A 150mL sample of NaOH has the pH of 9.5. On a Bunsen burner, 100mL of water was boiled from the NaOH solution. What is the expected pH of the solution?

Find the [OH-] in the solution. The pH is 9.5, so the pOH is 14 - 9.5 = 4.5.
[OH-] = 10^-4.5 M
Now use the dilution equation to find the new [OH-] after the volume is reduced from 150 mL to 50 mL:
M1V1 = M2V2
M1 = 10^-4.5 M
V1 = 150 mL
M2 = ?
V2 = 50 mL

(10^-4.5)(150) = M2(50)
M2 = 9.5 x 10^-5 M ≈ 1 • 10^-4 (We can only use one sig fig, because the pH was given to one decimal place.)

Now use this [OH-] to find pOH:
pOH = -log(1 x 10^-4) = 4.0

14 - pOH = pH, so the expected pH for the new solution is 10.

Ithinks its false. tell me if im wrong