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:
[tex]K=710=\frac{[I_{3}^{-}]}{[I^{-} ].[I_{2}]} =\frac{x}{(0.600-x).(0.0600-x)}[/tex]
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:
[tex]K=710=\frac{[I_{3}^{-}]}{[I^{-} ].[I_{2}]} =\frac{x}{(0.0600-x).(0.00600-x)}[/tex]
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.
