Chemistry Question Thread

Bibliii

Ka = the equilibirum constant of an acid dissociation reaction eg for the equilibrium system HCOOH + H2O <-> HCOO- + H3O+, Ka = [H3O+][HCOO-] / [HCOOH] (remember you exclude water for aqueous equilibria since its concentration change is negligible)

pKa = pH above which half of all acid molecules are deprotonated. Given by -log10(Ka), so higher pKa = weaker acid. If you inspect a titration curve, they're the half way points between equivalence points. They're really flat since when pH = pKa, an acid buffer system is most resistant to change according to the Henderson-Hasselbach equation (I don't know to what extent banana benders need to know this, I only learned Ka in uni chem in Victoria). You can memorise the property of pKa by the fact that when pH is above pKa, the equilibrium system is deficient in H+. According to Le Chat's principle, the system will want to partially oppose that, which it will by having your acid donate its proton.

Kb = the equilibrium constant of a base ionisation reaction eg the Kb for NH3 + H2O <--> NH4+ + OH-, Kb = [OH-][NH4+] / [NH3]. Kb and Ka can be generalised as [ionised species] / [unionised species]

pKb = the pH above which a base will be protonated. Similar story with pKa, when pH is above pKb, the system knows there's not enough H+. To compensate, you'll want less OH- in solution, forcing the base ionisation reaction to proceed in reverse

H3O+ = the protonated form of water, ie its conjugate acid. Abundant in acidic conditions since the acid has donated its H+ to the solvent molecule (H2O).

OH- = the deprotonated form of water ie its conjugate base. Abundant in basic conditions since the base has stolen H+s from H2O.

One small correction to jinx's excellent response, technically strong acids also have Ka and pKa values despite the fact that strong acid ionisation is pretty irreversible. For example, we are required to know in uni that pKa of HCl is -5 for purposes of comparing leaving group strength in organic reactions. Ditto for strong bases with regards to Kb and pKb

I don't know how QCE structures their data tests so someone else can help with that

jinx_58

Thanks! Could they ask about conjugate acid base pairs in a data test?

Hi guys,

I need some help. So I'm supposed to work out an equation for Sodium thiosulfate reacting with Hydrochloric acid. My teacher told me to search it up and so I did. When I did, it came up with Na2S2O3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + SO2(g) + S(s). Then, in another question, it said to write a balanced equation but THAT IS THE BALANCED EQUATION!!! I think. So what do I do? Do I just write down the balanced equation only?

Thanks so much, Leeshi!

Nvm, I figured it out but just want to check. Is the equation for Sodium thiosulfate reacting with Hydrochloric acid Na2S2O3(aq) + HCl --> NaCl+H2O+SO2+S

Please don't mind the numbers, they're supposed to be subscript ;-;

Thanks, Leeshi.

Leeshi you have 2Na on LHS but only 1Na on RHS, so the first equation you gave was correct

Billzene
Thank you for the feedback!

Heyoo,

I have my student experiment and my group decided to see what happens to conductivity when the temperature of ethanoic avid solution is increased. I've just got a tiny section of sample calculations to add before I move in my report. Would calculating Ka knowing the concentration H3O+ ions look like this (I'll try and type it out as best I can):
Ka = [H3O+][CH3COO-]/[CH3COOH]
Apparently CH3COO- concentration is equal to H3O+ so I think that explains the next step...
= [H3O+][H3O+]/[CH3COOH]
= [H3O+]²/[CH3COOH]

I have the concentration of CH3COO and also the concentration of H3O+ (via our experimental pH values) so substituting that in should be all good to go?

chocomilkenthusiast Good thinking with regards to the formulas, you can also use an ICE table if you’re unsure. I’d like to draw your attention to the fact that Ka is an equilibrium constant, which means it’s dependent on the endo/exo status of the ionisation reaction, so you may need some literature enthalpy values (theoretical accurate values) to assess your accuracy. To complicate things further, the enthalpy value changes depending on the ambient temperature. Just some things to think about here

lucy__ FYI the changing enthalpy based on changing temperature thing requires integral calculus (Kirchoff’s Law), so for the sake of your sanity, just look up some values of heat of ionisation for ethanoic acid at different temperatures.

The reason why I had to talk about it is because normally temperature changes don’t incur a sign change in the enthalpy change as in high school you assume heat capacities to be constant regardless of temperature. However, heats of ionisation (enthalpy change for an acid dissociation reaction) are so small that even the smallest temperature change may result in a new deltaH with an opposite sign.

Also, most literature refers to ethanoic acid as “acetic acid”, you’d be surprised how you can’t find anything with the proper IUPAC name but find a million things with its common name.

Leeshi usually with risk assessments you need to give the experiment title as well

Billzene

My title is safety and ethical considerations. However, the teacher wanted a caption below so I have now named it 'Table 1: Risks and management assessment'. I probably didn't do very well

-Leeshi

how do we balance the positive ions migrating from anode to cathode in electroylsis. since there isnt a salt bridge
if someone could explain

does anyone have any good diagrams to explain galvanic cells because im abit confused on how it works. like how do we know which is anode and cathode based on equations

chemistry1111

I 100% agree with watching those videos @jinx_58 sent. Tyler really does have good visual diagrams. As for the salt bridge, electrolytic cells don't have a salt bridge because it isn't divided into two separate half-cells like galvanic cells. In electrolytic cells, both electrodes are in the same solution which is the electrolyte. The electrolyte facilitates the balance of the charge for electrolytic cells. The videos should explain what I am saying as well.

Hope that makes sense.

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