Argh! Chemistry Homework help.

Zhavey

Hello, I am in 5th year doing chemistry and we are doing Volumetric Anylasis: Acid-Bases at the moment I have a most likely nooby question :P.

I have a fairly ok grasp with the content but I'm completely stumped with one question. (It's Chemistry Live! Page 173, 13.12)

"In carrying out an experiment, 2.5g of an impure sample of NaOH, was dissolved in water and made up to 250 cm^3 in a volumertric flask. 25 cm^3 portions of the base were titrated aganist 0.15 M H2SO4 solution. The titration figures obtained were 17.1cm^3, 16.7cm^3 and 16.8cm^3. Calc the % purity of the sample of NaOH/Caustic soda."

I know to get an average of the titre results and use that as one of your v's in the equtation (or maybe im wrong..). But otherwise I'm stumped as to what to do. The answer at the back of the book says 80.4% but I can't get anywhere near it.

Any help is apericated!! (Hoping these types of thread are allowed!)

EuropeanSon

2(NaOH)+H2SO4=Na2SO4+2(H2O)
x(25)=0.15(16.75)
x=0.1005
NaOH=40
40x0.1005=4.02
4.02/2=2.01
2.01/2.5=0.804
0.804x100=80.4%

rainbowtrout

Solution below:

Chemistry Ftw

Zhavey wrote: »

Hello, I am in 5th year doing chemistry and we are doing Volumetric Anylasis: Acid-Bases at the moment I have a most likely nooby question :P.

I have a fairly ok grasp with the content but I'm completely stumped with one question. (It's Chemistry Live! Page 173, 13.12)



I know to get an average of the titre results and use that as one of your v's in the equtation (or maybe im wrong..). But otherwise I'm stumped as to what to do. The answer at the back of the book says 80.4% but I can't get anywhere near it.

Any help is apericated!! (Hoping these types of thread are allowed!)

Hey there.
These problems can be tricky at first but with practice they become alot easier.
First of all you need to find the amount of moles of NaOH in 250cm^3. Simply use the formula that you should know by now to work this out: (v1xm1 over n1 = v2xm2 over n2).
v1=25
m1=x (what we are looking for)
n1=2

v2=16.75
m2=0.15
n2=1

Simply substitute these values into the formula and you should get the anwser: x= 0.201 moles/litre NaOH

Next calculate the Mr of NaOH : 23+16+1= 40g
Multiply 40 x 0.20244 and you get 8.04 per litre
We need to get the amount in 250cm^3 so we divide 8.04 by 4 and you get 2.01 grams per 250cm^3

The formula to calculate % purity is mass of pure NaOH divided by mass of impure sample multiplied by 100 over 1.

Your anwser should be 80.4%

Btw you should know that the balanced equation for the reaction between NaOH and H2SO4 is H2SO4 + 2NaOH --> Na2SO4 + 2H2O

rainbowtrout

Chemistry Ftw wrote: »

Hey there.
These problems can be tricky at first but with practice they become alot easier.
First of all you need to find the amount of moles of NaOH in 250cm^3. Simply use the formula that you should know by now to work this out:(v1xm1 over n1 = v2xm2 over n2).
v1=25
m1=x (what we are looking for)
n1=2

v2=16.87
m2=0.15
n2=1

Simply substitute these values into the formula and you should get the anwser: x= 0.20244 moles/litre NaOH

Next calculate the Mr of NaOH : 23+16+1= 40g
Multiply 40 x 0.20244 and you get 8.0976 per litre
We need to get the amount in 250cm^3 so we divide 8.0976 by 4 and you get 2.0244 grams per 250cm^3

The formula to calculate % purity is mass of pure NaOH divided by mass of impure sample multiplied by 100 over 1.

