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November 01, 2025, 10:56:16 am

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shinny

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Re: help with questions
« Reply #105 on: November 09, 2008, 10:01:18 pm »
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2. pH calculations are most accurate for strong monoprotic acids with concentrations greater than or equal to 10-7 M at 25 degrees. Can someone explain this to me as well.

This reminded me of a case before midyears where I had some trippy question (can't remember it exactly) involving a case sort of like chucking an extremely small amount of HCl into pure water such that you get . However, this would be illogical because now that implies you've raised the pH from 7 to 8 by adding ACID. Seems like equilibrium explains this now, but using raw pH calculations, I guess this is a reason why you can't use solutions less than

EDIT: Actually wait, how does this work? I guess it's feasible to have a solution of acid less than (think about grabbing 1 acid particle and chucking it in water), but is it just that in these cases, ? I'm assuming its something to do with equilibrium perhaps. Someone explain?
« Last Edit: November 09, 2008, 10:16:33 pm by shinjitsuzx »
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Mao

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Re: help with questions
« Reply #106 on: November 09, 2008, 11:09:10 pm »
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in this case, you have added M of HCl, which completely ionises so [1]

this drives the to the left, assuming it is 25 degrees [assuming 1 litre, no other sources of these ions and no other impurities, letting x be the number of moles consumed in the backwards reaction]:



hence, the new concentration of is , and the pH is 6.98

so as you see... :P

[in the exam, we'd just say it's still 7]

normal calculations are just [1], and since we normally deal with 0.1M or concentrations of that like [a lot higher than 10^-7], we can usually ignore the subsequent calculations because it will have insignificant impact.
but in this case, there is significant impact [of pH ~0.02!!!]

if the acid was to the power of -9 [or less], then we can safely say that the pH will still be 7, the change will be too insignificant to do anything.
« Last Edit: November 09, 2008, 11:14:17 pm by Mao »
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shinny

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Re: help with questions
« Reply #107 on: November 09, 2008, 11:12:32 pm »
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So is this the reason why you can't use direct pH calculations for solutions less than ?
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Mao

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Re: help with questions
« Reply #108 on: November 09, 2008, 11:16:16 pm »
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So is this the reason why you can't use direct pH calculations for solutions less than ?

yes

you stated the reason yourself: pH cannot be raised by addition of acid.
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bec

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Re: help with questions
« Reply #109 on: November 10, 2008, 07:58:36 am »
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in this case, you have added M of HCl, which completely ionises so [1]

Why doesn't [H+]=[HCl] since it ionises completely?

Collin Li

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Re: help with questions
« Reply #110 on: November 10, 2008, 12:46:23 pm »
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in this case, you have added M of HCl, which completely ionises so [1]

Why doesn't [H+]=[HCl] since it ionises completely?

In all acid-base calculations, we usually ignore the fact that of protons sit around in neutral solutions. However, if we add something of a similar magnitude to that, we shouldn't ignore it.

We can definitely ignore it if we add say, hydrochloric acid, because if we accounted for the , it would only be seen at the 7th decimal place or so.

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Re: help with questions
« Reply #111 on: November 10, 2008, 02:41:21 pm »
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Ok, more questions:

1. The Earth's ocean's contain significant amounts of dissolved carbon dioxide. The dissolving process can be described by the following equilibria:

CO2(g) <----> CO2(aq)
CO2(aq) + H2O(l) <-----> H+(aq) + HCO3-(aq)

What is the likely effect of increasing the concentration of atmospheric CO2 on the pH of seawater at the ocean surface and the final concentration of dissolved CO2?

The answer is: pH decreases and the final dissolved [CO2(aq)] will be lower than the previous equilibrium concentration.
Can someone explain to me why this is so.


Let's just go back. Mao did you say that the answer i suppplied is wrong. Is the dissolved CO2 supposed to be higher than the previous equlibrium concentration because this is what i believe as well. Can anyone explain this to me?

Collin Li

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Re: help with questions
« Reply #112 on: November 10, 2008, 03:41:15 pm »
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The answer is wrong.

Even if we add more CO2, although the system will try to remove some of it, the final result is that there will still be more CO2 than there was before the injection of CO2 (i.e.: previous equilibrium concentration).

Rosie

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Re: help with questions
« Reply #113 on: November 11, 2008, 11:11:19 am »
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Ok. Thanks for clearing that up.

Q. The pH of a 10-9 M solution of HCl is closest to:
The answer is 7.

However, isn't it supposed to be 9 or does it have something to do with HCl being a strong acid.


Q. The actual pH of a 1.5*10-4 M solution of CH3COOH at 25 degrees is best described by: 3.8 < pH < 7
Can someone show me relevant working out to get the answer
 

Collin Li

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Re: help with questions
« Reply #114 on: November 11, 2008, 11:25:48 am »
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No problem!

Quote
Q. The pH of a 10-9 M solution of HCl is closest to:
The answer is 7.

However, isn't it supposed to be 9 or does it have something to do with HCl being a strong acid.

Nah, it's got to do with the fact that neutral water is more acidic than this!



Or, in general:



The concentration should be something like , and you'll just get 7 (or close) for the pH.

Quote
Q. The actual pH of a 1.5*10-4 M solution of CH3COOH at 25 degrees is best described by: 3.8 < pH < 7
Can someone show me relevant working out to get the answer

Not sure what's going on with the inequality, but the lower bound (3.8) is assuming that all of the protons deprotonate (strong acid), and the upper bound (7) is assuming that none of the protons deprotonate (neutral).