how to identify strong acids and alkalis???

Calculate*the*pH*of*a*buffer*solution*formed*by*mixing*20.0cm3*of*1.20moldm-3*of*methanoic*acid*(Ka=1.78*10-4)*with*20.0cm3*0.500moldm-3*of*sodium*methanoate
for this question, you are supposed to know that sodium methanoate fully dissosciates but i'm not sure how you're supposed to know which compounds are strong alkalis, which are weak or which are strong acids, weak acids, etc. it says knwoing it fully dissosciates gives it a 1:1 ratio between the ions and sodium methanoate but i just want to know how i should figure this out?
are there a alist of all the strong acids/strong alkalis i should know?
btw the answer to this is 3.37 but i keep getting 3.6. i asked chatgpt and it also got 3.6 so im not sure how to do it...

I hope this helps I’ve labelled the steps for you !

Calculate*the*pH*of*a*buffer*solution*formed*by*mixing*20.0cm3*of*1.20moldm-3*of*methanoic*acid*(Ka=1.78*10-4)*with*20.0cm3*0.500moldm-3*of*sodium*methanoate
for this question, you are supposed to know that sodium methanoate fully dissosciates but i'm not sure how you're supposed to know which compounds are strong alkalis, which are weak or which are strong acids, weak acids, etc. it says knwoing it fully dissosciates gives it a 1:1 ratio between the ions and sodium methanoate but i just want to know how i should figure this out?
are there a alist of all the strong acids/strong alkalis i should know?
btw the answer to this is 3.37 but i keep getting 3.6. i asked chatgpt and it also got 3.6 so im not sure how to do it...

Also just to let you know that sodium ethanoate is a conjugate base so it’s unable to dissociate let alone fully ! For weak acid and it’s conjugate base you’re able to just apply the concentrations straight into the ka expression ! Don’t forget to use the new volume also ! If u have any more questions you can message me

Answering your second question first:
At A level, you only ever have a few examples of strong acids and bases you ever need to worry about.
Strong acids: HNO3, H2SO4, HCl, HClO3, HClO4 (and of course the bromine and iodine equivalents of the last 3 acids in that list).
Strong bases: hydroxides of the group 1 metals (e.g NaOH), Ba(OH)2.
At least at A level, if an acid or base doesn’t come up on either of the above lists, it almost certainly weak.
If they either give you a Ka/pKa or get you to calculate one, the acid used is weak (since strong acids dissociate too much for Ka and pKa to be meaningfully measured).
It’s also worth learning that the salts formed by weak acids and strong bases tend to be weak bases themselves as you get the following equilibrium:
A^- (aq) + H2O (l) <=> HA (aq) + OH^- (aq)
Where HA is a weak acid with conjugate base A^-
Similarly, salts formed when strong acids and weak bases are usually weak acids themselves:
HB^+ (aq) <=> H^+ (aq) + B (aq)
Where B is a weak base with conjugate acid HB^+
If a salt forms in the reaction of a strong acid with a strong base, or in the reaction of a weak acid with a weak base, the salt is likely neutral.
You need to know that ionic compounds dissociate upon dissolution in a solvent, such as water. In the case of sodium methanoate, HCOONa, you would get 1 Na^+ for every HCOO^- formed as implied by the formula. Something (marginally) more complicated such as MgI2 would form 1 Mg^2+ and 2 I^-, for example.
As for the exam question - this is an instance where you are mixing the conjugate acid (methanoic acid) and conjugate base (a source of methanoate ions, HCOONa) and so you don’t actually need to treat it as though any neutralisation reactions are taking place. This is why you don’t subtract anything.
So starting by finding a Ka expression:
Ka = [H^+][HCOO^-]/[HCOOH]
Calculating the relevant moles:
n(HCOOH) = 20/1000 dm^3 x 1.20 mol/dm^3 = 0.024 mol
n(HCOONa) = 20/1000 dm^3 x 0.50 mol/dm^3 = 0.010 mol
Plugging these and the Ka value given in the question directly into the Ka expression (since the 1/V terms cancel if you convert them to concentrations, meaning there is no point):
1.78 x 10^-4 = [H^+] x 0.010/0.024
=> 4.272 x 10^-4 = [H^+]
Finally using pH = -log[H^+]:
pH = -log[4.272 x 10^-4] = 3.3693… ≈ 3.37 (pH is quoted to 2 decimal places in most cases)

Hi, wow. thank you so much this explanation is so detailed. im just confused now on how to identify when its a neutralisation reaction. for example for ths question : A buffer solution contains 0.025 mol of sodium ethanoate dissolved in 500 cm3 of 0.0700 mol dm-3 ethanoic acid at 25 °C. A sample of 5.00 cm3 of 2.00 mol dm-3 hydrochloric acid is added to this buffer solution.Calculate the pH of the solution formed.
I subtracted 0.025 and 0.01 and added 0.035 and 0.01. what makes this different from the question i posted? the fact that it says 'dissolved' hints that its a neutralisation reaction right? so could i go off on that for other questions? mix= no neutralisation
dissolve= neutralisation
thank you, i appreciate your help

