[steph_r]

Hello everyone..
I have been attempting this question for a while now and still havn't got the right answer.. is there something silly that i'm doing wrong?

Here is the question:

Calculate the pH of the mixture: 10mL of 0.10M NaOH is diluted to 100mL of solution

Here is my attempt:

NaOH  (Na+) + (OH-)

[OH-] = 0.1M
[H30+] x [OH-] = 10^-14
[H30+] x 0.1 = 10^-14
[H30+] = 1.0 x 10^-13 M

V = 10mL = 0.01L
n = c x v = (1.0 x 10^-13) x 0.01 = 1.0 x 10^-15 mol

Total volume = 10mL + 100mL = 110mL  (which is 0.11L)

Therefore,
C(H30+) = n/v = (1.0 x 10^-15)/0.11 = (9.09 x 10^-15)

Therefore,
-log(9.09 x 10^-15) = 14?                  (ANSWER IN BOOK: 12)

HELP WOULD BE GREATLY APPRECIATED..
Cheers

[macman104]

You're making it far too difficult.

Are you familiar with the formula, M₁V₁=M₂V₂?

This is Molarity_initial*Volume_initial=Molarity_final*Volume_final

Also, if a solution is diluted TO 100mL, it doesn't mean you add 100mL.  It means the final volume solution is 100mL (so in your case, you would add 90mL of water).  Solve the above equation for your final molarity, and solve for the pH.

[AWK]

Diluted to 100 mL means an addition enough water to reach a final volume 100 mL (not 110)

Even then

C(H30+) = n/v = (1.0 x 10^-15)/0.11 = (9.09 x 10^-15)

you make error in calculations. Calculate dilution of NaOH, then convert to H₃O⁺ concentration.

[steph_r]

Thank you both for your *delete me*!
I now understand where i went wrong..
Cheers

[Guitarmaniac86]

Hello everyone..
I have been attempting this question for a while now and still havn't got the right answer.. is there something silly that i'm doing wrong?

Here is the question:

Calculate the pH of the mixture: 10mL of 0.10M NaOH is diluted to 100mL of solution

Here is my attempt:

NaOH  (Na+) + (OH-)

[OH-] = 0.1M
[H30+] x [OH-] = 10^-14
[H30+] x 0.1 = 10^-14
[H30+] = 1.0 x 10^-13 M

V = 10mL = 0.01L
n = c x v = (1.0 x 10^-13) x 0.01 = 1.0 x 10^-15 mol

Total volume = 10mL + 100mL = 110mL  (which is 0.11L)

Therefore,
C(H30+) = n/v = (1.0 x 10^-15)/0.11 = (9.09 x 10^-15)

Therefore,
-log(9.09 x 10^-15) = 14?                  (ANSWER IN BOOK: 12)

HELP WOULD BE GREATLY APPRECIATED..
Cheers

If 10ml of 0.1M NaOH is diluted in 100ml you have diluted it by a factor of 10 and hence you have 0.1 / 10 = 0.01 M NaOH

Therefore you have a pH of 12

[yukiheart]

I think it can be calulate easier :
PH of NaOH 0,1 M = 13
because : [OH-]=0,1 ----> [H+]=10^-13--->pH=-log[H+]=13
when diluted to 100ml mean [OH-]=0,1/10 =0,01----->[H+]=10^-12--->PH=12

have a quick formular to calculate : (can apply to mixture acid or bazo)

  n(times need to diluted)= 10^(delta pH)
in this case :
 n=100/10=10^(13-x)
13-x=1
x=12
Ok ?Im not good at English hehe

[steph_r]

THANK YOU!