Question

A student mixes 5.9 mL of 0.0019 M Fe(NO3)3 with 4.4 mL of 0.0045 M KSCN. What is the initial concentration of Fe3 in the final solution?

Answer 1

C Fe³⁺ = 0.00109 M

Step-by-step explanation:

The reaction between Fe(NO₃)₃ and KSCN is:

Fe³⁺(aq) + SCN⁻(aq) → FeSCN²⁺(aq)

5.9mL 4.4mL

0.0019M 0.0045M

The moles (η) of Fe(NO₃)₃ are:

`\eta_(Fe(NO_3)_3) = C_(Fe(NO_3)_3) \cdot V_(Fe(NO_3)_3) = 0.0019 mol/L \cdot 5.9 \dot 10^(-3) L = 1.121 \cdot 10 ^(-5) moles`

Since in 1 mol of Fe(NO₃)₃ we have 1 mol of Fe³⁺, the moles of Fe³⁺ are:

`\eta_(Fe^(3+)) = \eta_(Fe(NO_3)_3) = 1.121 \cdot 10 ^(-5) moles`

Now, to calculate the initial concentration of Fe³⁺ we need first to find the total volume in the final solution:

`V_(T) = V_(Fe(NO_3)_3) + V_(KSCN) = 5.9 mL + 4.4 mL = 10.3 mL = 0.0103 L`

Finally, the the initial concentration of Fe³⁺ is:

`C_(Fe^(3+)) = (\eta_(Fe^(3+)))/(V_(T)) = (1.121 \cdot 10 ^(-5) moles)/(0.0103 L) = 0.00109 M`

I hope it helps you!