Question

A stock solution will be prepared by mixing the following chemicals together:
3.0 mL of 0.00200 M KSCN
10.0 mL of 0.200 M Fe(NO3)3
17.0 mL of 0.5 M HNO3
Determine the molar concentration of Fe(NO3)3 in the stock solution.
Determine the molar concentration of Fe3+ in the stock solution

Answer 1

A stock solution will be prepared by mixing the following chemicals together:

3.0 mL of 0.00200 M KSCN

10.0 mL of 0.200 M Fe(NO3)3

17.0 mL of 0.5 M HNO3

Determine the molar concentration of Fe(NO3)3 in the stock solution.

Determine the molar concentration of Fe3+ in the stock solution

Solution. For a start, let's decide on the total volume of the solution. Since the concentrations of substances are small, then we define the total volume of the solution as the sum of the volumes of all the mixed solutions, that is: $V(\text{sol.}) = 3.0\,\text{ml} + 10.0\,\text{ml} + 17.0\,\text{ml} = 30\,\text{ml}$.

Now we define the quantities of substances:

$v_{\text{KSCN}} = V(\text{sol.KSCN}) \times c(\text{KSCN}) = 0.003 \times 0.002 = 0.000006\,\text{moles},$

$v_{\text{Fe(NO3)3}} = V(\text{sol.Fe(NO3)3}) \times c(\text{Fe(NO3)3}) = 0.01 \times 0.2 = 0.002\,\text{moles},$

$v_{\text{HNO3}} = V(\text{sol.HNO3}) \times c(\text{HNO3}) = 0.017 \times 0.5 = 0.0085\,\text{moles}.$

The following reaction will take place in the solution:

$\mathrm{Fe(NO_3)_3 + KSCN = [Fe(SCN)]^{2+} + 3NO_3^- + K^+}$ or $\mathrm{Fe^{3+} + SCN^{-} = [Fe(SCN)]^{2+}}$.

We get a decrease in the amount of substance nitrate iron on the amount of 0.000006 moles, and $v_{\text{Fe(NO3)3}} = 0.002 - 0.000006 = 0.001994\,\text{moles}$. $c(\text{Fe(NO3)3}) = \frac{0.001994}{0.03} = 0.0665\,\text{M} = c(\text{Fe}^{3+})$.

Answer: 0.0665 M; 0.0665 M.

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