pretty sure its the half neutralisation thing so u go at the equivalence point which is the middle of the steep verticle line and just read off the volume
(not 100% sure but this is how i do it)

Reply 84

mhm.0987

5

hows everyone revising for the practicals

Reply 85

TypicalNerd

18

Original post by mhm.0987

pretty sure its the half neutralisation thing so u go at the equivalence point which is the middle of the steep verticle line and just read off the volume
(not 100% sure but this is how i do it)

Half neutralisation is reading the pH off the curve at the point that is midway between not adding any alkali at all and the vertical section in order to find the pKa.

Reading volume of alkali that reaches the equivalence point is a valid method, but because that is difficult to do precisely, so using the vertical section as a whole is sufficient.

Reply 86

mhm.0987

5

Original post by TypicalNerd

Half neutralisation is reading the pH off the curve at the point that is midway between not adding any alkali at all and the vertical section in order to find the pKa.

Reading volume of alkali that reaches the equivalence point is a valid method, but because that is difficult to do precisely, so using the vertical section as a whole is sufficient.

so am i wrong? bc isnt [HA] = [A-] at half neutralisation?

Reply 87

ifif4

13

Can someone please help me with this question?

Original post by mhm.0987

hows everyone revising for the practicals

Writing reciting and answering PPQs
And keeping a record of those weird one markers

Reply 89

mhm.0987

5

Original post by ifif4

Can someone please help me with this question?

is it B?

Reply 90

ifif4

13

Original post by mhm.0987

is it B?

yes. How did you work it out?

Reply 91

cerealsnatcher

3

Original post by ifif4

Can someone please help me with this question?

I think cuz the rate falls by factor of 4 concentration of hydrogen ions also falls by a factor of 4, therefore find the conc of hydrogen ions originally from ph = 0.7 and divide it by 4 then find ph again

Reply 92

mhm.0987

5

(Original post by ifif4)yes. How did you work it out?

Original post by ifif4

yes. How did you work it out?

i found the [H+] from 0.7 then divided by 4 and then did the -log[H+]

Reply 93

boomsicles48

15

How are people revising for this as there is a lot of content to cover?

Reply 94

ifif4

13

Original post by mhm.0987

(Original post by ifif4)yes. How did you work it out?

i found the [H+] from 0.7 then divided by 4 and then did the -log[H+]

Original post by cerealsnatcher

I think cuz the rate falls by factor of 4 concentration of hydrogen ions also falls by a factor of 4, therefore find the conc of hydrogen ions originally from ph = 0.7 and divide it by 4 then find ph again

Thank you

Reply 95

mhm.0987

5

Original post by boomsicles48

How are people revising for this as there is a lot of content to cover?

u kinda dont need to cover all the content bc u should know most of it anyway bc of the other exams so now jus doing pps

Reply 96

FatalArcher

11

If anyone has spare time, maybe look at OCR B PPQs? My teachers have recommended completing them if i have spare time - they are mainly practicals based.

Reply 97

TypicalNerd

18

Original post by mhm.0987

so am i wrong? bc isnt [HA] = [A-] at half neutralisation?

Half neutralisation is the point that is midway between adding absolutely nothing to the solution and the vertical section itself. It is used exclusively for finding the pKa.

Equivalence is mid way through the vertical section. This is used to determine suitable pH indicators, but the volume corresponding to equivalence is how much you need to add in order to fully neutralise the solution in the conical flask.

So your method in your first post was essentially correct, but your choice of terminology was off.