Stankovic123's chem q's

[lzxnl]

For the first point, water is the solvent. As the number of moles of water drop, so does the volume of water. Also, the concentration of water in water is so ridiculously high in comparison that you can neglect water in aqueous equilibria generally.

There are two factors at play here. Firstly, the Ka value needs to be sufficiently high. Secondly, the concentration can't be too low, otherwise the percentage ionisation will be too high.

I'll give an example. Let's assume that we have a solution of concentration M molar, acidity constant K.
Assume that the concentration of H+ and conjugate base formed is x.
So M-x molar acid left.
Subbing into K expression
K = x^2/(M-x)
x^2 = MK - xK
x^2 + xK - MK = 0
x = 1/2*(-K + sqrt(K^2+4MK))                                    quadratic formula, plus sign because x > 0

We want the case when K << M as that is what normally happens

x = 1/2*(-K + 2*sqrt(MK)*sqrt(1 + K/4M))  by bringing out the sqrt(4MK) from the square root

People are going to hate me for doing this, but I'm going to go ahead anyway.
(1+a)^n, for small a, is roughly equal to 1 + an + n(n-1)/2 * a^2

So apply this to the square root term
sqrt(1+K/4M) = (1+K/4M)^1/2, and assuming K/4M is small, which it generally is:
sqrt(1+K/4M) ~ 1+K/8M + 1/2*-1/2*(K/4M)^2 = 1+K/8M - K^2/64M^2

Subbing this back into the original expression for x
x~1/2*(-K+2sqrt(MK)*(1+K/8M - K^2/64M^2))

OK. Let's see what this means. If M is sufficiently larger than K, then K^2/64M^2 is meaninglessly small. We can discard that term.
If K/8M is also sufficiently small, then we can remove that term as well.
We are then left with x~1/2*(-K + 2sqrt(MK) = -K/2 + sqrt(MK)
The term sqrt(MK) is what you would normally get by assuming M>>x in K = x^2/(M-x)
But if you assumed that K/8M was tiny, you wouldn't care about the individual value of K, so the -K/2 term is also negligible in this case.

What does all this mean? You need to look at the ratio K/M and consider how significant it is to the concentration in M of acid produced. Generally, if you have a concentration of around 0.1 M, and you have a normal weak acid with a Ka of 10^-5, then K/M = roughly 10^-4, which is tiny. But if you had a solution of 10^-3 M and your Ka was 10^-4...then K/M is 0.1, which can't be neglected.

Yeah. This thing. I hope what I did was vaguely accurate. For a monoprotic acid, the ratio acidity constant/initial concentration has to be small. As for how small, that's for you to judge.

[Mao]

That does make sense, but then how does the equilibrium position shift for reactions where the mol ratios on each side of the equation are not equal? Whats the mechanism behind this? Your concentrations would still be changed by the same factor with a decrease/increase in volume wouldnt they?

Remember that equilibrium is maintained only if the reaction rates stay the same. In equimolar reactions, the reaction rates are scaled by the same amount when volume/pressure is changed, so there is no shift in equilibrium position.

In non-equimolar reactions, the reaction rates are NOT scaled by the same factor. e.g. A+2B <--> 2C, where [A]=[B ]=2 M, [C]=10 M. If we dilute everything by half, then the forward reaction will be 1/8 of its original rate, but the backwards reaction will be 1/4, so the back reaction is faster and pushes the equilibrium position to the left.

Remember that the equilibrium position is not always at the same fixed position for a reaction. It is a consequence of the thermodynamics, and is a purely mathematical construct. The equilibrium position can be quite different if the starting amounts are different. If you do Specialist Mathematics, you would find it easier to understand from differential equations: the same DE can lead to many different answers, depending on your initial condition. Same is true for equilibria.

That actually leads in nicely to my next question lol. When can we not assume that initial conc. and equilibrium conc. are equal?

Only if the questions states the system is initially in equilibrium.

[zvezda]

In non-equimolar reactions, the reaction rates are NOT scaled by the same factor. e.g. A+2B <--> 2C, where [A]=[B ]=2 M, [C]=10 M. If we dilute everything by half, then the forward reaction will be 1/8 of its original rate, but the backwards reaction will be 1/4, so the back reaction is faster and pushes the equilibrium position to the left.

Would you be able to expand on this a little bit? How did you calculate the reduction in the tae of the forward and back reactions?


Also, in relation to cells in redox, where does the actual energy from a cell come from? I know that chemical energy is converted to electrical, but if a battery were to supply energy to power something, isnt there some sort of net loss of energy in the reaction system?
Im struggling to imagine the processes. Would anyone know of any ressource which may clear things up a little bit; a link or some sort of video?
Thanks

[lzxnl]

Would you be able to expand on this a little bit? How did you calculate the reduction in the tae of the forward and back reactions?


Also, in relation to cells in redox, where does the actual energy from a cell come from? I know that chemical energy is converted to electrical, but if a battery were to supply energy to power something, isnt there some sort of net loss of energy in the reaction system?
Im struggling to imagine the processes. Would anyone know of any ressource which may clear things up a little bit; a link or some sort of video?
Thanks

In the forward reaction, you have 3 moles of reactants. So in the equilibrium law, you have [A]^2. If you halve both A and B, you multiply the original forward rate by 1/8. The back reaction only has two particles; halve that concentration, and you'll only have a drop to 1/4.


