/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 From the value of \(K_{f}\) list... [FREE SOLUTION] | 91影视

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Chapter 17: Problem 65

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 \(\mathrm{MN} \mathrm{H}_{3}(a q) .\)

Short Answer

Expert verified
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 by step solution

01

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
02

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)
03

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脳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.

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