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Offline Sophia7X

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couple questions {pH}
« on: October 20, 2012, 07:05:31 PM »
What is the pH of a 5e-8 M HCl solution?

My attempt at solution:
I don't think you can -log that because that gives you a basic pH and adding a strong to water however little in amount shouldn't make pH basic.
ionization of water: H2O  ::equil:: H+ + OH-
Difference between H+ and OH- concentration must be 5e-8 since HCl is a strong acid so assume 100% dissociation into H+.
[H+] - [OH-] = 5e-8 ([OH-] = -5e-8+[H+])
Kw = 1e-14 = [H+][OH-]

I substituted [OH-] into Kw, 1e-14 = [H+][-5e-8+ [H+]]
[H+] = 1.28e-7
pH = 6.89 but apparently the answer is 7.00... how can it be completely neutral?

________________________________________________________________

A + B --> AB is 1st order with respect to A, zero order with respect to B. Initial concentration of A and B = both 0.100 M. After 1.5 hours, conc of B = 0.060 M. What is the specific rate constant?

ln 0.06 = -k(1.5 hr) + ln 0.100
k = 0.34 1/hr
but the answer is 0.61 1/hr, how do I get this?
« Last Edit: October 20, 2012, 09:37:16 PM by Arkcon »
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Offline curiouscat

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Re: couple questions {pH}
« Reply #1 on: October 21, 2012, 12:05:50 AM »
A + B --> AB is 1st order with respect to A, zero order with respect to B. Initial concentration of A and B = both 0.100 M. After 1.5 hours, conc of B = 0.060 M. What is the specific rate constant?

ln 0.06 = -k(1.5 hr) + ln 0.100
k = 0.34 1/hr
but the answer is 0.61 1/hr, how do I get this?

I get 0.34 hr-1 too.

Offline Borek

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Re: couple questions {pH}
« Reply #2 on: October 21, 2012, 05:29:46 AM »
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Offline Sophia7X

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Re: couple questions {pH}
« Reply #3 on: October 22, 2012, 02:11:53 PM »
Entropy happens.

Offline Rutherford

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Re: couple questions {pH}
« Reply #4 on: October 23, 2012, 07:26:31 AM »
A + B --> AB is 1st order with respect to A, zero order with respect to B. Initial concentration of A and B = both 0.100 M. After 1.5 hours, conc of B = 0.060 M. What is the specific rate constant?

ln 0.06 = -k(1.5 hr) + ln 0.100
k = 0.34 1/hr
but the answer is 0.61 1/hr, how do I get this?
I got their answer when replacing ln0.06 with ln0.04 but 0.04 is the concentration change, not the amount that left (which is put in the rate law). I can't tell right away if it is a mistake or not.

Offline curiouscat

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Re: couple questions {pH}
« Reply #5 on: October 23, 2012, 07:40:33 AM »
I got their answer when replacing ln0.06 with ln0.04 but 0.04 is the concentration change, not the amount that left (which is put in the rate law). I can't tell right away if it is a mistake or not.

Unless the question meant to say: "After 1.5 hours, conc of B AB = 0.060 M"

Offline Aegis6

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Re: couple questions {pH}
« Reply #6 on: October 24, 2012, 05:57:32 PM »
I got a pH of 7.3

My math:

pH = -log [H+]

5x10-8 M HCl = 5x10-8 M H+

Once again, pH = -log [H+]

-log(5x10-8) = 7.3

Using sig figs, 7.3 becomes just 7.

QED, pH=7

Wolframalpha for verification: http://www.wolframalpha.com/input/?i=pH+5x10%5E-8+M+HCl

Offline Rutherford

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Re: couple questions {pH}
« Reply #7 on: October 25, 2012, 08:21:58 AM »
At such low concentrations water dissociation takes an important role. Sophia wrote it nicely in the first post.

For the second one, it has to be the concentration change of AB, because the concentrations of B and A aren't proportional and they can't be compared.

Offline curiouscat

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Re: couple questions {pH}
« Reply #8 on: October 25, 2012, 08:25:19 AM »
For the second one, it has to be the concentration change of AB, because the concentrations of B and A aren't proportional and they can't be compared.

You are confusing stoichiomety and kinetics. Conc. of A and B must remain equal at all times.

Offline Borek

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Re: couple questions {pH}
« Reply #9 on: October 25, 2012, 08:47:29 AM »
I got a pH of 7.3

So the solution of an acid is slightly basic?

Quote
Wolframalpha for verification: http://www.wolframalpha.com/input/?i=pH+5x10%5E-8+M+HCl

Several years ago WA was told their results are wrong - and they were told it many times, on many occasions. They slightly modified the engine so that when the result becomes nonsensical it is displayed as 7. Highly disappointing.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline Aegis6

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Re: couple questions {pH}
« Reply #10 on: October 25, 2012, 06:41:49 PM »
I got a pH of 7.3

So the solution of an acid is slightly basic?


The method for determining the pH of a strong acid or base is to take the negative logarithm of the concentration of H+ ions.

As you have 5x10-8 M HCl, and you have one mol of H+ per mole of HCl, it logically follows that you have 5x10-8 M H+.

pH = -log[H+]
pH = -log[5x10-8]
pH = 7.30

I couldn't possibly tell you why the value is basic, but that is the value you get when you calculate for the pH of the HCl solution.

Offline Borek

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Re: couple questions {pH}
« Reply #11 on: October 26, 2012, 04:16:50 AM »
As you have 5x10-8 M HCl, and you have one mol of H+ per mole of HCl, it logically follows that you have 5x10-8 M H+.

No, it doesn't follow. Sophia explained the correct approach in the very first post of this thread.
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Offline Rutherford

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Re: couple questions {pH}
« Reply #12 on: October 26, 2012, 09:28:41 AM »
For the second one, it has to be the concentration change of AB, because the concentrations of B and A aren't proportional and they can't be compared.

You are confusing stoichiomety and kinetics. Conc. of A and B must remain equal at all times.
How do you mean?
The rate law for the concentration of A is: ln[A]=ln[A]0-kt
For B it is: [B0]=[B0]0-kt
After time t, they can't be equal (added 0 above B because without it I get bold letters).

Offline curiouscat

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Re: couple questions {pH}
« Reply #13 on: October 26, 2012, 01:42:42 PM »
How do you mean?
The rate law for the concentration of A is: ln[A]=ln[A]0-kt
For B it is: [B0]=[B0]0-kt
After time t, they can't be equal (added 0 above B because without it I get bold letters).

Your rate expressions are wrong.

To make the fallacy evident: Assume you had 1 mol of both A and B to start with. If at some later time if you end up with (say) 0.4 mol A and 0.5 mol B that means more of A reacted than B.  Correct?

What did that excess A react with if not B?  If A+B-->C then dA=dB at all t.

Makes sense?

Offline Sophia7X

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Re: couple questions {pH}
« Reply #14 on: October 26, 2012, 10:42:29 PM »

The method for determining the pH of a strong acid or base is to take the negative logarithm of the concentration of H+ ions.


Hmm, not when the strong acid's concentration is less than the ~1*10-7 since at such a dilute concentration, ionization of water will play a role



I got their answer when replacing ln0.06 with ln0.04 but 0.04 is the concentration change, not the amount that left (which is put in the rate law). I can't tell right away if it is a mistake or not.

In that case, I guess the question just had a typo.
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