Calculate the pH of 0.005 M NaOH solution.

Explanation

`NaOH overset(aq)rarr Na^(+)(aq) + OH^(-)(aq)`
Now `[H^(+)][OH^(-)] = 1 xx 10^(-14)`
`[H^(+)] = (1 xx 10^(-14))/([OH^(-)]) = -log (2 xx 10^(-12))`
`= -log 2 - log 10`
`= -3.010 + 12 = 11.6990`