- #1

Safwat z

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The formula I am using is

pH=1/2(pKa-log molar concentration of the acid).

Thanks!

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- #1

Safwat z

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The formula I am using is

pH=1/2(pKa-log molar concentration of the acid).

Thanks!

- #2

Borek

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Compare http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-acid-base

- #3

Safwat z

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Compare http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-acid-base

Thank you Borek for your valuable explanation.

Compare http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-acid-base

- #4

epenguin

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Your equation is

pH = (pK_{a} - log molar concentration of the acid)/2 - let there be no misunderstanding.

You don't tell us what you need it for. For any practical purposes, calculation will serve just to give you a good idea for - whatever you need a good idea for. But when you come to the practice - measure it!

It's true it's rather complicated in general. But likely you can simplify. In particular if your acid is fairly concentrated. If it's fairly concentrated, then the acid is mostly totally associated - what you have in the solution is mostly H_{3}A and a little bit of H_{2}A^{-} and an equal amount of H^{+} (actually your formula is an approximation that *depends on* this) and pretty little of the other species HA^{2-} and A^{3-}. So you would not have to take into account these last two.

You can easily calculate whether you're in this condition or not: Say your citric acid is 1 M.

Your formula then gives you a pH of about 1.55. [H_{2}A^{-}] (= [H^{+}]`) Is only about 0.028 M. Repeat, the acid is mostly associated.

Under these conditions the HA^{2-} and A^{3-} concentrations are going to be negligible. You can work them out approximately from formula in Borek's link applied to the second dissociation step assuming pH is 1.55 and a total acid concentration 0.028 M. No need really - with a pK_{a} that is more than the 3 units below the pH, [HA^{2-}] is going to be only 10^{-3} of the [H_{2}A^{-}], and the [A^{3-}] even more than 100 times less. (These approximate calculations may seem circular but if you do them you see that they are consistent).

As you dilute the acid it dissociates, so this calculation won't work if it is 10^{-4} M but will be fairly okay at !0^{-1} even 10^{-2} M.

pH = (pK

You don't tell us what you need it for. For any practical purposes, calculation will serve just to give you a good idea for - whatever you need a good idea for. But when you come to the practice - measure it!

It's true it's rather complicated in general. But likely you can simplify. In particular if your acid is fairly concentrated. If it's fairly concentrated, then the acid is mostly totally associated - what you have in the solution is mostly H

You can easily calculate whether you're in this condition or not: Say your citric acid is 1 M.

Your formula then gives you a pH of about 1.55. [H

Under these conditions the HA

As you dilute the acid it dissociates, so this calculation won't work if it is 10

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- #5

Safwat z

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Thanks for your input :)Your equation is

pH = (pKa-log molar concentration of the acid)/2 - let there be no misunderstanding.

You don't tell us what you need it for. For any practical purposes, calculation will serve just to give you a good idea for - whatever you need a good idea for. But when you come to the practice - measure it!

It's true it's rather complicated in general. But likely you can simplify. In particular if your acid is fairly concentrated. If it's fairly concentrated, then the acid is mostly totally associated - what you have in the solution is mostly H_{3}A and a little bit of H_{2}A^{-}and an equal amount of H^{+}(actually your formula is an approximation thatdepends onthis) and pretty little of the other species HA^{2-}and A^{3-}. So you would not have to take into account these last two.

You can easily calculate whether you're in this condition or not: Say your citric acid is 1 M.

Your formula then gives you a pH of about 1.55. [H_{2}A^{-}] (= [H^{+}]`) Is only about 0.028 M. Repeat, the acid is mostly associated.

Under these conditions the HA^{2-}and A^{3-}concentrations are going to be negligible. You can work them out approximately from formula in Borek's link applied to the second dissociation step assuming pH is 1.55 and a total acid concentration 0.028 M. No need really - with a pK that is more than the 3 units below the pH, [HA^{2-}] is going to be only 10^{-3}of the [H_{2}A^{-}], and the [A^{3-}] even more than 100 times less. (These approximate calculations may seem circular but if you do then you see that they are consistent).

As you dilute the acid it dissociates, so this calculation won't work if it is 10^{-4}M but will be fairly okay at !0^{-1}even 10^{-2}M.

- #6

Safwat z

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This means at higher concentration of a weak acid I can predict it's pH using the formula. Thanks again for your explanation.

- #7

Borek

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Epenguin, you are right when I checked the pH of 0.1 M Citric acid its value was exactly equal the value I got using the monotropic formula when using tlhe first pKa1of Citric acid.

Which - sadly - doesn't mean the formula is OK. To make things even more complicated, pH is a measure not of the concentration of H

Ah, the joys of dealing with the real world

- #8

epenguin

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It's nice to see someone come back, especially with a relevant experimental result. Especially one that corresponds exactly to prediction, though as Borek says there may be an element of fluke in its coming out so exact. That's why I said at the start - measure it.

Main influences I know why measured value would deviate from calculated are ionic strength of the solution, and temperature. You'd have to go back to the original work that is referenced in the tables of pK's to see how closely the conditions where they were measured resemble yours. For some other factors such as the difference between concentration and activity, well they could well have been present in in the original measurements, and therefore the 'errors' cancel out. Anyway although I would not be much troubled by some discrepancy between predicted and measured the prediction ought to be quite better than ballpark.

