Gen Chem question

[Twitch 22]

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What is the molarity of K2SO4 in teh aqueous solution created by adding 112 grams of solid KOH to 500 mL of 0.50 M H2SO4?

 

[Twitch 22]

Is it similar to how one would work 117 in the destroyer?

 

[PooyaH]

This is asking for molarity and Q #117 is asking for the pH of the solution!

 

[klpennin]

Im not gonna solve it but basically:
Write equation out and Balance - ?KOH + ?H2SO4 ---> ?K2SO4 + ?H2O
Get moles of KOH and H2SO4 and find limiting reagent
Use limiting reagent to calculate moles of K2SO4 produced & L of water (density of water = 1)
Then put moles of K2SO4/L of water to get molarity

 

[Tina324]

is the answer .5M? i did that from what the above poster said.

 

[Tina324]

oh wait a minute i didn't do it correctly. i didn't put the moles of K2So4 produced (.25mol) over the L of water produce, I put it over the ml of H2So4 that we put in haha..is the answer .25/36M?

 

[td4azklz]

is the answer .5M? i did that from what the above poster said.

I got 1 M 😕

this is how i did it...

112 gm KOH (1 mol KOH/ 56gm KOH) * (0.5 mol H2SO4/2 mol KOH) * (1000mL / 1L) *1/500

What am I missing?

 

[Twitch 22]

The answer that was give was 0.5 Moles...

 

[futuredent22]

The answer that was give was 0.5 Moles...

You mean .5 moles/liter

Here's how you do it:
1. balance equation
2. find limiting reactant
3. convert moles of limiting reactant to moles of K2SO4
4. divide the moles of K2SO4 by .5 L

The fourth step might be messing some of you up. the volume is still 500ml
Edit: now that i think about it the 4th step is confusing me. I feel like the volume would have changed, but using the volume of acidic solution gives the correct answer

 

[Twitch 22]

Thanks I get it now I was using 0.5M and not converting it to Moles!!!

 

[Tina324]

You mean .5 moles/liter
Edit: now that i think about it the 4th step is confusing me. I feel like the volume would have changed, but using the volume of acidic solution gives the correct answer

yea .5M is what i originally got, but it WOULD make sense that the volume changes..maybe its negligible ughhh i hate gchemmm lol

where did u get this q from??

 

[yakuza]

I think this is more of a conceptual problem than math..? Someone correct me if I'm wrong but this is how you do it.

For aqueous solutions, Molarity = Molality so use moles/L

Balance the equation, find the limiting reagent, and you find that..
1mole H2SO4 --yields--> 1mole K2SO4


SO, the molarity(or molality) will be the same as H2SO4 which is 0.5M


Think of the problem as asking: How many moles of K2SO4 can be made from 1mole H2SO4? ..which is 1mole

2KOH + H2SO4 ---> K2SO4 + H2O

 

[doc toothache]

What is the molarity of K2SO4 in teh aqueous solution created by adding 112 grams of solid KOH to 500 mL of 0.50 M

H2SO4?

The answer that was give was 0.5 Moles...

This appears to be a simple stoichiometric problem which has been complicated by the interchangeable use of molar and moles.

2KOH + H2SO4 = K2SO4 + 2H2O

Since there are only 0.25 moles of H2SO4 the max yield of K2SO4 is 0.25 moles. The molarity of the K2SO4 will be 0.5 since we have 0.25 moles of the compound in roughly 500 ml of solution.