1. The following equilibrium is established when solutions of
and
are mixed.
 + SCN^- (aq) \leftrightharpoons Fe(SCN)^{2-} (aq))
is colourless.
is colourless.
^{2-})
is deep red.
The solution is diluted with an equal volume of water. The colour becomes paler red, the addition of the water has caused "
concentration and the number of moles of to decrease.[The bolded is the answer]
Now I understand this perfectly but it's just when I start to think more deeply (and perhaps more than I need to) that I get confused.
First is adding water will dilute the solution hence concentration of the entire solution decreases but shouldn't the system NOT partially oppose this change? Because the left hand side has 2 molecules but can't the right hand side also be considered to have 2 molecules? Since
is in (aq) it is the same as saying
(aq) and
^{4-})
(aq)? [Just like how in water
(aq) is the same as
(aq)
(aq) but it is just we are lazy so we don't need to write them separately?] So since there are the same number of molecules, the system would do nothing to oppose the change?
Secondly, how does the mole of
decrease? I know the concentration decreases because the solution is diluted by adding more water. So let
be the respective mol, concentration and volume of
when the water is not added. Now let
be the respective mol, concentration and volume of
when the water is added.
Looking
, the concentration is a number and
is the initial volume which again is a number, but once the water is added
decreases to
)
and
increases to
)
, so how do you determine whether now
is smaller than
or larger? You don't know how much
decrease or
increased so you can't find out whether the final mol is larger or smaller than the original?
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