Board of studies, fail ? (1 Viewer)

Zydra · Oct 29, 2011

I have tried 1000 times to understand the answer they give for this question and i just cant get it. I have tried going backwards from their answer but nothing works !!1?!

Please Help

http://www4.boardofstudies.nsw.edu....wer=D&courseID=15050&testQuestionID=274256109

Bobbo1 · Oct 29, 2011

gahh, so much work for a 1 marker but yes it works out fine - i got like 41.56g/L

make sure to apply the mole ratio (1/3) and it should work out

Zydra · Oct 29, 2011

Ahh i see, i had the x3 on the wrong side. I was thinking that cause it was triprotic it will need 3 x as much NaOH to reach the end point but it should have been it is 'worth' 3 x as much.

AzlanlovesSNSD · Oct 29, 2011

hey guys, i keep getting 125 (c) but dont understand why i have to divide by three to get b) i mean i know its got to do with the triprotic thing but can some please explain in a bit more detail ?

thanks!

roryclifford · Oct 29, 2011

AzlanlovesSNSD said:
hey guys, i keep getting 125 (c) but dont understand why i have to divide by three to get b) i mean i know its got to do with the triprotic thing but can some please explain in a bit more detail ?

thanks!

I'm not a chemistry master, but from what I did to get the right answer;

I wrote out the equation, so since Citric is triprotic it will donate Hydrogen 3 times to NaOH, so

$3NaOH + H_{3}C_{6}H_{5}O_{7} --------> 3H_{2}O + C_{6}H_{5}O_{7}Na$

Now, calculate moles of NaOH, so from the data, n =C x V = 0.0295 x 0.550 = 0.016225 (save in calculator to keep exact)

Then, from equation, mole ratio of NaOH to Citric acid is 1 : 3, therefore divide 0.016225 by 3.

Then to get the mass / litre answer, times this answer by the molar mass of Citric acid which it gives, and then divide this by the volume of Citric;

0.016225 / 3 = Ans x M = Ans / 0.025 = 41... (sorry dont have a calculator on me, this is just the stuff I wrote down on my paper lol)

And you should arrive at the answer. Unless I made a mistake, sorry if i did

b3kh1t · Oct 29, 2011

The equation for the reaction of citric acid with sodium hydoxide is: $C_{6}H_{8}O_{7}_{(aq)}+3NaOH_{(aq)}\rightarrow 3H_{2}O_{(l)}+Na_{3}C_{6}H_{5}O_{7}_{(aq)}$

The molar ratio of citric acid : NaOH ---> water : salt, can therefore be seen as: 1 : 3 ---> 3 : 1

now number of moles of NaOH = CxV = 0.0295x0.55 = 0.016225 moles
from the ratios it can be determined that the number of moles of NaOH is 3 times the number of moles of citric acid used.
therefore number of moles of citric acid = 0.016225/3 = 5.4083x10^-3 moles

therefore concentration of citric acid = n/V = 5.4083x10^-3/0.025 = 0.2163 mol/L
to change to grams per litre, multiply by the molar mass. 0.2163x192.12 = 41.562 g/L

therefore the answer must be (B).

AzlanlovesSNSD said:
hey guys, i keep getting 125 (c) but dont understand why i have to divide by three to get b) i mean i know its got to do with the triprotic thing but can some please explain in a bit more detail ?

thanks!

If you need more information just say.

b3kh1t · Oct 29, 2011

roryclifford said:
I'm not a chemistry master, but from what I did to get the right answer;

I wrote out the equation, so since Citric is triprotic it will donate Hydrogen 3 times to NaOH, so

$3NaOH + H_{3}C_{6}H_{5}O_{7} --------> 3H_{2}O + C_{6}H_{5}O_{7}Na$

Now, calculate moles of NaOH, so from the data, n =C x V = 0.0295 x 0.550 = 0.016225 (save in calculator to keep exact)

Then, from equation, mole ratio of NaOH to Citric acid is 1 : 3, therefore divide 0.016225 by 3.

