Consider the following reaction , Fe^(3+) (aq) + SCN^(-) (aq) hArr [Fe (SCN)]^(2+) (aq) A solution is made with initial Fe^(3+) , SCN^(-) concentration of 1 xx 10^(-3) M and 8 xx 10^(-4) M respectively . At equilibrium [Fe(SCN)]^(2-) concentration is 2 xx 10^(-4) M . Calculate the value of equilibrium constant.

Consider the following reaction , Fe^(3+) (aq) + SCN^(-) (aq) hArr [F

1.	Consider the following reaction , Fe^(3+) (aq) + SCN^(-) (aq) hArr [Fe (SCN)]^(2+) (aq) A solution is made with initial Fe^(3+) , SCN^(-) concentration of 1 xx 10^(-3) M and 8 xx 10^(-4) M respectively . At equilibrium [Fe(SCN)]^(2-) concentration is 2 xx 10^(-4) M . Calculate the value of equilibrium constant.
Answer» SOLUTION : `K_(eq) = ([FE (SCN)]^(2+))/([Fe^(3+)] [SCN^(-)]) = (2 xx 10^(-4) M)/(8 xx 10^(-4) M xx 6 xx 10^(-4) M) = 0.0416 xx 10^(4)` `K_(eq) = 4.16 xx 10^(2) M^(-1)`