Chemistry Question

2007-05-17 06:11:02

#1

Coolbeano: Level 13.5 BF2S Ninja Penguin Sensei; +378|7227

Since everyone here seems to be an expert on everything...

help me out with this:

How much (in grams) sodium acetate has to dissolved into a .5 dm³ 0.2 molar solution of acetic acid to create a buffer solution with a pH of 5?

Assume K_a of acetic acid to be 1.8*10^-5

Last edited by Coolbeano (2007-05-17 06:11:20)

2007-05-17 06:11:54

#2

Adams_BJ: Russian warship, go fuck yourself; +2,058|7086|Little Bentcock

7

2007-05-17 06:13:54

#3

Retalliation[1337]: Robin Hood ---> "u got arrownd"; +51|6938|Belgium

lol that was fast ...

2007-05-17 07:01:33

#4

Flecco: iPod is broken.; +1,048|7128|NT, like Mick Dundee

He's bullshitting tbh.

Beano, I have some mates who are doing that sort of chem. Sucks to be you.

Ask Surgeon, he might be able to help, was yammering on about analytical chem on IRC earlier.

Whoa... Can't believe these forums are still kicking.

2007-05-17 07:06:30

#5

Shem: sɥǝɯ; +152|6991|London (At Heart)

Edit.

Actually, cba, too tired, bed time methinks

Last edited by Shem (2007-05-17 07:06:56)

2007-05-17 07:08:21

#6

stef10: Member; +173|6946|Denmark

Which kind of level is this? Senior year of high school or what?

2007-05-17 07:08:59

#7

Ender2309: has joined the GOP; +470|7034|USA

fuuuck beano. i can do that for you later today, but i have to go to school.
and flecco, buffers are easy. i imagine his biggest problem is the db^3. thats uncommon (HINT: but converts straight to DCs).

nac2h3o2
h2c2h3o2
^correct?

edit: 1.0x10^-5 is your H+ value. we're working with a 1-1 product/reactant ratio.
beano, i just took my AP test tues, so i'm still fresh.

Last edited by Ender2309 (2007-05-17 07:11:57)

2007-05-17 07:29:57

#8

Ottomania: Troll has returned.; +62|6985|Istanbul-Turkey

Ender2309 wrote:
fuuuck beano. i can do that for you later today, but i have to go to school.
and flecco, buffers are easy. i imagine his biggest problem is the db^3. thats uncommon (HINT: but converts straight to DCs).

nac2h3o2
h2c2h3o2
^correct?

edit: 1.0x10^-5 is your H+ value. we're working with a 1-1 product/reactant ratio.
beano, i just took my AP test tues, so i'm still fresh.

so if we continue that, ch3coona would be 0,1 moles. and that makes 8,2 grams?(if the first post says 0.5 dm^3)

also it seem like I forget buffer solutions, and my chemistry book is at school.

Last edited by Ottomania (2007-05-17 07:31:29)

2007-05-17 07:59:11

#9

CommieChipmunk: Member; +488|7033|Portland, OR, USA

Well, i have like 10 minutes before I have to be at school....

But I could help you when I get back.. Though my brain is a little... dead after that AP chem test.

2007-05-17 12:38:04

#10

Ottomania: Troll has returned.; +62|6985|Istanbul-Turkey

CH3COOH -> CH3COO + H
START 0,2M 0 0

CHANGES 0,2-10^-5 10^-5 10^-5

CH3COONa-> CH3COO + Na
X X X

Ka = 1,8* 10^-5= [CH3COO]*[H]
------------------ = (X+10^-5)(10^-5)
CH3COOH (0,2-10^-5)

(10^-5 is a little number so it wouldnt be cared) (I mean X + 10^-5 = X))

= X*10^-5
-----------------= 1,8*10^-5 X=0,36 MOLAR
0,2

so: I think we need the capacity of ch3coona to find its mass.

