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Topic: pH/Ka  (Read 32863 times)

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rebeccak

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pH/Ka
« on: May 15, 2004, 02:15:47 PM »
Hello,

I was wondering if someone could remind me how to find the pH of a solution given the Ka value and the concentration of the solution?  (I know that pH is the -log of the [H+], and that Ka=[H+][A-]/[HA], but I don't know what to plug in for [A-]..?)

Thank you!
« Last Edit: May 15, 2004, 02:17:21 PM by rebeccak »

Offline jdurg

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Re:pH/Ka
« Reply #1 on: May 15, 2004, 05:53:04 PM »
All you'll need is some basic algebra, so it's not that difficult.   ;D  Having the formulas correct is the most important part, and I see that you've gotten that correct already.  If you've got the Ka value and the concentration of the solution, then you've got everything you need.  The Ka value is pretty much the equillibrium constant of the equation: HA -> H+ + A-.  The next thing we need to know is whether or not this is a monoprotic acid or not.  If the acid only has one proton it can donate, then it's VERY easy to figure out the pH.  If it's got more than one it becomes a bit more complicated.

So for now we'll just assume it's a monoprotic acid.  For every H+ ion you have, you'll have the exact same number of A- ions.  Since we don't know the concentration of the ions at eq, we'll substitute an x for them in the equation you listed above.  This will make the equation become Ka=(x2)/[HA].  We also know that at equillibrium, the concentration of the undissociated acid will be the initial concentration minus the concentration of the ions formed.  This turns the equation into Ka=(x2)/([HA]init-x).  Now you just need to solve for x which will be the H+ concentration for that acid at equillibrium.

Ka = (x2)/([HA]init-x)
([HA]init-x)(Ka) = x2
([HA]initKa)-(Ka)x = x2
([HA]initKa) = x2 + (Ka)x

All you need to do is solve for x and take the -log of it to find the pH.   ;D
« Last Edit: May 15, 2004, 05:54:13 PM by jdurg »
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ll7080

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Re:pH/Ka
« Reply #2 on: May 24, 2004, 07:47:27 PM »
How do I calculate the pH of a solution that has a [OH-]=0.01M?

Offline jdurg

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Re:pH/Ka
« Reply #3 on: May 24, 2004, 07:51:32 PM »
First you would need to calculate the pOH, which is the same thing as pH only the concentration of OH- is used in place of the H+ concentration.  Once you have the pOH, you need to remember that pH + pOH = 14.  It should be pretty easy, actually.   ;D
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