NSG 2020 Q31 - no calculation for answers PLEASE HELP (1 Viewer)

[SadCeliac]

My answers were:

b) pH = 8.6 @ 22.5mL

c)
[NaOH] = 0.15M
v(NaOH) = 0.0225L (from graph) at equivalence point
therefore n(NaOH) = 0.003375mol
therefore since 1:1 ration
n(acid) = 0.003375mol
v(acid) = 0.025L aliquot
c=n/v
therefore [acid] = 0.138 M

is this correct??


edit looking at my own graph which was slightly less accurate than theirs the eq point looked like it was at 8.6pH at 22.5mL, obviously their graph is a little different but it's still around about that pH=8 mark and I'd assume you just use the eq point values you took from part b for part c

 

[carrotsss]

n(acid) should be 0.003375mol

 

[SadCeliac]

n(acid) should be 0.003375mol

ah yes but that was just when i was typing it out lol

 

[SadCeliac]

but otherwise it's correct??

 

[carrotsss]

but otherwise it's correct??

yeah it’s a pre simple q

 

[SadCeliac]

yeah it’s a pre simple q

yeah lol idk why i stressed so much over it

letsgooo got 89 in the paper but it was really easy

 

[wizzkids]

Obviously this student was in a bit of a hurry and missed the equivalence point by a substantial margin.
If you add the last drops of NaOH slowly you should be able to achieve a precision of 0.1 mL in the aliquot.
Look at the results at 20.0 mL and 24.0 mL. This is where the inflection points occur.
I estimate the equivalence point at between 22.0 and 22.1 mL.
That would give [CH₃COOH] = 0.133 m/L but since we are limited to 2 significant figures, say 0.13 m/L

 

[Luukas.2]

The data provided here are deeply dodgy. If the pH of the acetic acid solution is 3.8 without any base having been added, then the acetic acid solution has a concentration of something like 0.002 mol/L.

 

[synthesisFR]

@Luukas.2 surely ur a uni student or teacher right (please dont be in our grade then we are screwed)

 

[SadCeliac]

Obviously this student was in a bit of a hurry and missed the equivalence point by a substantial margin.
If you add the last drops of NaOH slowly you should be able to achieve a precision of 0.1 mL in the aliquot.
Look at the results at 20.0 mL and 24.0 mL. This is where the inflection points occur.
I estimate the equivalence point at between 22.0 and 22.1 mL.
That would give [CH₃COOH] = 0.133 m/L but since we are limited to 2 significant figures, say 0.13 m/L

so I was basically right with 0.138M ???

 

[carrotsss]

@Luukas.2 surely ur a uni student or teacher right (please dont be in our grade then we are screwed)

he’s a student

 

[Luukas.2]

so I was basically right with 0.138M ???

From the data, the pH of the equivalence point is around 8... so long as the volume is consistent with the graph drawn and a sensible pH, whatever concentration is calculated is reasonable. I see no reason that you should have been penalised... though the examiners / markers might have a different view.

 

[Luukas.2]

@Luukas.2 surely ur a uni student or teacher right (please dont be in our grade then we are screwed)

he’s a student

Aren't we all students? Learning new things every day, as students of life?

... Though, I do believe that there is not much about HSC chemistry left for me to learn.

 

[synthesisFR]

Aren't we all students? Learning new things every day, as students of life?

... Though, I do believe that there is not much about HSC chemistry left for me to learn.

Nvm guys he’s an English teacher

 

[Luukas.2]

Nvm guys he’s an English teacher

As we all know, English teachers are noted for their proficiency in Sciences and Mathematics.

After all, Shakespeare wrote so eloquently of the unparalleled beauty of quartic polynomials with complex coefficients in his famous Sonnet 18 (Shall I compare thee to a summer's day? ...)

 

[SadCeliac]

As we all know, English teachers are noted for their proficiency in Sciences and Mathematics.

After all, Shakespeare wrote so eloquently of the unparalleled beauty of quartic polynomials with complex coefficients in his famous Sonnet 18 (Shall I compare thee to a summer's day? ...)

Bro cool it with the 99.95