Water, water everywhere and not a drop to put into my equilibrium constant calculation...

Kc, Kw, Ka, Kp...

When do you add water to the equation and when do you not?

This is more straight forward than you think.

It comes down to how much water have you got. If your whole reaction is happening in water then you have got so much of it that the little bit that you are making is irrelevant and should be left outt of the calculation of Kc. Effectively it comes down to the fact you can't dissolve water in water.

So any aqueous reaction you leave out water, this is most reactions including all Kw and Ka's and most Kc's.

There will be some Kc calculations where you are reacting liquids and you deliberately don't add water. Esterification is the most obvious of these. To get the reaction to work you make sure water is not present (as it is a product and the equilibrium will stay to the left). There are very few others.

Kp is different as this is all about pressure and water, just like everything else, has a pressure as long as it is a gas.

All neutralisations are equal, but some neutralisations are more equal than others.

Have you ever thought about this fact...

If you add an equal amount of acid to an equal amount alkali, the solution you make isn't necessarily neutral.

i.e. when you add the same amount of OH- to H+ the solution isn't always pH 7.

i.e. neutralisation  isn't (necessarily) neutral.

Makes no sense, until you start to think about what is in the "neutral" solution.

Lets use an example,

adding ethanoic acid to sodium hydroxide

ethanoic acid + sodium hydroxide --> Sodium ethanoate + water

Ionically, this is -

ethanoic acid + hydroxide ions --> ethanoate ions + water.

So at "neutralisation", the only things in the conical flask are ethanoate ions and water.

Water is obviously neutral so it must be the ethanoate that is making it (in this case) basic.

Ethanoate ions will absorb protons from water making hydroxides, turning the solution basic.

Why does ethanoate do this? Well, that comes down to the fact that ethanoate ions and ethanoic acid are conjugate acid/base pairs. If one is rubbish, the other is awesome.

Ethanoic acid is a weak acid (rubbish) so ethanoate is a relatively strong base (awesome).

So neutralisation is not the best name for this situation where acid cancels out base. A better name is equivalence as we have an equivalent amount of acid and base (but it isn't neutral).

We shall keep the name neutralisation for when the pH is 7. Which is fairly irrelevant from a chemical point of view. Which is a shame considering your chemistry teacher has been obsessing on it since year 7.

(for an explanation of the pig reference - google, animal farm)

Liberté. Egalité. Acidité.

Ever been confused about the line in the spec...


"...appreciate that carboxylic acids liberate CO2 from carbonates and hydrogen carbonates but phenol does not..."

Appreciate?
Liberate?

In plain English, know that carboxylic acids give off carbon dioxide with carbonates and hydrogen carbonates but phenol doesn't.

That's better,

Lets first establish the reason that carboxylic acids give off CO2 with carbonates and phenols don't is because carboxylic acids are more acidic.

Want to know why?

It comes down to the stability of the thing (anion) formed when the proton has gone. If the anion formed is stable then the proton will happily leave. 

The more stable it is, the more likely the proton is to leave, the more acidic the molecule.

Lets look at the contenders

When carboxylic acids give off a proton they form the carboxylate ion which is stable because the negative charge (lone pairs) forms a miniature delocalised system across the two oxygens, like this...

When phenols give off a proton they form the phenoxide ion which is stable because the negative charge (lone pairs) becomes part of the benzene delocalised system, like this...

The delocalisation is more effective in carboxylic acids then phenol so carboxylic acids are more acidic.

C'est très simple.

Liberté
Egalité
Acidité

Power of Love

Here is a long (and a little mathsy explanation) of what an order of a reaction is.

The maths here is pretty basic, as long as you understand POWERs (e.g. squaring)  then it is LOVEly. Pretty awful justification of the blog post title. Lets move on.

You know from GCSE (or earlier) that as you increase concentration, rate increases. Greater chance of collision, more succesful collision, blah, blah, blah... You might even have learnt that concentration and rate are directly proportional. 

At A level you learn that both of these facts are only true some of the time. In reality, you will now learn there are three different options (in reality there are more than that but lets not get into that now)

Option A - Some reactants as you increase their concentration, the rate is not changed, these are called zero order reactants. In other words put in as much of it as you want, it ain't going any faster.

Option B - Some reactants as you increase their concentration, the rate increase proportionally (i.e. double conc  = double rate), these are called first order reactants. In other words, pretty much what a GCSE student would've predicted

Option C - Some reactants as you increase their concentration, the rate changes by that number squared (i.e. double conc = quadruple rate, triple conc = 9 times increase in rate etc.) , these are called second order reactants.  In other words more of this stuff really gets reactions going.

You need to understand where these numbers (zero, first, second) come from. These numbers (or orders) are needed as powers in the equations so that they works mathematically

So, (in case you've forgotten) the rate equation is

Rate = k x [Conc].

If it is zero order, that gives us, Rate = k x [Conc]⁰, 

and as anything to the power zero = 1 

that gives us

^{Rate = k x 1 or Rate = k} 

^{in other words, Concentration of the reactant is not in the equation so it doesn't affect rate. (Just like I said in Option A above)}

If it is first order, that gives us, Rate = k x [Conc]¹, 

and as the power 1 doesn't change anything

that gives us

^{Rate = k x [Conc]}  

^{in other words, Concentration of the reactant is directly proportional to rate. (Just like I said in Option B above)}

If it is second order, that gives us, Rate = k x [Conc]², 

^{in other words, If you double Concentration of the reactant the rate goes up by that number squared. So a doubling of a [Conc] would lead to a quadrupling of the rate as 22 = 4. (Just like I said in Option C above)}

So what is the order? 

It is the power, to which you raise the concentration in the rate equation. 

If you read through the whole post, and understand it that last statement is almost pointlessly obvious