If you mix an acid and a base that are not a conjugate pair, you get neutralisation.
A conjugate pair is basically just a fancy way of describing an acid and a salt of that same acid. If you think about it, the following reaction would
take place if you mix an acid with one it’s salts:
HA (aq) + A^- (aq) <=> A^- (aq) + HA (aq)
So for every “neutralisation” that takes place, you end up re-making the acid and making the same amount of both acid and base as you started with. This is why you don’t add or subtract anything when you have a conjugate pair.
In the question you’ve now posted, there is neutralisation because you are mixing a strong acid with the salt of a weak acid (if you recall from my first post, this salt should be a weak base itself). This is not a conjugate pair, as the acid used (HCl) doesn’t dissociate to form ethanoate ions - it forms Cl^-. As such, the CH3COO^- will neutralise the HCl to form CH3COOH and Cl^-. As such, the moles of CH3COOH increase (so you add on the amount that is formed in the reaction) and the moles of CH3COO^- decrease (so you subtract the amount that react).
Dissolution isn’t necessarily an indicator of neutralisation taking place - it just means that any ionic compounds or strong acids are fully dissociated into ions. Weak acids will only be dissociated to a tiny extent, however.

thank you, i understand it perfecty now 🙂 i appreicate you time in writing these explanations out, thanks alot!
If you could please explain this last question to me: a) 2.00 cm3 of 0.100 mol dm-3 NaOH is added to 100.0 cm3 of water. Calculate the change in pH of the water.
I'm not sure on how this question works. i know i'd have to find the [H] of both NaOH and water. i know water is 7.00.. and then i'd do 0.002 multiplied by 0.1 to get mol of NaOH but this is where i seem to be going wrong?

If the temperature is 25°C, then yes, the pH of water initially is just 7 as Kw is 10^-14 mol^2/dm^6.
You need to be aware of Kw = [H^+][OH^-] for this question.
Since you are using 2 cm^3 of 0.100 mol/dm^3 NaOH, you have 0.0002 mol of NaOH and therefore 0.0002 mol of OH^- ions.
Since you are mixing 2 cm^3 of solution with 100 cm^3 of water, the total volume is 102 cm^3.
So [OH^-] = (0.0002 mol)/(0.102 dm^3) = 0.00196… mol/dm^3
Plugging this into the Kw expression:
Kw = 10^-14 mol^2/dm^6 = [H^+] x 0.00196
=> [H^+] = 5.10 x 10^-12
So pH = -log(5.10 x 10^-12) = 11.29 (2 dp)
But since it wants the change in pH, you take off 7 and so you get 4.29.

how come for this one, the volumes are added to form 0.102 whereas for the others, the volume isn't added? also I can't thank you enough, you're a life saver.

You are mixing two solutions, so you add the volumes together to get the total.
For the above examples with the Ka expressions, you could add the volumes and calculate the concentrations at equilibrium - and it wouldn’t be wrong to do so - just there isn’t any point as the 1/volume terms cancel as you have a concentration on the top and a concentration on the bottom of the Ka expression.
To see what I mean, let’s say we have
Ka = [H^+][A^-]/[HA]
If we let [A^-] be n(A^-)/V and [HA] be n(HA)/V, (where n means moles and V means the volume of the solution) we can substitute these into the expression above:
Ka = [H^+] x (n(A^-)/V) / (n(HA)/V)
Re-writing the above:
Ka = ([H^+] x n(A^-) x 1/V)/(n(HA) x 1/V)
See how there is a 1/V on the top of the fraction and a 1/V on the bottom? These cancel and so the fraction reduces as follows:
Ka = [H^+] x n(A^-) / n(HA)
But if we compare this with the Kw expression, we find that there is no denominator with a 1/V term to cancel the 1/V in [OH^-]. As such, the volume does matter for Kw, but not for Ka.

Calculate*the*pH*of*a*buffer*solution*formed*by*mixing*20.0cm3*of*1.20moldm-3*of*methanoic*acid*(Ka=1.78*10-4)*with*20.0cm3*0.500moldm-3*of*sodium*methanoate
for this question, you are supposed to know that sodium methanoate fully dissosciates but i'm not sure how you're supposed to know which compounds are strong alkalis, which are weak or which are strong acids, weak acids, etc. it says knwoing it fully dissosciates gives it a 1:1 ratio between the ions and sodium methanoate but i just want to know how i should figure this out?
are there a alist of all the strong acids/strong alkalis i should know?
btw the answer to this is 3.37 but i keep getting 3.6. i asked chatgpt and it also got 3.6 so im not sure how to do it...

I'm doing a chemistry degree and this tends to work for me. If a H is attached to a heteroatom like S,O,N (any top right of p block or halogen) it'll tent to be a strong acid. Weak acids at alevel tent to be H atoms bonded to less electronegative atoms like C (think things like carboxylic acids). Strong bases are most things with s-block elements with the negative charge being on the heteroatom (S, N, O often) e.g., KOtBu. Weak bases are harder to spot but often use a lone pair to pick up a proton like in ammonia. Take this with a pinch of salt as there are outliers such as NEt3 which can also be strong bases