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Topic: How many moles of HF (Ka = 6.8x10^-4) must be in 0.25 L to form a 2.6 pH soln?  (Read 5608 times)

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Offline sjbyrne

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How many moles of HF (Ka = 6.8x10^-4) must be in 0.25 L to form a 2.6 pH soln?
« on: April 08, 2010, 08:00:58 AM »
I am using the formula Ka = x^2/(y/.250L) and I found x by 10^-2.6=[H+].  I got x to be 2.5x10^-3 and put it in my Ka equation and got y = 2.3*10^-3 but for some reason this answer is incorrect.  Can someone tell me what I did wrong an what I need to do to fix it?
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Offline Borek

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Re: How many moles of HF (Ka = 6.8x10^-4) must be in 0.25 L to form a 2.6 pH soln?
« Reply #1 on: April 08, 2010, 10:14:56 AM »
What is y?
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Offline sjbyrne

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Re: How many moles of HF (Ka = 6.8x10^-4) must be in 0.25 L to form a 2.6 pH soln?
« Reply #2 on: April 08, 2010, 01:44:14 PM »
y is the number of moles of HF.  It is what I am looking for.  The equation for Ka is expressed in terms of concentration of products over concentration of reactants and concentration is molarity.  M=moles/liter and so y/.250 liters should help me find the number of moles of HF....I think
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Offline Borek

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Re: How many moles of HF (Ka = 6.8x10^-4) must be in 0.25 L to form a 2.6 pH soln?
« Reply #3 on: April 08, 2010, 03:04:26 PM »
I have not asked what you THINK y is, I asked what it really is...

Think: you have x2 in the nominator. What two ions does it stand for? How one of them is related to y?
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Offline sjbyrne

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Re: How many moles of HF (Ka = 6.8x10^-4) must be in 0.25 L to form a 2.6 pH soln?
« Reply #4 on: April 08, 2010, 04:33:53 PM »
Hahaha, well if I didn't think, then I wouldn't be able to get anything done so my thinking is actually a good thing even if you didn't specifically ask for it.  But let's see, x stands for the ions H3O+ and F-.  F- is the conjugate base of HF, but I am not sure how knowing the conjugate base would help.  I could find Kb through the equation Kw/Ka, but I still don't have x.  I feel like I should be able to use the equation Ka = x^2/(y/.250L) and then find x through 10^-2.6 and then solve for y.  I don't understand why this shouldn't work.  It has all the right variables and I don't see anything missing.  I understand that you are trying to drive me towards something to do with the conjugate base, but I am just not sure what exactly that is. 
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Offline Borek

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Re: How many moles of HF (Ka = 6.8x10^-4) must be in 0.25 L to form a 2.6 pH soln?
« Reply #5 on: April 08, 2010, 05:13:41 PM »
y is not ANALYTICAL concentration of hydrofluoric acid. It is EQUILIBRIUM concentration.
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Offline sjbyrne

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Re: How many moles of HF (Ka = 6.8x10^-4) must be in 0.25 L to form a 2.6 pH soln?
« Reply #6 on: April 08, 2010, 08:10:37 PM »
Well, I went through and did my ice tables and found I needed to add x to the equilibrium concentration of HF (y) and got 0.0048, but this is wrong.  My professor is having a help session tomorrow so even though I already used all my submissions for this question (using MasteringChemistry), I really need to figure this out for future reference and will just ask him tomorrow.  Thanks for trying to help though!
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