When maths meets chemistry

[illuminati]

Q3. Two solutions of NaOH and KOH were mixed together to form 200ml of a combined solution. The concentration of the NaOH is three times as concentrated as the KOH. The molarity of the KOH in mol/L is 5 times its volume in litres. A 20ml aliquot of the solution was taken and diluted to 100ml. From there another aliquot of 20ml was taken and then titrated against 0.154M HCl. The average titre was (976/55)(approx 17.74545) ml. Determine the volume of NaOH solution mixed to form the 200ml solution to the nearest millilitre.

HAVE FUN

EDIT: yeah guys, this isn't on the course - ONLY IF YOU GET BORED

[Nobby]

Only you would set a question like this you evil bastard

[illuminati]

Only you would set a question like this you evil bastard

yeah man
for chemistry!
but cmon, surely someone with methods/chem knowledge can work this baby out?

[Nobby]

Only you would set a question like this you evil bastard

yeah man
for chemistry!
but cmon, surely someone with methods/chem knowledge can work this baby out?

I've only done gallery of graphs...

[illuminati]

only need the quadratic formula here

[b^3]

Ok here goes nothing........ (Also screw sig figs)
v(NaOH)+v(KOH)=200ml
v(NaOH)+v(KOH)=0.200 L ........[1]
c(NaOH)=3c(KOH) .....................[2]
c(KOH)=5v(KOH) .......................[3]
From [3] ....... [3a]

Now the final amount of HCL will be

n(OH^-)final=n(HCl)=0.0027328

HCl_(aq)+OH^-_(aq)--->H₂O_(aq)+Cl^-_(aq)
So.......
n(KOH)+n(NaOH)=0.13664 mol .............[4]
as c=n*v
c(KOH)*v(KOH)+c(NaOH)*v(NaOH)=0.13664
5[v(KOH)]²+3c(KOH)*(0.2-v(KOH))=0.13664 from [3], [2] and [1]
5[v(KOH)]²+3(5v(KOH))*(0.2-v(KOH))=0.13664 from [3]
Expand
5[v(KOH)]²+3v(KOH)-15[v(KOH))]²=0.13664
10[v(KOH)]²-3v(KOH)+0.13644=0


BUT 0<v(KOH)<0.2 L
SOO

v(NaOH)=0.200-0.05589
=0.1441106 L
=144.111 mL
=144 mL



There is a slim chance that I didn't make a mistake in there somewhere but lets hope its right (I doubt it will be though, but oh well)

EDIT: Plugging numbers back in and checking with the rules, it looks ok, so we'll see

EDIT2: Fixed to included the initial aliquot, which illuminati pointed out, thanks illuminati

[ggxoxo]

^ Woah

[illuminati]

Ok here goes nothing........ (Also screw sig figs)
v(NaOH)+v(KOH)=200ml
v(NaOH)+v(KOH)=0.200 L ........[1]
c(NaOH)=3c(KOH) .....................[2]
c(KOH)=5v(KOH) .......................[3]
From [3] ....... [3a]

Now the final amount of HCL will be


HCl_(aq)+OH^-_(aq)--->H₂O_(aq)+Cl^-_(aq)
n(OH^-)=n(HCl)=0.013664 mol
So.......
n(KOH)+n(NaOH)=0.013664 mol .............[4]
as c=n*v
c(KOH)*v(KOH)+c(NaOH)*v(NaOH)=0.013664
5[v(KOH)]²+3c(KOH)*(0.2-v(KOH))=0.013664 from [3], [2] and [1]
5[v(KOH)]²+3(5v(KOH))*(0.2-v(KOH))=0.013664 from [3]
Expand
5[v(KOH)]²+3v(KOH)-15[v(KOH))]²=0.013664
10[v(KOH)]²-3v(KOH)+0.013644=0


BUT 0<v(KOH)<0.2 L
SOO

v(NaOH)=0.200-0.004619
=0.195381 L
=195.538 mL
=196 mL

There is a slim chance that I didn't make a mistake in there somewhere but lets hope its right (I doubt it will be though, but oh well)

EDIT: Plugging numbers back in and checking with the rules, it looks ok, so we'll see

Close - but you missed my initial aliquot

[Nobby]

144mL is the answer. Rather elementary to be honest...

[b^3]

Ok here goes nothing........ (Also screw sig figs)
v(NaOH)+v(KOH)=200ml
v(NaOH)+v(KOH)=0.200 L ........[1]
c(NaOH)=3c(KOH) .....................[2]
c(KOH)=5v(KOH) .......................[3]
From [3] ....... [3a]

Now the final amount of HCL will be


HCl_(aq)+OH^-_(aq)--->H₂O_(aq)+Cl^-_(aq)
n(OH^-)=n(HCl)=0.013664 mol
So.......
n(KOH)+n(NaOH)=0.013664 mol .............[4]
as c=n*v
c(KOH)*v(KOH)+c(NaOH)*v(NaOH)=0.013664
5[v(KOH)]²+3c(KOH)*(0.2-v(KOH))=0.013664 from [3], [2] and [1]
5[v(KOH)]²+3(5v(KOH))*(0.2-v(KOH))=0.013664 from [3]
Expand
5[v(KOH)]²+3v(KOH)-15[v(KOH))]²=0.013664
10[v(KOH)]²-3v(KOH)+0.013644=0


BUT 0<v(KOH)<0.2 L
SOO

v(NaOH)=0.200-0.004619
=0.195381 L
=195.538 mL
=196 mL

There is a slim chance that I didn't make a mistake in there somewhere but lets hope its right (I doubt it will be though, but oh well)

EDIT: Plugging numbers back in and checking with the rules, it looks ok, so we'll see

Close - but you missed my initial aliquot

AH YES I did (dam, I was close )
Should be this then:
Now the final amount of HCL will be

n(OH^-)final=n(HCl)=0.0027328

HCl_(aq)+OH^-_(aq)--->H₂O_(aq)+Cl^-_(aq)
So.......
n(KOH)+n(NaOH)=0.13664 mol .............[4]
as c=n*v
c(KOH)*v(KOH)+c(NaOH)*v(NaOH)=0.13664
5[v(KOH)]²+3c(KOH)*(0.2-v(KOH))=0.13664 from [3], [2] and [1]
5[v(KOH)]²+3(5v(KOH))*(0.2-v(KOH))=0.13664 from [3]
Expand
5[v(KOH)]²+3v(KOH)-15[v(KOH))]²=0.13664
10[v(KOH)]²-3v(KOH)+0.13644=0


BUT 0<v(KOH)<0.2 L
SOO

v(NaOH)=0.200-0.05589
=0.1441106 L
=144.111 mL
=144 mL



I'll fix up the post above in an edit and acknowledge the error

[illuminati]

And this is what happens
when maths meets chemistry
HAHA i should post a question like this every week just for personal lolz
but congrats b^3 =D