Your equation is an approximation valid at high concentrations. Getting on top of this subject around pH's which generates such a large fraction of the homework questions involves approximations. That is I want to say: it's not like there is The Theory and then, oh, here you can use a convenient approximation and there another one. Rather the approximations are an essential part of the (limited) subject the student has to master. In fact there are several different kinds of approximation, of varying goodness and generality used, and you need to have an understanding of them and how good they are, any limitations.

For that of your equation you don't have to guess, you can easily see at what point it breaks down. I'll just treat it like a monobasic acid, which for present purposes we have seen is good enough. When the total concentration of your acid is equal to K_{a} then from the equilibrium equation.

$$ K_a = \frac{[H^+][A^-]}{[HA]} $$

It follows that if ##[A^-] = [HA]## then ##[H^+] = K_a ## or ##pH = pK_a## and vice versa.

So in your case when K_{a} = 7.4×10^{-4}, [H^{+}] = [A^{-}] = 7.4×10^{-4} so total citrate molarity is about 1.5×10^{-3}. At that concentration then half of it is A^{-} contrary to the assumptions giving your equation which are that the acid is concentrated enough that nearly 100% of it is HA, so approximately equal to the total acid molarity, C, used in your formula.

So you expect the formula to be noticeably failing around 1.5×10^{-3 } M. But if you increase the citrate concentration tenfold you expect A^{-} to be only about a tenth of the total acid, and your equation to hold moderately well. By 5×10^{-2}M it should be good enough for all practical purposes. But if on the other hand you go to concentrations 10 times lower than K_{a} the acid will be nearly all dissociated and [H^{+}] = [A^{-}] ≅ C, so then plot of pH against - log C will be approximately linear with slope 1. This slope will thus be: close to at ½ high concentrations as predicted by your equation changing to 1 at lower, changeover happening around pH = pK_{a} ≈ - log 2C.

This is all illustrated in an overall equation

$$ [H^+] = \frac{K_a + \sqrt{K_a(K_a + 4C)}}{2} $$

plotted here:

Red pH (ordinate) against - log C (abcissa) for monobasicn acid K_{a} = 7.4×10^{-4}

Blue pH = (3.13 + C)/2

Green pH = C

Even that equation is not the complete answer, obviously the linearity cannot continue forever - it has to level off by pH 7 (you cannot make a solution and alkaline by diluting acid!). In other words we have to take the dissociation of the water molecules into consideration at high dilutions. And all this is just for a monobasic acid; with citric when you get much below 10^{-4} M its second dissociation is significant (you could use your original formula though with the second pK_{a} because then the first dissociation is approximately 100%.) It must be a rare that such calculations are useful however.

(Sorry this answer has been delayed; this was due to difficulties doing the plot which turned out to be nothing but silly mistakes in inputting the formula. And it was unnecessary to solve the equation for [H^{+}] either to plot the curve, or to see the different linear approximations at high and low concentrations - I could have just used

$$C = \frac{[H^+]^2}{K_a} + [H^+] $$

)

Main influences I know why measured value would deviate from calculated are ionic strength of the solution, and temperature. You'd have to go back to the original work that is referenced in the tables of pK's to see how closely the conditions where they were measured resemble yours. For some other factors such as the difference between concentration and activity, well they could well have been present in in the original measurements, and therefore the 'errors' cancel out. Anyway although I would not be much troubled by some discrepancy between predicted and measured the prediction ought to be quite better than ballpark.

Your equation is an approximation valid at high concentrations. Getting on top of this subject around pH's which generates such a large fraction of the homework questions involves approximations. That is I want to say: it's not like there is The Theory and then, oh, here you can use a convenient approximation and there another one. Rather the approximations are an essential part of the (limited) subject the student has to master. In fact there are several different kinds of approximation, of varying goodness and generality used, and you need to have an understanding of them and how good they are, any limitations.

For that of your equation you don't have to guess, you can easily see at what point it breaks down. I'll just treat it like a monobasic acid, which for present purposes we have seen is good enough. When the total concentration of your acid is equal to K

$$ K_a = \frac{[H^+][A^-]}{[HA]} $$

It follows that if ##[A^-] = [HA]## then ##[H^+] = K_a ## or ##pH = pK_a## and vice versa.

So in your case when K

So you expect the formula to be noticeably failing around 1.5×10

This is all illustrated in an overall equation

$$ [H^+] = \frac{K_a + \sqrt{K_a(K_a + 4C)}}{2} $$

plotted here:

Red pH (ordinate) against - log C (abcissa) for monobasicn acid K

Blue pH = (3.13 + C)/2

Green pH = C

Even that equation is not the complete answer, obviously the linearity cannot continue forever - it has to level off by pH 7 (you cannot make a solution and alkaline by diluting acid!). In other words we have to take the dissociation of the water molecules into consideration at high dilutions. And all this is just for a monobasic acid; with citric when you get much below 10

(Sorry this answer has been delayed; this was due to difficulties doing the plot which turned out to be nothing but silly mistakes in inputting the formula. And it was unnecessary to solve the equation for [H

$$C = \frac{[H^+]^2}{K_a} + [H^+] $$

)

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