Question

​A 65 mL solution of 0.40M Al(NO₃)₃ ​is mixed with 125 mL of 0.17M iron (II) Nitrate. Solid sodium hydroxide is then added with out any changes in volume. What is the Molarity of Al³⁺​ and Fe^​2+​ in the solution before addition of sodium hydroxide?

Answer 1

We shall use the dilution equation

M₁*V₁ = M₂*V₂ where M₁ = molarity of the stock (concentrated) solution; M₂ = molarity of the dilute solution; V₁ = volume of the stock (concentrated) solution and V₂ = final volume of the solution.

We have V₂ = (65 + 125) mL = 190 mL.

Do the calculation for Al³⁺:

(0.40 M)*(65 mL) = M₂*(190 mL)

===> M₂ = (65*0.40)/(190) M = 0.1368 M ≈ 0.14 M

Do the calculation for Fe²⁺:

(0.17 M)*(125 mL) = M₂*(190 mL)

===> M₂ = (125*0.17)/(190) M = 0.1118 M ≈ 0.11 M

The molarity of Al³⁺ is 0.14 M while the molarity of Fe²⁺ is 0.11 M (ans).