The copper(l) ion forms a complex ion with \(\mathrm{CN}^{-}\) according to the following equation: $$\mathrm{Cu}^{+}(a q)+3 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Cu}(\mathrm{CN})_{3}^{2-}(a q) \quad K=1.0 \times 10^{11}$$ a. Calculate the solubility of $\mathrm{CuBr}(s)\left(K_{\mathrm{sp}}=1.0 \times 10^{-5}\right)\( in \)1.0 \mathrm{L}\( of \)1.0 \mathrm{M} \mathrm{NaCN}$ . b. Calculate the concentration of \(\mathrm{Br}^{-}\) at equilibrium. c. Calculate the concentration of \(\mathrm{CN}^{-}\) at equilibrium.

We are given the following equilibrium expressions: $$a. \ \mathrm{CuBr}(s) \rightleftharpoons \mathrm{Cu}^{+} (aq) + \mathrm{Br}^{-}(aq) \quad K_{sp} = 1.0 \times 10^{-5}$$ $$b. \ \mathrm{Cu}^{+}(a q)+3 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Cu}(\mathrm{CN})_{3}^{2-}(a q) \quad K=1.0 \times 10^{11}$$ We need to find the solubility of CuBr in 1.0 L of 1.0 M NaCN. Let the solubility of CuBr in 1.0 L of NaCN be 'x' moles. Thus, at equilibrium, we have: - [Cu+] = x moles - [Br-] = x moles - [CN-] = (1.0 M - 3x) moles since 3 moles of CN- react with 1 mole of Cu+.

Using the given equilibrium constants, we have the following equilibrium expressions: $$a. \ K_{sp} = [\mathrm{Cu}^{+}][\mathrm{Br}^{-}]$$ $$b. \ K = \frac{[\mathrm{Cu}(\mathrm{CN})_{3}^{2-}]}{[\mathrm{Cu}^{+}][\mathrm{CN}^{-}]^{3}}$$ For the \(K_{sp}\) expression, we know that: $$1.0 \times 10^{-5} = (x)(x)$$ $$x = \sqrt{1.0 \times 10^{-5}} = 1.0 \times 10^{-2.5}$$ Now, for the K expression, we have: $$1.0 \times 10^{11} = \frac{[\mathrm{Cu}(\mathrm{CN})_{3}^{2-}]}{[(1.0 \times 10^{-2.5})(1.0 - 3(1.0 \times 10^{-2.5}))]^3}$$ We can substitute [\(\mathrm{Cu}(\mathrm{CN})_{3}^{2-}\)] as 'x' in the equation above: $$1.0 \times 10^{11} = \frac{x}{[(1.0 \times 10^{-2.5})(1.0 - 3(1.0 \times 10^{-2.5}))]^3}$$

Re-arrange the equation and solve for x, which represents the solubility of CuBr: $$x = 1.0 \times 10^{11} [(1.0 \times 10^{-2.5})(1.0 - 3(1.0 \times 10^{-2.5}))]^3$$ $$x \approx 5.63 \times 10^{-26}$$ a. The solubility of CuBr in 1.0 L of 1.0 M NaCN is approximately \(5.63 \times 10^{-26}\ \mathrm{mol}\).

As we calculated the solubility 'x' of CuBr in 1.0 L of 1.0 M NaCN, the concentration of Br- ions at equilibrium will be the same: b. The concentration of Br- ions at equilibrium is approximately \(5.63 \times 10^{-26}\ \mathrm{M}\).

Since 3 moles of CN- react with 1 mole of Cu+ as per the equation, the change in the concentration of CN- ions will be: Decrease in CN- ions = Moles of Cu+ reacted x 3 = (1.0 x 10^{-2.5}) x 3 The initial concentration of CN- ions is given as 1.0 M, so the final concentration of CN- ions at equilibrium will be: Final concentration of CN- ions = Initial concentration - Decrease in concentration = 1.0 - (1.0 x 10^{-2.5} x 3) c. The concentration of CN- ions at equilibrium is approximately \(0.9997\ \mathrm{M}\).