[xtheunknown0]

A solution of sodium hydroxide has pH of 10. 10 mL of this solution is mixed with 990 mL of water. The pH of the diluted solution is closest to...?

I seem to be stuck because I need to know the volume of the original solution (which I denote as v)

n_total(H3O+) = 10^(-10) * v mol
n_used(H3O+) = 10^(-12) * v mol
pH = -log(10^(-12) * v / 1.000)

What is your advice?

TIA,
xtheunknown0

[rabolisk]

Hint: You don't need to know the volume of the original solution.

[AWK]

10 + 990 = 1000
You diluted your solution 1000/10 = 10² time, hence pH = 10 + 2

[sjb]

10 + 990 = 1000
You diluted your solution 1000/10 = 10² time, hence pH = 10 + 2

Sure? Aren't you going the wrong way here?

[DevaDevil]

yes it is 10-2 (dilution brings it closer to neutral)

if you want to calculate using the concentrations:L

10 ml with pH 10; what is the concentration of OH- (or H+, whichever way you would like to calculate it)?
then concentration times volume (10 ml) gives the number of moles of your 10 ml sample.
then you know the new volume (1000 ml) and thus the new pH

[Vidya]

let me make it more clear
pH = 10
conc of H+ = 10^-10
M1 = 10^-10 M
V1= 10 ml
M2 =?
V2= 1000ml
use dilution law
M1V1=M2V2
take out new molarity
thats going to be H+ ion conc of new diluted soln
take out pH of the new soln 
new pH = 12

[sjb]

let me make it more clear
pH = 10
conc of H+ = 10^-10
M1 = 10^-10 M
V1= 10 ml
M2 =?
V2= 1000ml
use dilution law
M1V1=M2V2
take out new molarity
thats going to be H+ ion conc of new diluted soln
take out pH of the new soln 
new pH = 12

In the main, true, but don't forget the concentration of H+ in water is not 0.

[rabolisk]

pH is close to 8 not 12.

[AWK]

10 + 990 = 1000
You diluted your solution 1000/10 = 10² time, hence pH = 10 + 2

Sure? Aren't you going the wrong way here?

Sorry, of course should be 10 - 2