Calculate pH of 0.0050M NaOH

How do I calculate the pH of 0.0050M NaOH ?
I understand how to do it when there is H in the substance, however this I am not sure about.
I thought that I could replace the H with Na, is that wrong ?
All help greatly appreciated,
xxx

NaOH is a strong base, meaning it will dissociate completely in aqueous solution.

You're familiar with pH = -log[H+] for acids

However, for bases, it's: pOH = -log[OH-]

To convert from pOH to pH, use:

pH = 14 - pOH

Hence.

[OH-] = 0.0050 M

pOH = -log(0.0050) = 2.30

pH = 14 - 2.30 = 11.7 ~ 12

[EDIT] You can't replace the H with Na. Don't do that in any other questions please.

NaOH <-----------> Na+ + OH-
1:1
Conc = 0.005M = 5 x 10^-3M
Kw = KaKb
p[H2O] = p[H+] + p[OH-]
Calculate p[OH-] first noting that p[H2O] = 14:
p[OH] = -log[OH] = -log[5 x 10^-3] = -0.6990 + 3 = 2.30103
Thus, p[H] = 14 - 2.30103 = 11.6990
pH = 11.7//