TomRiddle369
Jan 29, 2011, 03:41 AM
The pH of a dissolution of NaOH is 13, which Volume of water do I have to add to make the pH 12?
Unknown008
Jan 29, 2011, 09:42 AM
You need to write down the relevant equations.
NaOH \rightarrow Na^+ + OH^-
H_2O \rightleftharpoons H^+ + OH^-
You know the pH of the NaOH, you can find the concentration of OH^- in the solution.
Remember that:
pH + pOH = 14
pH = -\log[H^+]
pOH = -\log [OH^-]
K_w = 1.0 \times 10^{-14} = [H^+][OH^-]
Can you give your problem a try now? :)