Your anwser should be 80.976%

Btw you should know that the balanced equation for the reaction between NaOH and H2SO4 is H2SO4 + 2NaOH --> Na2SO4 + 2H2O

That's incorrect. When getting the average titre, you take the average of two titres that agree to within 0.1cm3 of each other, so it's the average of 16.7 and 16.8 giving 16.75

Your titre knocks out all your figures for the rest of the calculation

rainbowtrout

EuropeanSon wrote: »

2(NaOH)+H2SO4=Na2SO4+2(H2O)
x(25)=0.15(16.75)
x=0.1005
NaOH=40
40x0.1005=4.02
4.02/2=2.01
2.01/2.5=0.804
0.804x100=80.4%

The solution at the end is correct, but your molarity is incorrect. You just got rid of the No. of moles of NaOH for no reason, which is 2. This should have been part of the calculation.

You then went on to divide by two which gave you the right answer in the end but did not give you the correct molarity at the start.

Chemistry Ftw

rainbowtrout wrote: »

That's incorrect. When getting the average titre, you take the average of two titres that agree to within 0.1cm3 of each other, so it's the average of 16.7 and 16.8 giving 16.75

Your titre knocks out all your figures for the rest of the calculation

Well spotted, I'll fix that now.

rainbowtrout

Chemistry Ftw wrote: »

Well spotted, I'll fix that now.

I've posted the correct solution above as an attachment - I'm a chemistry teacher by the way

Chemistry Ftw

rainbowtrout wrote: »

I've posted the correct solution above as an attachment - I'm a chemistry teacher by the way

Cheers I'm studying chemistry in sixth year myself.

Zhavey

Wow thanks for all the replies!!

I didn't realise there was *2*NaOH, where did the two come from? Again, most likely a horribly easy question but I'm not sure..

EDIT: I read the replies more carefully and I understand now with the product it gives, which I wasn't aware of. Thank you all so much! If only my own chemistry teacher was as helpful as you all!

rainbowtrout

Zhavey wrote: »

Wow thanks for all the replies!!

I didn't realise there was *2*NaOH, where did the two come from? Again, most likely a horribly easy question but I'm not sure..

EDIT: I read the replies more carefully and I understand now with the product it gives, which I wasn't aware of. Thank you all so much! If only my own chemistry teacher was as helpful as you all!

The 2 in front of the NaOH is needed in the orginal equation to balance it. If you look at the right hand side of the equation you'll see that Na2SO4 contains two sodiums, so two are also needed on the left etc. Always check that the equation is balanced before you go ahead with the calculation. Normally you are given the equation but if you are not, you'll need to know it and be able to balance it.

Zhavey

Yeah not having the equation threw me off. But I understand it a bit better now.

EuropeanSon

rainbowtrout wrote: »

The solution at the end is correct, but your molarity is incorrect. You just got rid of the No. of moles of NaOH for no reason, which is 2. This should have been part of the calculation.

You then went on to divide by two which gave you the right answer in the end but did not give you the correct molarity at the start.

Didn't bother with the formula, I just do that crap at the end. It's ages since I've done LC Chemistry. x wasn't necessarily the molarity.

rainbowtrout

EuropeanSon wrote: »

Didn't bother with the formula, I just do that crap at the end. It's ages since I've done LC Chemistry. x wasn't necessarily the molarity.

Well it's better for the OP to learn the correct way to do it seeing as he will be doing the LC next year and marks are given for correct method/docked for incorrect in the exam.

EuropeanSon

rainbowtrout wrote: »

Well it's better for the OP to learn the correct way to do it seeing as he will be doing the LC next year and marks are given for correct method/docked for incorrect in the exam.

I'm sure he can spare 3 marks, if that. :P
I'm just kidding really, you were right to point that out.

RMD

EuropeanSon wrote: »

I'm sure he can spare 3 marks, if that. :P
I'm just kidding really, you were right to point that out.

I'm not sure what grade the OP is looking for, but if you put it this way, you can only "spare" 5 marks per question if you want an A1. I'd much rather pick-up 3 easy marks than piss it away for no reason.