Then to get the mass / litre answer, times this answer by the molar mass of Citric acid which it gives, and then divide this by the volume of Citric;

0.016225 / 3 = Ans x M = Ans / 0.025 = 41... (sorry dont have a calculator on me, this is just the stuff I wrote down on my paper lol)

And you should arrive at the answer. Unless I made a mistake, sorry if i did

Nah that should be fine but just in your equation, the salt on the product side, there is meant to be 3 sodium attached to the negative ion. Be careful because in the exam the marker may be picky and had a bad day lol

AzlanlovesSNSD · Oct 29, 2011

roryclifford said:
I'm not a chemistry master, but from what I did to get the right answer;

I wrote out the equation, so since Citric is triprotic it will donate Hydrogen 3 times to NaOH, so

$3NaOH + H_{3}C_{6}H_{5}O_{7} --------> 3H_{2}O + C_{6}H_{5}O_{7}Na$

Now, calculate moles of NaOH, so from the data, n =C x V = 0.0295 x 0.550 = 0.016225 (save in calculator to keep exact)

Then, from equation, mole ratio of NaOH to Citric acid is 1 : 3, therefore divide 0.016225 by 3.

Then to get the mass / litre answer, times this answer by the molar mass of Citric acid which it gives, and then divide this by the volume of Citric;

0.016225 / 3 = Ans x M = Ans / 0.025 = 41... (sorry dont have a calculator on me, this is just the stuff I wrote down on my paper lol)

And you should arrive at the answer. Unless I made a mistake, sorry if i did

THANKS SO MUCH I GET IT NOW! i had my chemical equation wrong!

AzlanlovesSNSD · Oct 29, 2011

b3kh1t said:
The equation for the reaction of citric acid with sodium hydoxide is: $C_{6}H_{8}O_{7}_{(aq)}+3NaOH_{(aq)}\rightarrow 3H_{2}O_{(l)}+Na_{3}C_{6}H_{5}O_{7}_{(aq)}$

The molar ratio of citric acid : NaOH ---> water : salt, can therefore be seen as: 1 : 3 ---> 3 : 1

now number of moles of NaOH = CxV = 0.0295x0.55 = 0.016225 moles
from the ratios it can be determined that the number of moles of NaOH is 3 times the number of moles of citric acid used.
therefore number of moles of citric acid = 0.016225/3 = 5.4083x10^-3 moles

therefore concentration of citric acid = n/V = 5.4083x10^-3/0.025 = 0.2163 mol/L
to change to grams per litre, multiply by the molar mass. 0.2163x192.12 = 41.562 g/L

therefore the answer must be (B).

If you need more information just say.

No, i completley get it now! i had my chemical equation wrong. THANKS SO MUCH.

i would "rep" you guys but i dont know how to, if u want me 2 then tell me how and i will

b3kh1t · Oct 29, 2011

AzlanlovesSNSD said:
No, i completley get it now! i had my chemical equation wrong. THANKS SO MUCH.

i would "rep" you guys but i dont know how to, if u want me 2 then tell me how and i will

That's good n lol it's aiight

roryclifford · Oct 29, 2011

b3kh1t said:
Nah that should be fine but just in your equation, the salt on the product side, there is meant to be 3 sodium attached to the negative ion. Be careful because in the exam the marker may be picky and had a bad day lol

Oh i had it haha, before I tried latex for the first time, must have accidentally deleted it.

Cheers though

b3kh1t · Oct 29, 2011

roryclifford said:
Oh i had it haha, before I tried latex for the first time, must have accidentally deleted it.

Cheers though

thats's aiight

best of luck with the rest of your exams

roryclifford · Oct 29, 2011

b3kh1t said:
thats's aiight best of luck with the rest of your exams

You to

nephh22 · Oct 29, 2011

n (NaOH) = 0.550 x 0.02950 = 0.016225 mol
c (citric acid) = 0.016225/0.0250 = 0.649 mol/L (however all the answers are in g/L)
So, c (citric acid) = 0.649 x 192.12 = 124.68588 g/L
since citric acid is a triprotic acid, c (citric acid) = 124.68588/3 = 41.6 g/L

Aclipse · Oct 30, 2011

Are we supposed to know the chemical equations??

b3kh1t · Oct 30, 2011

Aclipse said:
Are we supposed to know the chemical equations??

You need to know how to write them and how some reactions occur

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