Problem 65 - Chemistry: The Central Science

Question

From the value of \(K_{f}\) listed in Table \(17.1,\) calculate the concentration of \(\mathrm{Ni}^{2}(a q)\) and \(\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) that are present at equilibrium after dissolving 1.25 \(\mathrm{g} \mathrm{NiCl}_{2}\) in 100.0 \(\mathrm{mL}\) of 0.20 \(\mathrm{MN} \mathrm{H}_{3}(a q) .\)

Step-by-Step Solution

Answer

The equilibrium concentrations are found by solving the equation \(2.5 \times 10^8 = \frac{x}{(0.0964-x)(0.20-6x)^6}\). The value of \(x\) represents the equilibrium concentration of Ni(NH3)6^2+. To obtain the equilibrium concentration of Ni^2+(aq), subtract the value of \(x\) from the initial concentration of 0.0964 M. Solving this equation may require numerical methods or approximation techniques.

Step 1: Calculate the initial concentration of Ni^2+ from NiCl2

Calculate the number of moles and the initial concentration of Ni^2+ formed by dissolving 1.25 g of NiCl2 in 100.0 mL of NH3(aq). First, determine the molar mass of NiCl2: Ni: 58.69 g/mol Cl: 35.45 g/mol Molar mass of NiCl2 = 58.69 + 2 * 35.45 = 129.59 g/mol Now, calculate the number of moles of NiCl2: moles of NiCl2 = (mass of NiCl2) / (molar mass of NiCl2) moles of NiCl2 = 1.25 g / 129.59 g/mol = 0.00964 mol Since 1 mol of NiCl2 gives 1 mol of Ni^2+ ions, the initial moles of Ni^2+ are also 0.00964 mol. Now, calculate the initial concentration of Ni^2+: Concentration of Ni^2+ = (moles of Ni^2+) / (volume in L) Concentration of Ni^2+ = 0.00964 mol / 0.100 L = 0.0964 M

Step 2: Set up the reaction and expression for the equilibrium constant (Kf)

Ni^2+(aq) + 6NH3(aq) <=> Ni(NH3)6^2+(aq) The stability constant (Kf) for the complex formation reaction is given. We need to set up an equation using Kf that relates the equilibrium concentrations of Ni^2+, NH3, and Ni(NH3)6^2+. Kf = [Ni(NH3)6^2+] / ([Ni^2+] * [NH3]^6)

Step 3: Set up the ICE table and find the equilibrium concentrations

Using the ICE (Initial, Change, Equilibrium) table, set up the equilibrium equations for each species: Ni^2+(aq) + 6NH3(aq) <=> Ni(NH3)6^2+(aq) Initial: 0.0964 M 0.20 M 0 M Change: -x -6x +x Equilibrium: 0.0964-x 0.20-6x x Now, substitute the equilibrium concentrations into the Kf expression: Kf = x / ((0.0964-x) * (0.20-6x)^6) Given that Kf=2.5×10^8, we can solve this equation to find the value of x, which represents the equilibrium concentration of Ni(NH3)6^2+. To obtain the equilibrium concentration of Ni^2+ (aq), subtract the value of x from the initial concentration (0.0964 M). It's important to note that solving the equilibrium equation may require numerical methods or approximation techniques, as the equation has a high exponent and cannot be easily solved directly.