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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: DannyN on July 11, 2011, 07:04:43 pm

Title: Need help with a question
Post by: DannyN on July 11, 2011, 07:04:43 pm

hi, guys  i dont know the answer to this multiple choice question( dont have the answer too), can you help me out? :)

Which of the following mixtures would have a pH of 2?
a)100mL of HCL with pH=3 + 10mL of HCL with pH=1
b)100mL of HCL with pH=3 + 100mL of HCL with pH=1
c)100mL of NaOH with pH=13 + 100mL of HCL with pH=1
d)100mL of HCL with pH=1 + 100mL of water

 can you please provide a brief explanation, why you've chose the answer too. thanks!

Title: Re: Need help with an equilibirum question
Post by: omgwtf123 on July 11, 2011, 07:20:37 pm

is the answer B? just to make sure first :P

Title: Re: Need help with an equilibirum question
Post by: omgwtf123 on July 11, 2011, 07:23:29 pm

aaaaaaaaaaa wait, did a silly mistake >.<, ignore my above post , sorry

Title: Re: Need help with an equilibirum question
Post by: omgwtf123 on July 11, 2011, 07:25:42 pm

i change my answer to A ~~~~

Title: Re: Need help with an equilibirum question
Post by: tony3272 on July 11, 2011, 07:26:37 pm

I got A
All you do is you determine the concentration of each HCL through their pH. That is, one has a concentration of 10^-3, and other has 10^-1.
Then you work out the amount in mol of each, which is 10^-4 and 10^-3 respectively.
Then just add them together to work out the total amount in mol of 0.0011mol.
therefore new concentration is 0.0011mol/0.11L = 0.01M
which has a pH of 2

Title: Re: Need help with an equilibirum question
Post by: b^3 on July 11, 2011, 07:31:51 pm

Wait isn't it d?
ph of water is 7 so the amount of H3O+ ions is 10^-7*.1=10^-8 mol
then for the hcl, the amount of H3O+ ions is .1*10^-1=10^-2 mol
add them together and you get 0.01000001 mol
then -log10(0.01000001)= 1.99999999 which is 2?

Title: Re: Need help with an equilibirum question
Post by: omgwtf123 on July 11, 2011, 07:32:12 pm

1)Firstly change the pH of each solution into conc.... using 10^-pH. U will get 10^-3 M, and 10^-1 M of solutions respectively.

2) Change each of the solutions into mole,; n=cv , to get 0.0001 mol and 0.001mol and 0.001 mol respectively.
3) Add the two mols together as u cannot simply add concentrations as they r like before :P
4) use c=n/v remembering to divide by the total volume . u will get 0.0011/0/11= 0.01
5) Use pH= -Log(0.01)= 2
Hope this is right :)

Title: Re: Need help with an equilibirum question
Post by: tony3272 on July 11, 2011, 07:33:25 pm

Quote from: b^3 on July 11, 2011, 07:31:51 pm

Wait isn't it d?
ph of water is 7 so the amount of H3O+ ions is 10^-7*.1=10^-8 mol
then for the hcl, the amount of H3O+ ions is .1*10^-1=10^-2 mol
add them together and you get 0.01000001 mol
then -log10(0.01000001)= 1.99999999 which is 2?

You forgot to divide by the volume for the new concentration.

Title: Re: Need help with an equilibirum question
Post by: DannyN on July 11, 2011, 07:36:11 pm

thanks guys! :D

Title: Re: Need help with an equilibirum question
Post by: b^3 on July 11, 2011, 07:36:17 pm

oh wait sorry, pressed enter twice on calc (im using graphics) that makes my 5th mistake on VN for tonight, what am I doing, I think I need to rest.

Title: Re: Need help with an equilibirum question
Post by: pi on July 11, 2011, 07:41:33 pm

hmmmmmm, how is this an equilibrium question?



(btw, it IS A, as mentioned above ;) )

Title: Re: Need help with a question
Post by: DannyN on July 11, 2011, 07:51:49 pm

oh hehe sorry :P