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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? :)