[limekiwi]

I need to find the pH of a solution obtained by 125mL of 0.606M NaOH diluted to 15L of water.
This is what I did...

Amount of OH:
(125 mL) x (0.606 mmol NaOH / 1 mL soln) x (1 mmol OH- / 1 mmol NaOH) = 75.75 mmol

The total amount of solution:
125 mL + 15 000 mL = 15 125 mL

The concentration of OH-:
[OH-] = (75.75 mmol OH- / 15 125 mL) = 5.008 x 10^-3 M

Do I plug this value into pOH = -log[OH-]? Then use 14 - pOH = pH?
I'm not sure if I approached the problem correctly... any suggestions?  :-)

[macman104]

Didn't double check the math, but the approach seems to be good.  Well done!

[DrCMS]

You have not read the question properly

I need to find the pH of a solution obtained by 125mL of 0.606M NaOH diluted to 15L of water.

I've highlighted the relevent word

So this next bit you did is wrong

The total amount of solution:
125 mL + 15 000 mL = 15 125 mL

Total amount is 15L apart from that you did it correctly.