Gchem people

[yakuza]

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This problem isn't kicking in, can someone explain please?

Whats the molarity of K2SO4 in an aq solution created by adding 112g of solid KOH to 500ml of H2SO4 aq?

answer: 0.5 M

 

[JBarr29]

ok so i think i know how to do this problem, however i get a molarity of one when i do it.
to start you off...
2KOH + H2SO4 ---> K2SO4 + 2H2O
112g of KOH = 1 Mole
this will yield .5 moles of K2SO4(mole ratio)
.5/.5L = 1Molar ??? not .5 so im not too sure.

 

[Danny289]

This problem isn't kicking in, can someone explain please?

Whats the molarity of K2SO4 in an aq solution created by adding 112g of solid KOH to 500ml of H2SO4 aq?

answer: 0.5 M

are you sure in this problem there is nothing about H2SO4 molarity?

 

[yakuza]

are you sure in this problem there is nothing about H2SO4 molarity?

I really have no idea..

The way I work it is all wrong

 

[keliao]

ok so i think i know how to do this problem, however i get a molarity of one when i do it.
to start you off...
2KOH + H2SO4 ---> K2SO4 + 2H2O
112g of KOH = 1 Mole
this will yield .5 moles of K2SO4(mole ratio)
.5/.5L = 1Molar ??? not .5 so im not too sure.

112g yield 2 moles???

 

[keliao]

so what's the correct way to solve this??
Molarity of K2SO4 . M=moles/Liter of solution
???????????

 

[unitix]

I actually brought out a piece of paper for this, I got 2M for the Potassium Sulfate. However Im dead tired probably screwed up, I'll do it again when I wake up from my nap. This is kinda bad this should be easy for us!!

 

[chessxwizard]

I think you need the molarity of the acid to see how much of the salt is created.

 

[klutzy1987]

Based on the information given i would say the answer is 1, however it is impossible to solve with the given information because you do not know the final volume. Obviously 112g of KOH will increase the volume but by how much??? Crappy question, where is it from,

Lemme guess kaplans or barrons.

 

[yakuza]

Based on the information given i would say the answer is 1, however it is impossible to solve with the given information because you do not know the final volume. Obviously 112g of KOH will increase the volume but by how much??? Crappy question, where is it from,

Lemme guess kaplans or barrons.

Kaplan's

A couple of these questions are driving me insane. They make me feel like I don't know sht in gchem!

 

[lln002]

i think the answer is 8 M.

(2moles of KOH)/(.5L of H2SO4) = 4 M

But since we want to know how the concentration of K2SO4 is, we would have to multiply by 2 Normality because for every KOH that reacts with the acid we get K'2'SO4 Molarity.

that's my take on it. your 2cents are welcome! =)

 

[keliao]

i think the answer is 8 M.

(2moles of KOH)/(.5L of H2SO4) = 4 M

But since we want to know how the concentration of K2SO4 is, we would have to multiply by 2 Normality because for every KOH that reacts with the acid we get K'2'SO4 Molarity.

that's my take on it. your 2cents are welcome! =)

i got 4M from 2moles/.5L. i thought Normality was based on # of hydrogen on the molecule.

 

[Doa110]

i got 4M from 2moles/.5L. i thought Normality was based on # of hydrogen on the molecule.

Answer: 0.5
Well....very tricky question!

This is a titration. Right? So think Normality!
we have H2SO4+2KOH. ..>K2SO4+2H2O

If we use the Volume of H2SO4 and Normality of H2SO4. In that case the answer is 4. However the question is asking about the salt molarity!
Answer: 0.5, because each H2SO4 provides 2 moles of H+.....> 2 normal,
but each H2SO4 dissociates to only 1 sulfateprecipitant (K2SO4), which would be 1 normal....> M*I=N...>M*2=1..>M=o.5