molarity Q from KBB

[datdat]

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Q: molarity of K2SO4 in aq. solution created by adding 112g of KOH(solid) to 500mL of 0.5M H2SO4(aq)

the answer is 0.5M but I don't know why....

-mole of KOH in 500mL = 112g x 1/56g x 1/.5 = 4M and?


could anyone can help me out?

 

[goffdent]

what's the question? "what is the molarity of K2SO4 in aq. soln.....? Only thing i come up with is that the molarity as written is 0.5M. It's molarity isn't going to change, it's still .5 moles/liter even if you add base to it. But then again i don't understand what the question is asking.

 

[toothdoc0121]

im not sure if im right on how to do this but this is what i did

112/56= 2 mol of KOH

.5L * .5M= .25mol H2SO4

2 + .25= 2.25

2.25mol/.5L= 4.5M

 

[watt]

i think the question is asking about the molarity of the products. so setup your reaction:

2KOH + H2SO4 --> K2SO4 + 2H2O

since u have
112g/56g/mol = 2mol of KOH
and
0.5M x 500mL = 0.25mol of H2SO4
H2SO4 is your limiting reagent.

moles of H2SO4 to K2SO4 = 1:1

therefore moles of H2SO4 = K2SO4
0.25mol H2SO4 x (1mol K2SO4)/(1mol H2SO4) = 0.25mol K2SO4

then convert to molarity
0.25mol/500mL = 0.5M

 

[doc toothache]

We are making this problem more complicated than it is. The limiting factor in this reaction is the concentration of H2SO4 and from the stoichiometry we know the product of K2SO4 will have the same concentration as the acid.

2KOH + H2SO4 --> K2SO4 + 2H2O