Question #05c6a

1 Answer
Jun 28, 2016

Answer:

#\mathbf((4))#

Explanation:

Here's my reasoning:

The equilibrium expression can be written:

#\mathbf(Fe(OH)_(2(s))rightleftharpoonsFe^(2+)(aq)+2OH^(-)(aq))#

For which:

#\mathbf(K_(sp)=[Fe_((aq))^(2+)][OH_((aq))^(-)]^2)#

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 #\mathbf10#.

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

#\mathbf(2x# to #\mathbf(2x)##\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 #\mathbf((4))# 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.