I mean I kind of understand, however personally for the first question, I wouldn't have rounded up so much and instead would give 1.29x10-3, but yeah maybe my round-up is bad
n equals concentration x volume. As the concentration we found was 0.3024 moldm-3, we need to convert this into moles hence we multiply this by 250cm3/1000
you don't need to work it out if you're doing the conc of acid divide the conc of salt because the moles of acid divide the moles of salt gives the same ratio
@@p.p6343 thank you! I don't suppose you know why sometimes the moles of acid or alkali added become the moles of the salt but other times the moles of the acid or alkai are taken away from or added onto the moles of the salt? I get very confused about it
@@Harry-es7zh I think you're referring to the initial moles compared to the final moles? The final moles of salt formed is equal to the moles of the limiting reactant either acid or alkali which is all used up. Hope I haven't misunderstood what you meant lol
In the first your working out the [h+] conc with a given ph so you use the equation: [h+] = 10^-ph. the ph is given in the question as 2.89 so: [h+] = 10^-2.89. giving the answer 1.3 x 10^-3 ( answer C) In the second your working out the ph but you need to initally work out the [h+] conc from the Ka. Use: [h+] = Ka x [acid] / [salt]. We know the [acid] is half that of the [salt] so you can just sub in 1 and 2 into the equation giving you: [H+] = ka x 1/2 Using the given ka value of 1.7 x 10^-4 [H+] = 1.7 x 10^-4 x 1/2 gives [H+] = 8.5x10^-5 now you can use the ph = -log(h+) ph = -log(8.5x10^-5) this gives the answer of ph 4.07 ( answer D) Basically one your given the ph in the question and the other your working it out. hope this kind of helped?
The concentration calculated initially is moles per dm³ but we need to know the moles needed for the 250cm³ solution. That's 1/4 of the volume so we divide by 4. Hope that helps
I mean I kind of understand, however personally for the first question, I wouldn't have rounded up so much and instead would give 1.29x10-3, but yeah maybe my round-up is bad
in part C why times by 0.8 to get it to be in equilibrium at 125cm3
In Q6 why did you divide the moles by 4 in the 250cm³?
This may be wrong but I am guessing it is because 250 x 4 = 1000 ml and 1000ml is 1dm3. SO essentially we have a quarter of it.
n equals concentration x volume. As the concentration we found was 0.3024 moldm-3, we need to convert this into moles hence we multiply this by 250cm3/1000
I dont understand why you divide by 4 in the last question @machemguy
@machemguy dont worry sir. Got it
Its because I do it as x1000/volume
is there ever a situation in which you don't need to work out the total volume present or must you do this for all buffer calcs
you don't need to work it out if you're doing the conc of acid divide the conc of salt because the moles of acid divide the moles of salt gives the same ratio
@@p.p6343 thank you! I don't suppose you know why sometimes the moles of acid or alkali added become the moles of the salt but other times the moles of the acid or alkai are taken away from or added onto the moles of the salt? I get very confused about it
@@Harry-es7zh I think you're referring to the initial moles compared to the final moles? The final moles of salt formed is equal to the moles of the limiting reactant either acid or alkali which is all used up. Hope I haven't misunderstood what you meant lol
wait, why dont i have to do the -log in the first but in the second? whats the difference between the two
In the first your working out the [h+] conc with a given ph so you use the equation:
[h+] = 10^-ph.
the ph is given in the question as 2.89 so:
[h+] = 10^-2.89.
giving the answer 1.3 x 10^-3 ( answer C)
In the second your working out the ph but you need to initally work out the [h+] conc from the Ka. Use:
[h+] = Ka x [acid] / [salt].
We know the [acid] is half that of the [salt] so you can just sub in 1 and 2 into the equation giving you:
[H+] = ka x 1/2
Using the given ka value of 1.7 x 10^-4
[H+] = 1.7 x 10^-4 x 1/2
gives [H+] = 8.5x10^-5
now you can use the ph = -log(h+)
ph = -log(8.5x10^-5)
this gives the answer of ph 4.07 ( answer D)
Basically one your given the ph in the question and the other your working it out.
hope this kind of helped?
@@samwilson7012 ah im sorry, i never replied 😐 exam season..
thank you so much for the help!!
On question 6 why do we have to divide the moles by 4?
The concentration calculated initially is moles per dm³ but we need to know the moles needed for the 250cm³ solution. That's 1/4 of the volume so we divide by 4. Hope that helps
@@MaChemGuy Thank you!
isn’t question 5 option C? I keep getting to the power of -4 and not -6
I’ve just done it again now and get option B still
What year are these question?
This particular one is Y13
@@MaChemGuy are these questions a mix of papers from a2?
@@crypticgod1134 Correct