Question

Calculate the concentrations of iron(iii) ions and scn- ions when 5.00 ml of 0.002m iron(iii) nitrate is mixed with 4.00 ml of 0.002 m nascn and dilute nitric acid is added to bring the total volume of the solution to 20.0 ml. What is the concentration of iron(iii) ions?

Answer 1

Volume of Fe(NO₃)₃ = 5.00 ml

Total voluem of the solution = 20.0 ml

Concentration of Fe(NO₃)₃ = 0.002 M

Assuming complete dissociation of  Fe(NO₃)₃ concentration of Fe³⁺  = 0.200 M

Using the dilution equation, M₁V₁ = M₂V₂

where M₁ and M₂ are the initial and final concentration of Fe³⁺  respectively and V₁ and V₂ are the initial and final volume of Fe³⁺  respectively

Plugging the data we get,

M₂ = $\frac{0.002 M x 5.00 ml}{20.0 ml}$ = 5 x 10⁻⁴

Therefore, the concentration of Fe³⁺ once diluted in the mixture is 5 x 10⁻⁴M