# help of a chem questionWatch

This discussion is closed.
#1
HA(s)+NaOH(l)= NaA(aq)+H20(l)

7.1g of acid was added to 250cm^3 of water in a 250cm^3 flask of water, then 25cm^3 of it is placed in the buiret via a pipette. On average it takes 23.27cm^3 of the HA to change the colour of 25cm^3 of 0.15M NaOH to change colour.
What is the concentration of the HA, also the manufacuture says the purity if 99.1%, find your value of the purity and comment on differece (if any)

i posted one like this b4 but had got some of the data mixed up, anyone help me?
0
13 years ago
#2
(Original post by papz_007)
HA(s)+NaOH(l)= NaA(aq)+H20(l)

7.1g of acid was added to 250cm^3 of water in a 250cm^3 flask of water, then 25cm^3 of it is placed in the buiret via a pipette. On average it takes 23.27cm^3 of the HA to change the colour of 25cm^3 of 0.15M NaOH to change colour.
What is the concentration of the HA, also the manufacuture says the purity if 99.1%, find your value of the purity and comment on differece (if any)

i posted one like this b4 but had got some of the data mixed up, anyone help me?
Moles NaOH: 25/1000 * 0.15 = 0.00375 mol
Moles HA also = 0.00375 mol
Concentration HA: 0.00375 / (23.27/1000) = 0.161151...M

Now you need to tell us the molecular mass of the acid you were using, otherwise you cannot work out the purity.
0
#3
Mr was 169 m8
0
13 years ago
#4
(Original post by papz_007)
Mr was 169 m8
Right so total moles acid from experiment: 0.161151... * 0.25 = 0.040287...mol
Mass this should equate to: 0.040287... * 169 = 6.8086...g
Purity calculated: 6.8086... / 7.1 *100 = 95.895..%
0
#5
can u give me a 100% comfirmation answer is right?

y is so total moles acid from experiment: 0.161151... * 0.25 = 0.040287...mol
??

shouldn it be 3.75x10^3/ 23.73x10^3 =0.16...M then do

n*M=m......so 3.75x10^3*169= 0.634g

then do 7.1-0.634/7.1=ans*100= 91.07%

what have i done wrong (if it is wrong)? i dont understand the way and why u did certain calcs??
0
13 years ago
#6
(Original post by papz_007)
can u give me a 100% comfirmation answer is right?

y is so total moles acid from experiment: 0.161151... * 0.25 = 0.040287...mol
??
This is the concentration of the acid * amount of acid that was formed from 7.1g

shouldn it be 3.75x10^-3/ 23.73x10^-3 =0.16...M then do
This is just a different way to do what I've done to find the concentration of the acid.

n*M=m......so 3.75x10^-3*169= 0.634g

then do 7.1-0.634/7.1=ans*100= 91.07%
This doesn't make sense. You are making a calculation that is independent of the experiment you just did!

Which other steps do you not get?
0
#7
Purity calculated: 6.8086... / 7.1 *100 = 95.895..%.........i was under the impression that u had to take away the orginal from the value i figured out , then divide by the givin value * 100.

hence i did it this way

7.1-0.634/7.1=ans*100= 91.07%
0
13 years ago
#8
you are using 0.634 in your calculation.

This is the number of grams of acid in your 23.27cm3 titre which is not the same as the grams of acid in the 7.1g you weighed out (as this was in 250cm3)

Scale your value up like this:

0.634/23.27 *250

=6.8113 = Golden Maverick's value (with slight rounding error)

now your method will work ...

(7.1 - 6.8113)/7.1 *100 = 4.1....

100 - 4.1 .... = 95.9%
0
#9
oh, i see now, so the way I did it i worked out the percentage differnce between the manufacture and my purity right?

method u 2 have shown is slightly different it shudnt matter which one i use right guys?
0
13 years ago
#10
No ... you're method would be wrong unless you used the correct value (ie 6.8 rather than 0.6)

The reason is ... you are working out the purity based on the number of moles you KNOW are contained in the original 7.1g you weighed out. You KNOW how many moles this contains from your titration calculations. You then calculate what percentage of that 7.1g is the acid (and assume the rest is impurities).

You can't use 0.6g for this calculation (ie the amount of acid contained in the titre ... as this is NOT the amount of acid contained in the original 7.1g of solid.

So basically your method is not wrong - but just incomplete. You needed to scale your 0.6 value up to the 250cm3 that you dissolved the 7.1g in.
0
#11
what would be the concentration in gdm^3 then, i got 28.4gdm^-3? is that correct?

by working working out my concentration and dividing by the manufacture concentration and *100...will that also get me the same answer?
0
13 years ago
#12
28.4 is the concentration of the entire solute in g/dm3 (using 7.1g)

27.23 is the concentration (in g/dm3) of the ACID in solution (using 6.8..... g)

yes you will get the same answer by going:

27.23 / 28.4 * 100

=95.9%
0
#13
thanks alot guys ........ u guys are 100% cool
0
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