A2 AQA Confusing find pH question

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Applicant75235
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#1
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KaylaB
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(Original post by koolgurl14)
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When water is added, does the solution become stronger or weaker?
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ghostmarston
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First calculate the moles of Hcl= 0.154*10/1000=1.54x10-3
then calculate the new concentration of Hcl after adding water= 1.54x10-3/(990+10)*1000=1.54x10-3
then caluclate pH= -log(1.54x10-3)=2.81
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Applicant75235
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(Original post by ghostmarston)
First calculate the moles of Hcl= 0.154*10/1000=1.54x10-3
then calculate the new concentration of Hcl after adding water= 1.54x10-3/(990+10)*1000=1.54x10-3
then caluclate pH= -log(1.54x10-3)=2.81
dividing with 0.1 deffo wont give u the same number?
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Applicant75235
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#5
(Original post by KaylaB)
When water is added, does the solution become stronger or weaker?
Weaker?
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ghostmarston
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where did you get 0.1 from?
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B_9710
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(Original post by KaylaB)
When water is added, does the solution become stronger or weaker?
I know what you mean but that question doesn't make much sense.
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KaylaB
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#8
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(Original post by koolgurl14)
Weaker?
Exactly, as the answers has been posted above, just remember that the general way to do these questions is to calculate the amount of moles of the acid in the beginning, then calculate the concentration of the acid for the total volume, then find the pH as you would normally
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KaylaB
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(Original post by B_9710)
I know what you mean but that question doesn't make much sense.
Care to elaborate?
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B_9710
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#10
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#10
You know the concentration of H+ in water at 298k is around  10^{-7} \text{M} , using n=cv you can work out the moles of H+ in the water.
Then you want to work out the moles of H+ added to the solution.
Then use pH = -\log[\text{H}^+] .
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B_9710
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#11
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(Original post by KaylaB)
Care to elaborate?
Well depending on what you add to a solution, the solution could get more strongly acidic, more strongly alkaline, or more weakly acidic or alkaline.
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KaylaB
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(Original post by B_9710)
Well depending on what you add to a solution, the solution could get more strongly acidic, more strongly alkaline, or more weakly acidic or alkaline.
I said about adding water a neutral substance, which can be used for dilution
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SubwayLover1
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#13
Use the formula :

H+ new =[H+] old x Old Volume / New Volume
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B_9710
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(Original post by KaylaB)
I said about adding water a neutral substance, which can be used for dilution
That's not what the question said though. But yeah cool I'm not saying your wrong though.
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KaylaB
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(Original post by B_9710)
That's not what the question said though. But yeah cool I'm not saying your wrong though.
It literally says in the question that the HCl is added to 990cm^3 of water
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#16
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#16
(Original post by KaylaB)
It literally says in the question that the HCl is added to 990cm^3 of water
So does adding acid to water make the solution stronger?
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KaylaB
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#17
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(Original post by B_9710)
So does adding acid to water make the solution stronger?
Well there isn't really a concentration of water, so adding some acid to water would still just make a dilute acid
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B_9710
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#18
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#18
(Original post by KaylaB)
Well there isn't really a concentration of water, so adding some acid to water would still just make a dilute acid
Yeah I know, this is why what you said doesn't make much sense, the acid gets more dilute, but you can't say it gets weaker. pH does not really tell you how strong an acid is unless you're comparing 2 acids with same conc.
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KaylaB
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#19
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#19
(Original post by B_9710)
Yeah I know, this is why what you said doesn't make much sense, the acid gets more dilute, but you can't say it gets weaker. pH does not really tell you how strong an acid is unless you're comparing 2 acids with same conc.
Ah okay so it was just my wording was a bit off, thanks for letting me know so I can stop sounding like a fool :hat2:
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#20
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#20
(Original post by KaylaB)
Ah okay so it was just my wording was a bit off, thanks for letting me know so I can stop sounding like a fool :hat2:
I'm not trying to say you're a fool or anything like that or come across as a knob, just being pedantic. :gthumb:
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