Equilibrium position -- a + b = c -- effect of dilution
iodine is sparingly soluble in pure water. however, it does `dissolve' in solutions containing excess iodide ion because of the following reaction: i-(aq) + i2(aq)= i3-(aq) k = 710. for each of the following cases calculate the equilibrium ratio of [i3-] to [i2]. 6.00×10-2 mol of i2 is added to 1.00 l of 6.00×10-1 m ki solution.
the solution above is diluted to

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:

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:

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.

the tides are influenced by both the moon and the sun, it's easy to see that when the sun lines up with the moon and the earth, as during a new moon or full moon, the tidal effect is increased. these are known as spring tides, named not for the season, but for the fact that the water "springs" higher than normal. on the other hand, if the sun and the moon are 90 degrees apart in relation to an observer on earth as during the first quarter moon or third quarter moon, then high tides are not as high as they normally would be. this is because despite its greater distance, the sun's mass allows it to exert enough gravitational force on the oceans that it can negate some of the effects of the moon's pull. this phenomenon of lower high tides is called a neap tide. if you need more information on this topic visit moonconnection.com.