91Ó°ÊÓ

We first need to determine the initial concentrations of \(\mathrm{Cd}^{2+}\), \(\mathrm{Cd}(\mathrm{CN})_{4}^{2-}\) and \(\mathrm{CN}^{-}\). \n\nGiven that 0.50 g of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) dissolves in \(5.0 \times 10^{2} \mathrm{~mL}\) of water, we can calculate the initial \(\mathrm{Cd}^{2+}\) ion concentration, using molar mass of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) approximately equal to 308.5 g/mol.\n\nSo the concentration of \(\mathrm{Cd}^{2+}\) = (0.50g / 308.5 g/mol) / (5.0 * 10^2 ml * 1L/1000 ml) = 0.00324 M The concentration of \(\mathrm{CN}^{-}\) is given to be 0.50 M and initial concentration of \(\mathrm{Cd}(\mathrm{CN})_{4}^{2-}\) is 0 because it has not yet formed.

The reaction between \(\mathrm{Cd}^{2+}\) and \(\mathrm{CN}^{-}\) is as follows: \[ \mathrm{Cd}^{2+} + 4\mathrm{CN}^{-} <=> \mathrm{Cd}(\mathrm{CN})_{4}^{2-} \] We can write the equilibrium expression for the reaction as: \[ [ \mathrm{Cd}(\mathrm{CN})_{4}^{2-} ] / ( [\mathrm{Cd}^{2+}] [\mathrm{CN}^{-}]^4 ) \]

Let us assume x mol/L of \(\mathrm{Cd}^{2+}\) reacted. At equilibrium, [ \(\mathrm{Cd}^{2+}\) ] = 0.00324 - x M, [ \(\mathrm{CN}^{-}\) ] = 0.50 - 4x M, and , [ \(\mathrm{Cd}(\mathrm{CN})_{4}^{2-}\) ] = x M. Since the \(\mathrm{Cd}(\mathrm{CN})_{4}^{2-}\) complex is very stable, we can generally assume reaction goes to completion, i.e. x = 0.00324 M. So, the equilibrium concentrations we find are: [ \(\mathrm{Cd}^{2+}\) ] ≈ 0 M, [ \(\mathrm{CN}^{-}\) ] = 0.50 - 4 * 0.00324 M ≈ 0.487 M and [ \(\mathrm{Cd}(\mathrm{CN})_{4}^{2-}\) ] = 0.00324 M.