4. You mix a 123.0 mL sample of a solution that is 0.0121 M in NiCl2 with a 185.

Question

4. You mix a 123.0 mL sample of a solution that is 0.0121 M in NiCl2 with a 185.0 mL sample of a solution that is 0.236 M in NH3. After the solution reaches equilibrium, what concentration of Ni2+(aq) remains? (Formation constant is Kf=2.0×108.)

5. A 130.0 mL sample of a solution that is 3.0×103 M in AgNO3 is mixed with a 230.0 mL sample of a solution that is 0.11 M in NaCN.After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

6. Calculate the solubility of silver chloride in a solution that is 0.200 M in NH3.

Express your answer using two significant figures.

Explanation / Answer

4) 0.0121 moles/liter NiCl2 x 0.123 liters = 1.49*10^-3 moles NiCl2

0.236 moles NH3/1liter x 0.185 liters = 0.0437 moles NH3

formula of the complex ion is: [Ni(NH3)6]+2

NiCl2 + 6NH3 == [Ni(NH3)6]+2 + 2Cl-
Assume all the NiCl2 reacts since the Kf is very large, assume all the NiCl2 reacts:

1.49*10^-3 moles NiCl2 x 6 moles NH3/1mole NiCl2 = 8.94*10^-3 moles of NH3 are consumed.

Moles of NH3 that remain = 0.0437 – 8.94*10^-3 = 0.0348

Moles of [Ni(NH3)6]+2 that are formed =
1.49*10^-3 moles NiCl2 x 1mole [Ni(NH3)6]+2/ 1 mole NiCl2 = 1.49*10^-3 moles ]Ni(NH3)6]+2.

Final volume = 123 mL + 185 mL = 308 mL = 0.308 Liters

[NH3] = 0.0348 moles/0.308 liters = 0.112 M
[Ni(NH3)6]+2 = 1.49*10^-3 moles/0.308 liters = 4.84*10^-3 M

[Ni(NH3)6{+2}]/ [Ni+2][NH3]^6 = Kf = 2.0 x 10^8

(4.84*10^-3)/[Ni+2](0.112)^6 = 2.0 x 10^8

[Ni+2] = 4.84*10^-3/(0.112)^6(2.0 x 10^8)

[Ni+2] = 1.23 x 10^-5

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