Question

Equilibrium Position -- A + B = C -- Effect of Dilution

Answer 1

Answer:

299

Explanation:

The initial concentrations are:

[I₂] = 6.00 × 10⁻² mol / 1.00 L = 6.00 × 10⁻² M

[KI] = [I⁻] = 6.00 × 10⁻¹ M

We can find the concentrations at equilbrium using an ICE chart.

       I⁻(aq)          +           I₂(aq)           ⇄           I₃⁻(aq)

I    6.00 × 10⁻¹            6.00 × 10⁻²                    0

C       -x                             -x                             x

E  6.00 × 10⁻¹-x         6.00 × 10⁻²-x                   x

The equilibrium constant (K) is:

$K=710=\frac{[I_{3}^{-}]}{[I^{-} ].[I_{2}]} =\frac{x}{(0.600-x).(0.0600-x)}$

Solving the quadratic equation, we get the feasible x = 0.0598

The ratio [I₃⁻] to [I₂] is 0.0598 / 2.00 × 10⁻⁴ = 299.

If we diluted the initial mixture to, for instance, 10.0 L (dilution 1:10),

The initial concentrations are:

[I₂] = 6.00 × 10⁻³ M

[KI] = [I⁻] = 6.00 × 10⁻² M

We can find the concetrations at equilbrium using an ICE chart.

       I⁻(aq)          +           I₂(aq)           ⇄           I₃⁻(aq)

I    6.00 × 10⁻²            6.00 × 10⁻³                    0

C       -x                             -x                             x

E  6.00 × 10⁻²-x         6.00 × 10⁻³-x                   x

The equilibrium constant (K) is:

$K=710=\frac{[I_{3}^{-}]}{[I^{-} ].[I_{2}]} =\frac{x}{(0.0600-x).(0.00600-x)}$

Solving the quadratic equation, we get the feasible x = 0.00585

The ratio [I₃⁻] to [I₂] is 0.00585 / 1.50 × 10⁻⁴ = 39. The solubility of I₂ decreases.