# Question 05c6a

Jun 28, 2016

$\setminus m a t h b f \left(\left(4\right)\right)$

#### Explanation:

Here's my reasoning:

The equilibrium expression can be written:

$\setminus m a t h b f \left(F e {\left(O H\right)}_{2 \left(s\right)} r i g h t \le f t h a r p \infty n s F {e}^{2 +} \left(a q\right) + 2 O {H}^{-} \left(a q\right)\right)$

For which:

$\setminus m a t h b f \left({K}_{s p} = \left[F {e}_{\left(a q\right)}^{2 +}\right] {\left[O {H}_{\left(a q\right)}^{-}\right]}^{2}\right)$

Now if \mathbf([Fe_((aq))^(2+)]=x

Then it follows that:

\mathbf([OH_((aq))^-]=2x

And

\mathbf(K_(sp)=x(2x)^2=4x^3

Initially the \mathbf(pH=8

Since

\mathbf(pH+pOH=14

Then

\mathbf(pOH=14-8=6

:.\mathbf([OH_((aq))^-]=10^(-6)" ""mol/l"

If the \mathbf(pH is raised to $9$ then:

\mathbf(pOH=14-9=5

:.\mathbf([OH_((aq))^-]=10^(-5)" ""mol/l"

This shows that the concentration of \mathbf(OH^- ions has increased by a factor of $\setminus m a t h b f 10$.

So the equilibrium moles of \mathbf(OH^- has gone from

\mathbf(2x to $\setminus m a t h b f \left(2 x\right)$\mathbf(xx10=20x

The value of \mathbf(K_(sp) must remain constant as this only depends on the temperature.

If you disturb a system at equilibrium by altering the concentrations of any of the species, then the system will adjust to return the same value of \mathbf(K which is in accordance with Le Chatelier's Principle.

To get the same value of \mathbf(K_(sp) you can see that the equilibrium moles of \mathbf(Fe^(2+) must decrease by a factor of \mathbf(100:

\mathbf(K_(sp)=x/(100)xx(20x)^2=x/cancel(100)xxcancel(400)x^2=4x^3

So the equilibrium moles of \mathbf(Fe^(2+) has gone from \mathbf(x to \mathbf(x/100 to conserve the value of \mathbf(K_(sp).

\mathbf(x/"Vol" represents the solubility of the iron(II) hydroxide so this also reduces by a factor of \mathbf(100 giving $\setminus m a t h b f \left(\left(4\right)\right)$ to be the correct response.

This is an example of "The Common Ion Effect", which, as I have hoped to demonstrate, is a consequence of the principle of Le Chatelier.

We have raised the concentration of the \mathbf(OH^(-)# ions so the system has responded by opposing this change and shifting to the left.

This has the consequence of reducing the solubility of the iron(II) hydroxide.