Gen chem question

[predentgal]

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Hey guys..I have a quick gen chem question. I posted the link to the question below as well as the explanation they offer. What I don't understand is why we don't use 500ml for the concentration of h2so4 and instead we use 1 liter to obtain molarity of K2SO4?😕

http://img64.imageshack.us/img64/7006/genchem.png

 

[predentgal]

This is how I went about to solve the problem. So there is .5M H2S04 IN 500ML(.5L) then moles of H2S04 in that 500 mls. is .25. Since H2S04 and k2S04 are present in a 1 to 1 ratio then k2s04 should be .25 too.

 

[UndergradGuy7]

We have the amount of moles by 0.5 M * 0.5 L = 0.25 moles of SO4.

We also have the amount of moles of KOH on 112 grams... 112/56 = 2 moles KOH = moles of K

So it is a limiting reactant. If we have 0.25 moles of SO4 and K2SO4 requires 2 moles of K for everyone 1 mol of SO4 we will need 0.50 moles of K since (0.25 * 2 = 0.50). We have 2 moles of K, so we have an excess of K, which is not important.

The question asks for the molarity of K2SO4. The molarity of K2SO4 is equal to the moles of SO4/Volume (0.5L) or (1/2 moles of K needed to react)/Volume (0.5L). So the molarity is 0.25 moles / 0.5 L = 0.5 Moles/Liter or [(1/2)(0.50)]/0.5L.

I assume the green highlighted box in the drawing is the answer, which matches my answer.

 

[ddslee]

This is how I went about to solve the problem. So there is .5M H2S04 IN 500ML(.5L) then moles of H2S04 in that 500 mls. is .25. Since H2S04 and k2S04 are present in a 1 to 1 ratio then k2s04 should be .25 too.

You are absolutely right....it will be .25 for 500ml though...

molarity is specifically mol/L.. so you would have to times by 2..