2007-05-17 23:35:34

#11

Coolbeano: Level 13.5 BF2S Ninja Penguin Sensei; +378|7227

Ottomania wrote:
CH3COOH -> CH3COO + H
START 0,2M 0 0

CHANGES 0,2-10^-5 10^-5 10^-5

CH3COONa-> CH3COO + Na
X X X

Ka = 1,8* 10^-5= [CH3COO]*[H]
------------------ = (X+10^-5)(10^-5)
CH3COOH (0,2-10^-5)

(10^-5 is a little number so it wouldnt be cared) (I mean X + 10^-5 = X))

= X*10^-5
-----------------= 1,8*10^-5 X=0,36 MOLAR
0,2

so: I think we need the capacity of ch3coona to find its mass.

I think you're missing something - is the acetate in the K_a equation solely from the sodium acetate added? I mean what about the acetate already in the solution (I know it isn't much, but is it small enough to be negligible?) from the acid dissolved? So it looks like you're just investigating how much acetate would be if the hydronium count increased....

Oh and not concentration, but just the salt - I can do the stoichiometry, that stuff is easy.

2007-05-18 00:27:33

#12

naightknifar: Served and Out; +642|7025|Southampton, UK

BOOOOOOOOOOOOOOOOOOOOOOOOOOM!

S0ri my h3ad jst 3xplodeded ova teh sm4rtn3ss of J00z!

2007-05-18 07:27:59

#13

Ottomania: Troll has returned.; +62|6985|Istanbul-Turkey

Coolbeano wrote:
Ottomania wrote:
CH3COOH -> CH3COO + H
START 0,2M 0 0

CHANGES 0,2-10^-5 10^-5 10^-5

CH3COONa-> CH3COO + Na
X X X

Ka = 1,8* 10^-5= [CH3COO]*[H]
------------------ = (X+10^-5)(10^-5)
CH3COOH (0,2-10^-5)

(10^-5 is a little number so it wouldnt be cared) (I mean X + 10^-5 = X))

= X*10^-5
-----------------= 1,8*10^-5 X=0,36 MOLAR
0,2

so: I think we need the capacity of ch3coona to find its mass.
I think you're missing something - is the acetate in the K_a equation solely from the sodium acetate added? I mean what about the acetate already in the solution (I know it isn't much, but is it small enough to be negligible?) from the acid dissolved? So it looks like you're just investigating how much acetate would be if the hydronium count increased....

Oh and not concentration, but just the salt - I can do the stoichiometry, that stuff is easy.

well, we learnt like that to not care that number, 10^-5. if you care that 0,2M-10^-5 would be= 0,19999. after that it would be be harder to solve that question.

also I find the concentration of salt solution. if we know its capacity, we can find its mass too. if its capacity was 0.5 dm^3 as you said in the first post, then it would be 82*18/100 grams.

I wonder if you add a salt solution or dry salt?

Last edited by Ottomania (2007-05-18 07:30:54)

2007-05-18 10:00:34

#14

Coolbeano: Level 13.5 BF2S Ninja Penguin Sensei; +378|7227

Yeah dry salt - a numerical to gram calculation is easy. And I understand the rounding part, the Ks have default uncertainty of like 5% so I learnt to ignore it as well at this small number.

But what I did what you did up there and I still feel there's something missing. I never really understood how different buffers were from simple H₂O equations... plus my answer doesn't match the markscheme.

2007-05-18 13:47:56

#15

Ottomania: Troll has returned.; +62|6985|Istanbul-Turkey

anyways, do you have the right answer? or solution?

2007-05-18 13:52:55

#16

Mekstizzle: WALKER; +3,611|7085|London, England

joker3327 wrote:
Enemy boat spotted!!

Hahahaha.....i don't know whether to feel happy remembering that. Or feel sad knowing that i can't play BF2 until i get a new computer and thus won't hear it for a long time.

Whatever the case, the boat was spotted.

2007-05-18 13:53:45

#17

stkhoplite: Banned; +564|6943|Sheffield-England

I can write it down and ask my science teacher if you like...

BF2S Forums Chemistry Question

Ender2309 wrote:

Ottomania wrote:

Coolbeano wrote:

Ottomania wrote:

joker3327 wrote:

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