As for redox, imagine you have two batteries, a 5 V and a 3 V battery. If you connect them the wrong way around, i.e. negative to negative and positive to positive, what happens? The 3 V battery becomes a 3 V voltage drop, while the 5 V voltage rise is still a 5 V voltage rise. The net result is a 2 V voltage rise.
If a battery supplies energy to power a reaction like that, it is sort of like using a 5 V battery connected the wrong way to a 3 V battery. As for the energy loss, you have an energy loss in the battery that is transformed into raising the chemical potential of the system; imagine pushing a car up a hill. The energy inside the pusher is going to the increased gravitational potential of the car. Let go of the car, and it'll fall down. Remove the battery, and the reaction will proceed backwards. Equilibrium is simple a case where the potential is lowest.

[Mao]

Also, in relation to cells in redox, where does the actual energy from a cell come from? I know that chemical energy is converted to electrical, but if a battery were to supply energy to power something, isnt there some sort of net loss of energy in the reaction system?

That would be generally converted to heat. Remember that mediums which don't conduct electricity (read: insulators, i.e. most non-metals) will act as resistors, and readily dissipate excess electrical energy.

As for redox, imagine you have two batteries, a 5 V and a 3 V battery. If you connect them the wrong way around, i.e. negative to negative and positive to positive, what happens? The 3 V battery becomes a 3 V voltage drop, while the 5 V voltage rise is still a 5 V voltage rise. The net result is a 2 V voltage rise.
If a battery supplies energy to power a reaction like that, it is sort of like using a 5 V battery connected the wrong way to a 3 V battery. As for the energy loss, you have an energy loss in the battery that is transformed into raising the chemical potential of the system; imagine pushing a car up a hill. The energy inside the pusher is going to the increased gravitational potential of the car. Let go of the car, and it'll fall down. Remove the battery, and the reaction will proceed backwards. Equilibrium is simple a case where the potential is lowest.

This analogy is incorrect. A better analogy would be instead of a hill, it is a series of steps/stairs, and you are pushing the car up one step at a time using some amount of energy. If the energy is greater than what would be required to reach a step, that means the car is half-way between steps (unstable transition state/intermediate), which then relaxes to the step (stable product) and the difference is released as thermal energy (intramolecular vibrations, etc.).

[zvezda]

What classifies as a good conductor of heat?
Ive come across a question from a trial paper where the solutions said that "glass js a poor consuctor of heat". I feel like this is basic knowledge but i just do not know it lol.
Help much appreciated

[Stevensmay]

Copper is considered quite good. I believe aluminium and water sit somewhere in between.

[zvezda]

Copper is considered quite good. I believe aluminium and water sit somewhere in between.

What makes them good at conducting heat?

[Stevensmay]

Generally the close lattice packing allows them to conduct heat better. De-localised electrons can also transfer kinetic energy around, allowing heat to be moved better.

I'm sure someone can come up with a better answer than this.

[zvezda]

The de-localisation of electrons somewhat makes sense.

Also, what is the chemical basis of polyprotic acids having more than one Ka value? I know that its because they can donate more than one H+, but is there something more detailed than thia?
Thanks

[SocialRhubarb]

I think the important thing to note is that K values don't apply to particular chemical species - they apply to reactions.

The ionisation of polyprotic acid,  H₃PO₄ for example, can be represented by several reactions.







Since there are multiple reactions, the acid will have multiple Ka values.

[lzxnl]

It CAN be argued that the second ionisation constant for phosphoric acid is really the ionisation constant for the dihydrogen phosphate ion.

Although in more general contexts, yes the K values do depend on the reactions. Hydrochloric acid isn't a strong acid when dissolved in ethanoic acid, for instance.

[zvezda]

Ok thanks guys.

2 questions:
Why is the acetate ion a weak base? Shouldnt it be a strong base given that the Ka for the reverse reaction of ethanoic acids ionisation is high??

Also, if an equilibrium solution of the weak base aniline is diluted with water, would the pH increase or decrease afterwards? The answers in one of the company exams says that the pH would decrease but why? Havent both the conc. of hydronium and hydroxide decreased overall??

[brightsky]

Ethanoic acid is a weak acid (you can look up the ka value if you want) so its conjugate base is a weak base.

Increase overall, but due to back reaction, ph would decrease a little bit after initial dramatic increase.

[lzxnl]

Ok thanks guys.

2 questions:
Why is the acetate ion a weak base? Shouldnt it be a strong base given that the Ka for the reverse reaction of ethanoic acids ionisation is high??

Also, if an equilibrium solution of the weak base aniline is diluted with water, would the pH increase or decrease afterwards? The answers in one of the company exams says that the pH would decrease but why? Havent both the conc. of hydronium and hydroxide decreased overall??

Let's answer the first question.
For an acid HA<=>H+ + A-, Ka=[H+][A-]/[HA]
For its conjugate base, A- + H2O <=> HA + OH-
Just as we can define an acidity constant, we can define a basicity constant Kb such that Kb=[HA][OH-]/[A-] which measures the amount of OH- from the ionisation.
What if we multiply these constants? Ka*Kb=[H+][A-][HA][OH-]/([A-][HA])=[H+][OH-] which at normal conditions is 10^-14
So take the acidity constant and divide 10^-14 by it. This is effectively your base strength.
As you can see, ethanoic acid, with a Ka of around 1.7*10^-5, has a conjugate base strength of 5.9*10^-10 ish. That's very weak.

Now second question. You have a base with a certain amount of OH-. It's a base, so [H+]<10^-7. But the solution (water) you're adding as [H+]=10^-7. You tell me, do you think the pH will increase or decrease?
Another way of looking at it. If you dilute the solution, [OH-] must decrease. The opposite would make no sense. Therefore, the pH must what?