Question

You mix a 140.0 −mL−mL sample of a solution that is 0.0124 MM in NiCl2NiCl2 with a 200.0 −mL−mL sample of a solution that is 0.350 MM in NH3NH3.

After the solution reaches equilibrium, what concentration of Ni2+(aq)Ni2+(aq) remains? The value of KfKf for Ni(NH3)62+Ni(NH3)62+ is 2.0×1082.0×108.

Express the concentration to two significant figures and include the appropriate units.

Answer 1

concentration of Ni2+(aq) = 8.8 x 10^-7 M

Explanation

Given : initial concentration of NiCl2 = 0.0124 M

volume of NiCl2 = 140 mL

moles of Ni2+ = (initial concentration of NiCl2) * (volume of NiCl2)

moles of Ni2+ = (0.0124 M) * (140 mL)

moles of Ni2+ = 1.736 mmol

Similarly, moles of NH3 = 70.0 mmol

moles of Ni(NH3)6^2+ formed = moles of Ni2+

moles of Ni(NH3)6^2+ formed = 1.736 mmol

total volume = (volume NiCl2) + (volume NH3)

total volume = (140 mL) + (200 mL)

total volume = 340 mL

concentration of Ni(NH3)6^2+ = (moles of Ni(NH3)6^2+ formed) / (total volume)

concentration of Ni(NH3)6^2+ = (1.736 mmol) / (340 mL)

concentration of Ni(NH3)6^2+ = 5.106 x 10^-3 M

moles of NH3 consumed = 6 * (moles of Ni(NH3)6^2+ formed)

moles of NH3 consumed = 6 * (1.736 mmol)

moles of NH3 consumed = 10.416 mmol

moles of NH3 left = (initial moles of NH3) - (moles of NH3 consumed)

moles of NH3 left = (70.0 mmol) - (10.416 mmol)

moles of NH3 left = 59.584 mmol

concentration of NH3 = (moles of NH3 left) / (total volume)

concentration of NH3 = (59.584 mmol) / (340 mL)

concentration of NH3 = 0.17525 M

concentration of Ni2+ left = [Ni(NH3)6^2] / (Kf * [NH3]^6)

concentration of Ni2+ left = (5.106 x 10^-3 M) / [(2.0 x 10^8) * (0.17525 M)^6]

concentration of Ni2+ left = 8.8 x 10^-7 M