Question

Consider the following reaction,
$F{e}^{3+}\left(aq\right)+SC{N}^{-}\left(aq\right)\rightleftharpoons \left[Fe\right(SCN\left){]}^{2+}\right(aq)$
A solution is made with initial $F{e}^{3+},SC{N}^{-}$ concentration of $1\times {10}^{-3}M$ and $8\times {10}^{-4}M$ respectively. At equilbrium $\left[Fe\right(SCN){)}^{2+}$ concentration is $2\times {10}^{-4}M$. Calculate the value of equilibrium constant.

Answer 1

$F{e}^{3+}+SC{N}^{-}\rightleftharpoons \left[Fe\right(SCN){]}^{2+}$

When $t=0$, $1\times {10}^{-3}$ $8\times {10}^{-4}$ $0$

At time $=t$ $1\times {10}^{-3}-x$ $8\times {10}^{-4}-x$ $x$

$x=2\times {10}^{-4}$

$\left[SC{N}^{-}\right]=(8-2)\times {10}^{-4}=6\times {10}^{-4}M$

$\left[F{e}^{3+}\right]=(10-2)\times {10}^{-4}=8\times {10}^{-4}M$

$\therefore K=\frac{\left[Fe\right(SCN\left){]}^{2+}\right]}{\left[F{e}^{3+}\right]\left[SC{N}^{-}\right]}=\frac{2\times {10}^{-4}}{6\times {10}^{-4}\times 8\times {10}^{-4}}=416.67M^{-1}$