Question

In: Chemistry

You mix a 131.0 mL sample of a solution that is 0.0115 M in NiCl2 with...

You mix a 131.0 mL sample of a solution that is 0.0115 M in NiCl2 with a 173.5 mL sample of a solution that is 0.256 M in NH3.

After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? (Formation constant is Kf=2.0×108.)

Solutions

Expert Solution

The reaction between Ni (II) and ammonia can be shown as

Ni2+ (aq) + 6 NH3 (aq) -------> [Ni(NH3)6]2+ (aq)

Kf = [Ni(NH3)62+]/[Ni2+][NH3]6 = 2.0*108 ……(1)

Since the formation constant of the Ni-amine complex is very high, we can assume that the entire amount of Ni present in the solution has been complexed as Ni(II)-amine complex. Consequently, the molar concentration of Ni(II)-amine complex at equilibrium is the same as the initial concentration of Ni(II), i.e, 0.0115 M.

Moles of NH3 added = (volume of NH3 added)*(concentration of NH3) = (173.5 mL)*(0.256 mol/L) = 44.416 mmole (1 M = 1 mol/L).

Moles of Ni2+ added (as NiCl2) = (131.0 mL)*(0.0115 mol/L) = 1.5065 mmole.

As per the balanced stoichiometric equation,

1 mole Ni2+ = 6 moles NH3.

Therefore,

1.5065 mmole Ni2+ = (1.5065 mmole Ni2+)*(6 moles NH3/1 mole Ni2+) = 9.039 mmole NH3.

Mmoles of NH3 retained at equilibrium = (44.416 – 9.039) mmole = 35.377 mmole.

Total volume of the solution = (131.0 + 173.5) mL = 304.5 mL.

Equilibrium concentration of NH3 = (35.377 mmole)/(304.5 mL) = 0.1162 M.

Let x be the concentration of Ni2+ at equilibrium. We can write

Kf = [Ni(NH3)62+]/[Ni2+][NH3]6 = 2.0*108

===> 2.0*108 = (0.0115 M)/(x).(0.1162)6

===> 2.0*108 = (0.0115 M)/x.(2.4617*10-6)

===> x = (0.0115 M)/(2.0*108).(2.4617*10-6) = 2.3358*10-5 M

The concentration of free Ni2+ in the solution is 2.3358*10-5 M (ans).


Related Solutions

You mix a 124.0 mL sample of a solution that is 0.0121 M in NiCl2 with...
You mix a 124.0 mL sample of a solution that is 0.0121 M in NiCl2 with a 183.5 mL sample of a solution that is 0.254 M in NH3. Part A After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? (Formation constant is Kf=2.0×108.) Express your answer with the appropriate units.
You mix a 116.0 mL sample of a solution that is 0.0102 M in NiCl2 with...
You mix a 116.0 mL sample of a solution that is 0.0102 M in NiCl2 with a 165.0 mL sample of a solution that is 0.232 M in NH3. After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? (Formation constant is Kf=2.0×108.)
You mix a 150.0 −mL sample of a solution that is 0.0140 M in NiCl2 with...
You mix a 150.0 −mL sample of a solution that is 0.0140 M in NiCl2 with a 175.0 −mL sample of a solution that is 0.500 M in NH3. After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? The value of Kf for Ni(NH3)62+ is 2.0×108.
You mix a 120.5 mL sample of a solution that is 0.0123 M in NiCl2 with...
You mix a 120.5 mL sample of a solution that is 0.0123 M in NiCl2 with a 183.0 mL sample of a solution that is 0.270 M in NH3. After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? (Formation constant is Kf=2.0×108.)
You mix a 133.0 mL sample of a solution that is 0.0100 M in NiCl2 with...
You mix a 133.0 mL sample of a solution that is 0.0100 M in NiCl2 with a 183.5 mL sample of a solution that is 0.220 M in NH3. After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? (Formation constant is Kf=2.0×108.)
A. You mix a 125.5 mL sample of a solution that is 0.0111 M in NiCl2...
A. You mix a 125.5 mL sample of a solution that is 0.0111 M in NiCl2 with a 183.0 mL sample of a solution that is 0.225 M in NH3 After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? (Formation constant is Kf=2.0×108.)Express your answer with the appropriate units. B. A 120.0 −mL sample of a solution that is 2.7×10−3 M in AgNO3 is mixed with a 230.0 −mL sample of a solution that is 0.10 M in NaCN....
A 130.0 mL sample of a solution that is 0.0126 M in NiCl2 is mixed with...
A 130.0 mL sample of a solution that is 0.0126 M in NiCl2 is mixed with a 190.0 mL sample of a solution that is 0.400 M in NH3. -After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? The value of Kf for Ni(NH3)62+ is 2.0×108.
You mix a 140.0 −mL−mL sample of a solution that is 0.0124 MM in NiCl2NiCl2 with...
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.
1. Add 0.5 mL of 0.10 M NiCl2 solution to each of 7 test tubes using...
1. Add 0.5 mL of 0.10 M NiCl2 solution to each of 7 test tubes using dropper provided. Label these tubes 1 through 7. 2. Add the following to the respective tubes containing the NiCl2 solution, Tube 1: 5 drops of water (used here as a control experiment for dilution) Tube 2: 1 mL of conc. HCl Tube 3: 5 drops of conc. ammonia Tube 4: 5 drops of dilute (5% v/v) ethylenediamine solution Tube 5: 5 drops of 0.25...
You mix together 25.0 mL of a 0.025 M aqueous solution of copper(II) sulfate and 50.0...
You mix together 25.0 mL of a 0.025 M aqueous solution of copper(II) sulfate and 50.0 mL of 0.01 M sodium phosphate and isolate 68.3 mg of a precipitate. 1.Write the balanced equation for this reaction. Label all states. 2.What is the identity of the precipitate? 3.What is the theoretical yield of the reaction? 4.What is the percent yield of the reaction? You mix together aqueous solutions of 100.0 mL of 0.05 M sodium nitrate and 10.0 mL of potassium...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT