**Answer:**

[I⁻] = 0.0352M

**Explanation:**

Based on the equilibrium:

I₃⁻(aq) ⇄ I₂(aq) + I⁻(aq)

Kc is defined as:

Kc = 0.25 = [I₂] [I⁻] / [I₃⁻]

*The system reaches the equilbrium when the ratio* [I₂] [I⁻] / [I₃⁻] *is equal to 0.25*

In the beginning, you add 0.0401M of both [I₂] [I⁻]. When the reaction reach the equilibrium, xM of both [I₂] [I⁻] is consumed producing xM of [I₃⁻]. That is written as:

[I₃⁻] = X

[I₂] = 0.0401M - X

[I⁻] = 0.0401M - X

*X is known as reaction coordinate.*

Replacing in Kc:

0.25 = [I₂] [I⁻] / [I₃⁻]

0.25 = [0.0401M - X] [0.0401M - X] / [X]

0.25X = 0.00160801 - 0.0802X + X²

**0 = 0.00160801 - 0.3302X + X²**

Solving for X:

**X = 0.0049M → Right solution**

X = 0.3252M → False solution. Produce negative concentrations

Replacing, equilibrium concentrations will be:

[I₃⁻] = X

[I₂] = 0.0401M - X

[I⁻] = 0.0401M - X

[I₃⁻] = 0.0049M

[I₂] = 0.0352M

### [I⁻] = 0.0352M