# 3.1 mol of sf(Fe^(3+) is mixed with 3.2 mol of sf(SCN^-) and allowed to reach equilibrium, after which 3 mol of sf([Fe(SCN)]^(2+)) is formed. What is the value of sf(K_c) ?

Jun 14, 2016

${K}_{c} = 150 \text{ ""mol".^(-1)."l}$

#### Explanation:

I am assuming that you mean $N {H}_{4} S C {N}_{\left(a q\right)}$. The $S C {N}_{\left(a q\right)}^{-}$ ion forms a 1:1 complex with $F {e}_{\left(a q\right)}^{3 +}$

Construct an $\text{ICE}$ table:

$\text{ "Fe_((aq))^(3+)" "+" "SCN_((aq))^(-)" "rightleftharpoons" } {\left[F e \left(S C N\right)\right]}_{\left(a q\right)}^{2 +}$

$\textcolor{red}{\text{I"" "3.1" "3.2" }} 0$

$\textcolor{red}{\text{C"" "-x" "-x" }} + x$

$\textcolor{red}{\text{E"" "(3.1-x)" "(3.2-x)" }} x$

The question tells us that $x = 3$

$\therefore n F {e}_{\left(a q\right)}^{3 +} = 3.1 - 3 = 0.1$

$n S C {N}_{\left(a q\right)}^{-} = 3.2 - 3 = 0.2$

and

$n {\left[F e \left(S C N\right)\right]}_{\left(a q\right)}^{2 +} = 3$

The expression for ${K}_{c}$ is :

${K}_{c} = \frac{\left[F e {\left(S C N\right)}^{2 +}\right]}{\left[F {e}^{3 +}\right] \left[S C {N}^{-}\right]}$

$\therefore {K}_{c} = \frac{3}{0.1 \times 0.2} = 150 \text{ ""mol"^(-1)."l}$

The question should have specified the temperature as ${K}_{c}$ depends on this.