Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Problem of the Week Archive => Topic started by: Borek on December 03, 2012, 05:59:39 AM
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Concentration of H+ in a solution of a weak acid can be calculated from the simplified equation
[tex][H^+] = \sqrt{C K_a}[/tex]
However, there is a condition that must be fulfilled for this equation to work. Can you name this condition?
Interestingly, when used for a 0.57 nM solution of the acetic acid this equation gives a correct answer, even if the condition is not fulfilled. Can you explain this result?
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Dissociation of water neglected?
[tex]
K_a.C >> K_w
[/tex]
Also,
[tex]
C >> K_a
[/tex]
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Probably C>>Ka as curiouscat said, or other way (maybe better because you can check it) to check if the dissociation degree is under 5% ([H+]/C*100).
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HA ::equil:: H+ + A-
At equilibrium CA0-x x x
H2O ::equil:: H+ + OH-
At equilibrium something y y
[tex]
K_w=[H^+][OH^-] \\
K_w=(x+y)y (Eq. a) \\
K_a=\frac{[H^+][A^-]}{[HA]} \\
Ka=\frac{(x+y).x}{C_{A0}-x} (Eq. b) \\
K_w=10^{-14} \\
K_a=1.8 \times 10^{-5} \\
C_{A0}=0.57 \times 10^{-9} \\
[/tex]
Solve (Eq. a) and (Eq. b) simultaneously in x and y to get
[tex]
x=5.6684 \times 10^{-10} \\
y=9.9717 \times 10^{-8} \\
[H^+] = x+y = 1.0028 \times 10^{-7} \\
[/tex]
By Borek's shortcut
[tex]
[H^+] = \sqrt{C_{A0} \times K_a} = 1.013 \times 10{-7} \\
[/tex]
QED?
Two wrongs make a right? The Kw effect compensates for neglecting the HA dissociation?
An analytical closed form solution might be more elegant though.....Still working on it.
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check if the dissociation degree is under 5% ([H+]/C*100).
That's the one I was thinking about.
Two wrongs make a right?
Something like that.
For sufficiently diluted solution pH is close to 7.0. For every Ka you can find such Ca, that [itex]\sqrt {K_aC_a}=10^{-7}[/itex]. It is a completely meaningless fact >:D
http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base
Scroll down to the image, explanation is in the text.
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As an aside, I think these days when numerical solvers are so easily available I rather set up the full problem than dealing with approximations and when they hold.
Two non-linear eqns. in two unknowns. Solve! Saves me the headaches. ;D
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At what point in your chemistry career are you suppose to be able to answer these?
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Hard to say, as a lot depends on the educational system and these are different in different countries.
I believe I was about 17 when I learned enough to be able to answer this question. 19 in the worst case.
Math and ideas behind are relatively simple, the only problem is most people don't get into the details.
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Hard to say, as a lot depends on the educational system and these are different in different countries.
I believe I was about 17 when I learned enough to be able to answer this question. 19 in the worst case.
Math and ideas behind are relatively simple, the only problem is most people don't get into the details.
I was probably doing this better at 21 than I am now. Being in touch helps. Most higher education and careers specialize you and one tends to get rusty on the things not in his domain. Sad, in a way. Well, Borek's problems are helping me sandpaper my brain. :) Thanks Borek!
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I realized I could have worded my question better. How many years of chemistry does it take to answer these questions? Should a first year student be able to answer these? (Honors Chem in high school, all the chem classes now are full before I get to register, so I have to study by myself to keep myself as sharp as I can.)
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I realized I could have worded my question better. How many years of chemistry does it take to answer these questions?
GenChem101 should be perfectly enough.