A solution was prepared with an initial concentrations of Fe (aq) of 0.00111 M a

Question

A solution was prepared with an initial concentrations of Fe (aq) of 0.00111 M and SCN (aq) of 0.000625 M. Calculate the equilibrium constant for the Fe^3+/SCN reaction if the equilibrium concentration of FeSCN^2+ (aq) was measured to be 0.000248 M. Fe^3+ (aq) doubleheadarrow SCN (aq) FeSCN^2+ (aq) When reaction (1) and (2) below are added together, the result is reaction (3). Given K_1 = 0.00047. and K_2 = 1.00 times 10^14; find the equilibrium constant. K_3 H_2O(I) + HNO_2(aq) doubleheadarrow H_2O (aq) + NO_2 (aq) H_2 O+(aq) + OH (aq) doubleheadarrow 2H_2O (I) HNO_2(aq) + OH (aq) doubleheadarrow NO_2(aq)+ H_2O(I)

Explanation / Answer

Fe^3+ + SCN- ==> Fe(SCN)^2+

according to the given problem, x = 0.000248 M

[Fe^3+]eq = 0.00111- 0.000248 = 8.62*10^-4 M

[SCN^-]eqm = 0.000625 -0.000248 = 3.77*10^-4 M

[Fe(SCN)^2+]eqm = 0.000248 M

Keq = [Fe(SCN)^2+]eqm/ [Fe^3+]eq * [SCN^-]eqm

       = 0.000248 M/ 8.62*10^-4 M *3.77*10^-4 M

       = 763.14

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HNO2 (aq) + OH- (aq) <---> NO2^- (aq) + H2O (l)

Keq = [NO2^-]/[HNO2][OH-]

       = K1 * K2 = 0.00047 *10^14 = 47*10^9

Fe^3+ SCN^- Fe(SCN)^2+ initial 0.00111 0.000625 0 change -x -x +x equilibrium 0.00111-x 0.000625-x x

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