Preparing 5.00 L of pH 10.7 NaOH solution from 0.250 M stock

[ally1h]

Homework Statement

How would you prepare 5.00 L if an aqueous solution having a pH = 10.70, if you had a supply of 0.250 M NaOH?

Homework Equations

I don't even know anymore...

The Attempt at a Solution

I've tried a million different ways but I never come up with the answer provided in the book, which is 10 mL.

Here are two ways I've tried:
pH = 10.7; so pOH = 14.0-10.7 = 3.3
antilog (-3.3) = 5.0x10^-4 M

m1v1 = m2v2
v2 = m1v1 / m2 = (0.250 M x 5000 mL) / (5.0x10^-4) = 2500000 mL
That's DEFINITELY not right...

Second attempt:
5000 mL has x moles NaOH
5.0x10^-4 M / 5 L = 1x10^-4 moles NaOH
1x10^-4 mol NaOH x (40.0 g/mol) = 0.004 g NaOH
density of NaOH = 2.13 g/mL; so 0.004 g / 2.13 g/mL = 0.0018 mL
And that's not right either...

I'm missing SOMETHING and I've been working on this problem forever. Please help?

 

[ally1h]

Ah... you know what, never mind. I flipped the equation for M1V1 = M2V2 incorrectly. I've got the right answer now.

Sorry for wasting space!

 

[Borek]

antilog (-3.3) = 5.0x10^-4 M

OK

m1v1 = m2v2
v2 = m1v1 / m2 = (0.250 M x 5000 mL) / (5.0x10^-4) = 2500000 mL
That's DEFINITELY not right...

Good, approach, but lousy realisation. Write which solution - volume & concentration - is 1, and which is 2.

density of NaOH = 2.13 g/mL; so 0.004 g / 2.13 g/mL = 0.0018 mL
And that's not right either...

That's completely off - 2.13 is a denstisty of solid NaOH, not solution.


Edit: my uplink went down for 10 minutes so you have answered your own question before I was able to post.