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Topic: ph & equilibrium.  (Read 10613 times)

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Offline mookxi

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ph & equilibrium.
« on: June 04, 2006, 10:21:05 AM »
Some toilet cleaners contain bisulfate salts as their active ingredients. The hydrogen sulfate ion reacts according to the equation

HSO4-(aq) <=> SO42-(aq) + H+(aq)

and the equilibrium constant for this reaction is 0.012mol dm-3. At pH 3.0 the value of the fraction [HSO42-]/[SO42-] is closest to..

the answer is 0.10. I'm not sure how the fraction relates to the equilibrium expression which is:

K = [SO42-][H+]
      ------------------
        [HSO4-]

thanks in advance. I seem to have a weakness for equilibrium calculations.

Offline Borek

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Re: ph & equilibrium.
« Reply #1 on: June 04, 2006, 10:39:23 AM »
Rearrange the equation to get ratio you are asked about on one side and everything else on the second.
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Offline mookxi

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Re: ph & equilibrium.
« Reply #2 on: June 04, 2006, 10:01:03 PM »
Rearrange the equation to get ratio you are asked about on one side and everything else on the second.

I'm still confused, sorry. I get keep getting the answer 0.08 but i'm sure that's wrong.

Offline anarchron

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Re: ph & equilibrium.
« Reply #3 on: June 04, 2006, 10:06:30 PM »
Nah you're right.

Offline Will

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Re: ph & equilibrium.
« Reply #4 on: June 04, 2006, 10:16:48 PM »
My calculation was: (((10-3)2)/0.012)/10-3 and I got 1/12 or 0.08333...

I would say you are right, mookxi, but the question does say closest to...; does it give you a choice of values to chose from or does it just want your answer closest to two decimal places?

Offline mookxi

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Re: ph & equilibrium.
« Reply #5 on: June 04, 2006, 10:19:51 PM »
My calculation was: (((10-3)2)/0.012)/10-3 and I got 1/12 or 0.08333...

I would say you are right, mookxi, but the question does say closest to...; does it give you a choice of values to chose from or does it just want your answer closest to two decimal places?

It's a mutiple choice question so it gave me 10, 0.10, 0.01, 0.001.

Thanks. ^^

Offline Will

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Re: ph & equilibrium.
« Reply #6 on: June 04, 2006, 10:25:32 PM »
It's a mutiple choice question so it gave me 10, 0.10, 0.01, 0.001.

In this case the answers are just trying to confuse you! The correct answer is 0.08, but the 'answer' to this question would be 0.10. :)

Offline AWK

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Re: ph & equilibrium.
« Reply #7 on: June 05, 2006, 03:25:18 AM »
Equation should be:

        [SO42-][H+]
K =  ---------------------
        [HSO4-]-[H+]
AWK

Offline Borek

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Re: ph & equilibrium.
« Reply #8 on: June 05, 2006, 04:06:05 AM »
Equation should be:

        [SO42-][H+]
K =  ---------------------
        [HSO4-]-[H+]

I don't think so. You have pH given, if part of the H+ gets consumed pH will change and you will be forced to solve different question. In fact that's a classical HH equation problem, I just refrained myself from calling it this way posting for the first time.

But perhaps question wording is confusing.
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Offline AWK

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Re: ph & equilibrium.
« Reply #9 on: June 05, 2006, 05:42:29 AM »
pH is much higher than pK2, and HSO4- will dissociate at this pH,  hence H+ will be largely due to a dissociation of HSO4-
AWK

Offline Borek

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Re: ph & equilibrium.
« Reply #10 on: June 05, 2006, 06:08:00 AM »
pH is much higher than pK2, and HSO4- will dissociate at this pH,  hence H+ will be largely due to a dissociation of HSO4-

Difference between pH and pKa2 is not that large - pH = 3, pKa2 = 2. 1 unit difference is hardly "much higher".

Besides, problem doesn't state anything about the source of pH - it only states pH = 3.0. Thus assuming that HSO4- dissociates further is equivalent to stating "although we are told that pH=3.0 we are going to ignore that and assume that pH is different". I can't see any reason for doing such thing.

As I read the question it states that pH was checked/measured at 3.0 - and we have to just calculate ratio of SO42-/HSO4-. Note that if the solution has measured pH=3.0 this value already takes HSO4- dissociation into account.

What you propose is a solution to a completely different question - what will be the change of pH in the solution that starts at pH = 3.0 when HSO4- is added. But as concentration of HSO4- is not known, such question can't be answered.
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Offline AWK

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Re: ph & equilibrium.
« Reply #11 on: June 05, 2006, 08:36:50 AM »
Quote
Besides, problem doesn't state anything about the source of pH - it only states pH = 3.0. Thus assuming that HSO4- dissociates further is equivalent to stating "although we are told that pH=3.0 we are going to ignore that and assume that pH is different". I can't see any reason for doing such thing.
We cannot ignore!. At pH 1.92 in NaHSO4 the ratio [SO42- ]/[HSO4- ] is close 1. At pH=3 this ratio will be even higher. Hence the closest value to the "so called true answer"  (namely 0.1) is to far to be accepted.

typo 2+ corrected
« Last Edit: June 05, 2006, 08:43:47 AM by AWK »
AWK

Offline Will

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Re: ph & equilibrium.
« Reply #12 on: June 05, 2006, 08:41:45 AM »
Quote
Besides, problem doesn't state anything about the source of pH - it only states pH = 3.0. Thus assuming that HSO4- dissociates further is equivalent to stating "although we are told that pH=3.0 we are going to ignore that and assume that pH is different". I can't see any reason for doing such thing.
At pH 1.92 in NaHSO4 the ratio SO42+/HSO4- is 1. At pH=3 this ratio will be higher.

Won't that ratio (SO42-/HSO4-) be lower at pH of 3, because there will be more HSO4- and less SO42-(not 2+)?

Offline AWK

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Re: ph & equilibrium.
« Reply #13 on: June 05, 2006, 08:45:52 AM »
[
Won't that ratio (SO42-/HSO4-) be lower at pH of 3, because there will be more HSO4- and less SO42-(not 2+)?
The higher pH, the more SO42-
AWK

Offline Will

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Re: ph & equilibrium.
« Reply #14 on: June 05, 2006, 08:55:27 AM »
Won't that ratio (SO42-/HSO4-) be lower at pH of 3, because there will be more HSO4- and less SO42-(not 2+)?
The higher pH, the more SO42-

ooops, my bad. ;D

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