ph acid/base question?

[riadnicolas]

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How much water must be added to a 1 litre soln of a strong base of ph 13 in order to decrease the ph to 12?

I understand questions that ask to increase the ph but this one seemed new to me.

[jumpthe]

C1V1=C2V2

C1=10^(-13) from ph=-log(H3O+)
V1=1

C2= 10^(-12)
V2?

V2=0.1

therfore you must add 900ml

i think thats right, plz correct me anyone if im wrong

[riadnicolas]

see thats what i thought but the answer says; a decrease of one ph unit requires a ten fold dilution of the base hence add 9 litres of water

[mayank.2802]

Originally Posted by riadnicolas

see thats what i thought but the answer says; a decrease of one ph unit requires a ten fold dilution of the base hence add 9 litres of water

lol ask pierre!!
i get tutoring from him too

[EvozRule]

Originally Posted by riadnicolas

How much water must be added to a 1 litre soln of a strong base of ph 13 in order to decrease the ph to 12?

I understand questions that ask to increase the ph but this one seemed new to me.

10^-14 = [OH-] [H3O+]

-log[H3O+] = pH = 13

Therefore [H3O+] = 10^-13

For pH = 12, [H3O+] = 10^-12

Since you're increasing the the concentration of H3O+ ions by 10^1, i.e you multiply 10^-13 by 10 which gives you 10^-12, you are increasing the volume 10x.

So so add 9L to it. Sorry if I cant explain properly not very good witht the theory side of chem

:-S

[riadnicolas]

lol what a coincidence, how did you no he tutors me

[riadnicolas]

Originally Posted by EvozRule

10^-14 = [OH-] [H3O+]

-log[H3O+] = pH = 13

Therefore [H3O+] = 10^-13

For pH = 12, [H3O+] = 10^-12

Since you're increasing the the concentration of H3O+ ions by 10^1, i.e you multiply 10^-13 by 10 which gives you 10^-12, you are increasing the volume 10x.

So so add 9L to it. Sorry if I cant explain properly not very good witht the theory side of chem

:-S

so when increasing the ph by a factor of one you multiply the volume by 10. so in this case cv=cv doesnt work

[EvozRule]

Originally Posted by riadnicolas

so when increasing the ph by a factor of one you multiply the volume by 10. so in this case cv=cv doesnt work

Id say it wouldnt because you're changing concentrations of H3O+ and OH- by altering pH, which is why i guess their concetrations multiplied always equal 10^-14.

Yeah and by increasing i mean increasing the H3O+ which would be a decrease in pH value.

[riadnicolas]

Originally Posted by EvozRule

Id say it wouldnt because you're changing concentrations of H3O+ and OH- by altering pH, which is why i guess their concetrations multiplied always equal 10^-14.

Yeah and by increasing i mean increasing the H3O+ which would be a decrease in pH value.

thanks, and would you no if pOH would work with the cv=cv rule, cbf doing calculations

[mayank.2802]

Originally Posted by riadnicolas

lol what a coincidence, how did you no he tutors me

i was just looking at an email he sent me and ur name was in the "send to:" details thing.

it kinda rung a bell when i saw it nxt ur question.

Pierres awesome tho....gav me bout 20 attachments in an email ...mostly organic questions from sum american university exam.....bit loony.

[riadnicolas]

Originally Posted by mayank.2802

i was just looking at an email he sent me and ur name was in the "send to:" details thing.

it kinda rung a bell when i saw it nxt ur question.

Pierres awesome tho....gav me bout 20 attachments in an email ...mostly organic questions from sum american university exam.....bit loony.

yeh can be a little crazy with the difficulty of the questions to but he prepares well for an exam. good luck

[EvozRule]

Originally Posted by riadnicolas

thanks, and would you no if pOH would work with the cv=cv rule, cbf doing calculations

pOH isnt part of the course...is it?

but its the same principle,

pOH = -log[OH-]

still a logarithmic scale but a measure of OH- concentration, so in this case a low pOH value means a basic solution and a high pOH would mean an acidic solution.

[OH-] [H3O+] = 10^-14

pOH + pH = 14

[jumpthe]

OPPS!! sorry i didnt see that it was a base, looked at ph and thought it was an acid straight away, my mistake buddy soz.