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Experiment $23$

Advance Study Assignment: Determination of the Equilibrium Constant for a Chemical Reaction

A student mixes $5\mathrm{.}00mL$ of $2\mathrm{.}00{10}^{-3}$MFe$(N{O}_3{)}_3$ with $3\mathrm{.}00mL$ of $2\mathrm{.}00{10}^{-3}MKSCN.$ She finds that in the equilibrium mixture the concentration of $FeSC{N}^{2+}$ is $1\mathrm{.}28{10}^{-4}M.$ Find $K_1$ for the reaction: $F{e}^{3+}(aq)+SC{N}^{-}(aq)FeSC{N}^{2+}(aq).$

Step $1$ Find the molarity of $F{e}^{3+}$ and $SC{N}^{-}$after the two solutions are mixed $($no reaction has occurred$)$ and the concentrations diluted.

$MF{e}^{3+}MSC{N}^{-}$

Step $2$ What is the molarity of the $FeSC{N}^{2+}$ at equilibrium?

$MFeSC{N}^{2+}$

Based on the stoichiometry of the reaction, what is the change in concentration for each of the two reactants?

$[F{e}^{3+}]=,$;$[SC{N}^{-}]=$

Step $3$ Adding the change in concentration to the original molarities, calculate the equilibrium concentrations of $F{e}^{3+}$ and $SC{N}^{-}.$

$MF{e}^{3+}$;

$MSC{N}^{-}$

Step $4$ Using the equilibrium concentrations and the expression for $K_1$ found in equation

$(1),$ calculate the value of the equilibrium constant $K_1$

$K_1=$