Question

A 130.0-mLmL sample of a solution that is 3.0×10−33.0×10−3 MM in AgNO3AgNO3 is mixed with a 225.0-mLmL sample of a solution that is 0.14 MM in NaCNNaCN. For Ag(CN)2−,Ag(CN)2−, Kf=1.0×1021Kf=1.0×1021. After the solution reaches equilibrium, what concentration of Ag+(aq)Ag+(aq) remains?

Answer 1

1.2x10⁻²³M = [Ag⁺]

Step-by-step explanation:

The Kf of Ag(CN)₂⁻ is:

Ag⁺(aq) + 2CN⁻ → Ag(CN)₂⁻(aq)

Kf = 1.0x10²¹ = [Ag(CN)₂⁻] / [Ag⁺] [CN⁻]²

Where [] are equilibrium concentrations.

The initial concentrations of the reactants are:

[Ag⁺] = 3.0x10⁻³M * (130mL / 355mL) = 1.10x10⁻³M

335mL is the total volume of the solution

[CN⁻] = 0.14M * (225mL / 355mL) = 0.089M

In equilibrium, the concentrations are:

[Ag⁺] = 1.10x10⁻³M - X

[CN⁻] = 0.089M - 2X

Ag(CN)₂ = X

Replacing in Kf expression:

1.0x10²¹ = [Ag(CN)₂⁻] / [Ag⁺] [CN⁻]²

1.0x10²¹ = [X] / [1.10x10⁻³ - X] [0.089 - 2X ]²

1.0x10²¹ = [X] / 8.7131×10⁻⁶ - 0.0083126 X + 0.3604 X² - 4 X³

8.7131×10¹⁵ - 8.3126×10¹⁸ X + 3.604×10^20 X² - 4x10²¹ X³ = X

8.7131×10¹⁵ - 8.3126×10¹⁸ X + 3.604×10^20 X² - 4x10²¹ X³ = 0

X = 0.0011M → Right solution.

X = 0.0445M → False solution. Produce negative concentrations

Replacing:

[Ag⁺] = 0

[CN⁻] = 0.0868M

Ag(CN)₂ = 0.0011M

The equilibrium will shift to the left just a little bit until produce a determined amount of Ag⁺ that we can write as:

[Ag⁺] = X

Thus:

1.0x10²¹ = [0.0011] / [X] [0.0868M ]²

1.2x10⁻²³M = [Ag⁺]