric coefficient is different. So, by modifying the equation you can use it for stoichiometrically different reactant)

2 moles of K₃ PO₄ reacts with 3 moles of NiCl₂ to give desire products. So, the required volume for neutralization is given by the equation:

  

n₁ = stoichiometric coefficient of K₃ PO₄

n₂= stoichiometric coefficient of NiCl₂

V₁= (146 x 0.0122/0.205) x (2/3)

V₁= 5.79 mL