Destroyer Gen.Chem #172

[UCSD1984]

I wasn't sure how to approach this problem, and I suppose the explanation makes sense to me, but I don't see why the answer utilizes the moles of HBr rather than the moles of KOH. Here's the question...

If 30ml of .02 M HBr solution is mixed with 40 ml of .03 M KOH, what is the pH of the mixture?

Edit: The only reason I can think of as to why the answer uses moles of HBr instead of KOH is because only the amount of the limiting reagent is utilized by the excess. But what confuses me is why pOH was derived from moles of acid over total volume. Is it because technically since KOH is in excess, we only use the amount of KOH as per what the limiting acid moles was? Sorry, I'm not wording this very well.

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[Jsaxon]

This one got me too the first time I went over it. They are actually dividing the moles of base left over after the neutralizing reaction, over the new total volume of the solution. This will give you the [OH] which is needed to derive the pH. Its weird because the number of moles of base left over after you subtract the acid from it is the same as the number of moles of acid you start with.. Hope that helps

 

[KyoPhan]

Noooo,

They aren't using the mol of the acid! Though I can see why you thought that.

They are using the mole of base left over after the acid/base react with each other. They got the excess base, which is 6x10^-4. They did so by 1.2x10^-3 subtracted by 6x10^-4.

The mole of acid used is also 6 x 10^-4 coincidently!

 

[dmdluffy]

pH will mean concentration of HBr
pOH will mean concentration of KOH

That's how I sorted it in my head.

 

[UCSD1984]

Noooo,

They aren't using the mol of the acid! Though I can see why you thought that.

They are using the mole of base left over after the acid/base react with each other. They got the excess base, which is 6x10^-4. They did so by 1.2x10^-3 subtracted by 6x10^-4.

The mole of acid used is also 6 x 10^-4 coincidently!

1.2x10^-3 - 6x10^-4 = 6x10^-4...I see what you're saying. That's the amount of base left over and that's what must be used to determine the pOH. I was having a hard time seeing that 1.2x10^-3 - 6x10^-4 = 6x10^-4. I see it now. Thanks everyone.

 

[jefff]

1.2x10^-3 - 6x10^-4 = 6x10^-4...I see what you're saying. That's the amount of base left over and that's what must be used to determine the pOH. I was having a hard time seeing that 1.2x10^-3 - 6x10^-4 = 6x10^-4. I see it now. Thanks everyone.

ya im not sure why they even converted it back to 1.2e^-3

when i do the math for moles HBr i did:
2e^-2 M x 3e^-2 L = 6e^-4 moles HBr

and for KOH same method:
3e^-2 M x 4e^-2 L = 12e^-4 moles KOH (normally id move the dec. to make it 1.2e^-3 but here it may just confuse me for the math lol)