A2 Chemistry

[thecandydoll]

Hey guys.
I am really getting stuck on this one table,and its annoying me.
Q2 Oct/Nov 2010 Paper 53.
How to fill in the data!
PLEASEEE HELP ME OUTT HERE!!!1

[da_rockstar]

what i did for this question was that:
for the concentration of acid in the water i did the following:
Moles of acid(moles of alkali): 0.1A*10^-3
concentration of acid= n/V= 0.1A*10^-3/ 10*10^-3 = 0.01A

for concentration of acid in the ether:
moles of acid: 0.02B*10^-3
concentration of acid= 0.02B*10^-3/ 25*10^-3= (8*10^-4)B
 
hope it helps. i did this and got the graph with correct gradient i.e 16.33.

[Sue T]

aaaammm - isnt the acid dibasic - therefore th moles of H+ are twice th moles of acid - so u should divide by 2!

[thecandydoll]

but the volume of NaOH in acidwater is 24cm3 ( jsut the first one)
wont amount of base = amount of acid in water
0.1*24/1000 = ((10/1000)*X)/2
where x is equal to  concentration of acid in the layer of wateR?

[THEIGBOY]

i'm getting confused about the dibasic thing... i also divided by 2, my gradient came out to be 32.7.... which is double what the marking scheme says... and yeah... the reaction is as follows...
2NaOH + HOOC-R-COOH --> Na^+-OOC-R-COO^-Na⁺ + 2H₂O

so number of moles of di ethyl ether would be 1/2 of number of moles of NaOH
have i got this all wrong or sumthing?? or is it an examiner's mistake