Simple pH question.

[duhherro]

Hey guys, just doing some questions and came across this.

calculate ph of 10ml of 0.1M naoh diluted to 100ml of solution at 25 degrees.

I did it in 2 methods, one with the dilution formula

c(NaOH) = (0.1 x 10) / 100
= 0.01 M
[H3O+] = 10^-14 / 0.01
= 10^-12

ph= -log10 (10^-12)
=12

And the other answer i got was 14 , by doing the standard method of working out that
[OH-] = 0.1 M
therefore [H3O+] = 10^-14 / 0.1
=10^-13 M
n(H3O+) = 10^-13 x (10 x 10^-3)
And we know that the total dilute solution is 100 ml which is 0.1 L
So c(H3O+) = 10^-15 / 0.1
=10^-14

and we use the ph formula to get 14.

Really confused on which methods to follow!
Thanks in advance!

[golden]

Hey guys, just doing some questions and came across this.

calculate ph of 10ml of 0.1M naoh diluted to 100ml of solution at 25 degrees.

I did it in 2 methods, one with the dilution formula

c(NaOH) = (0.1 x 10) / 100
= 0.01 M
[H3O+] = 10^-14 / 0.01
= 10^-12

ph= -log10 (10^-12)
=12

And the other answer i got was 14 , by doing the standard method of working out that
[OH-] = 0.1 M
therefore [H3O+] = 10^-14 / 0.1
=10^-13 M
n(H3O+) = 10^-13 x (10 x 10^-3)
And we know that the total dilute solution is 100 ml which is 0.1 L
So c(H3O+) = 10^-15 / 0.1
=10^-14

and we use the ph formula to get 14.

Really confused on which methods to follow!
Thanks in advance!

Umm the solution should be 12 because it was previously at a pH of 13 before the dilution but when you diluted it by factor of 10 it decreases by one pH down.

[duhherro]

Thanks for that! I'll just stick to the first method then