# A 2.50ml aliquot of 0.0103 M Fe(NO3)3 is added to 2.50ml of 0.00101 M NaSCN and the following equilibrium is reached Fe(3+) + SCN(-) =FeNCS(2+). The concentration of FeNCS(2+) was determined to be 3.50 multiplied by 10(-4) M at equilibrium. Calculate Kc?

Feb 8, 2015

So, the overall reaction between iron (III) nitrate and sodium thiocyanate is this (all the species are in aqueous solution)

$F e {\left(N {O}_{3}\right)}_{3} + N a S C N r i g h t \le f t h a r p \infty n s F e S C N {\left(N {O}_{3}\right)}_{2} + N a N {O}_{3}$

LIke you know, the net ionic reaction shows the equilibrium between the ferric ion, $F {e}^{3 +}$, and the thiocyanate ion, $S C {N}^{-}$, with the red-orange colored $F e S C {N}^{2 +}$ complex ion

$F {e}^{3 +} + S C {N}^{-} r i g h t \le f t h a r p \infty n s F e S C {N}^{2 +}$

So, you know the equilibrium concentration of the complex ion. Here's what will save you some work on this problem.

The important thing to notice is that you have equal volumes of iron (III) nitrate and sodium thiocyanate mixed together. This means that the final volume of the solution will be twice the initial volumes.

Now, think dilution calculations. If the final volume doubles, the final concentration will be half of the initial concentration. This is true for both compounds, so their final concentrations in solution will be

$\left[F e {\left(N {O}_{3}\right)}_{3}\right] = \left[F {e}^{3 +}\right] = \text{0.0103 M"/2 = "0.00515 M}$, and

$\left[N a S C N\right] = \left[S C {N}^{-}\right] = \text{0.00101 M"/2 = "0.000505 M}$

Now think of the formation of the complex ion. Initially, you had no complex ion formed in solution. Then, at equilibrium, the concentration is given to be $3.50 \cdot {10}^{- 4} \text{M}$. This means that the concentration of the complex ion went from zero to $3.50 \cdot {10}^{- 4} \text{M}$.

Automatically, the concentrations of the ferric and thiocyanate ions must have decreased by the same amount. This happens because of the stochiometry of the reaction - 1 mole of $F {e}^{3 +}$ reacts with 1 mole of $S C {N}^{-}$ to produce 1 mole of $F e S C {N}^{2 +}$.

So, the equilibrium concentrations for thes two ions are

${\left[F e\right]}^{3 +} = \text{0.00515 M" - "0.000350 M" = "0.0048 M}$, and

$\left[S C {N}^{-}\right] = \text{0.000505 M" - "0.000350 M" = "0.000155 M}$

Now just plug these values into the expression for the equilibrium constant

${K}_{c} = \frac{\left[F e S C {N}^{2 +}\right]}{\left[F {e}^{3 +}\right] \cdot \left[S C {N}^{-}\right]} = \frac{0.000350}{0.0048 \cdot 0.000155} = 470.4$

Therefore, ${K}_{c} = 470$ - rounded to three sig figs.