A student working in the laboratory prepared the following reactants: 5 ml of 0.

Question

A student working in the laboratory prepared the following reactants: 5 ml of 0.008M Cd2+(aq) 10 ml of 0.007M SCN-(aq) 10 ml of 0.5M HNO3(aq) These reagents were mixed and allowed to stand for 10 minutes. The concentration of Cd(SCN)+ in the resulting equilibrium mixture is found to be 5x10-4M. Calculate the initial concentration of Cd2+(aq). Calculate the initial concentration of SCN-(aq). Calculate the equilibrium concentration of Cd2+(aq). Calculate the equilibrium concentration of SCN-(aq). Calculate the Keq for the reaction mixture.

Explanation / Answer

Total volume of the initial reaction mixture is 5ml + 10 ml +10 ml = 25 ml, so initial concentrations are [Cd++]o = (5 ml / 25 ml) (0.008 M) = 0.001 M; and [SCN-]o = (10 ml / 25 ml) (0.006 M) = 0.0024 M. The reaction is (apparently): Cd++ + SCN- ? Cd(SCN)+ There is 1:1 stoichiometry, so the concentrations at equilibrium are: [Cd(SCN)+] = 0.0008 M (given) [Cd++] = 0.001 M - 0.0008 M = 0.0008 M [SCN-] = 0.0024 M - 0.0005 M = 0.00190 M Keq = [Cd(SCN)+] / [Cd++] [SCN-]; Keq = (0.0005 M) / (0.0005 M) (0.00190 M); Keq = 526.3 M^-1.

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