Calculate the pH value of the following
0.01 M NaOH.

Explanation

NaOH is a strong base, it is assumed to be completely ionised. Therefore, `[OH^(-)]= [NaOH] = 0.01M = 1.0 xx 10^(-2) mol L^(-1)`
Hence, `[H_(3)O^(+)] = (K_(w))/([OH^(-)])= (1.0 xx 10^(-14))/(1.0 xx 10^(-2))`
`=1.0 xx 10^(-12) mol L^(-1)`
`pH= -log_(10) [H_(3)O^(+)]`
`=-log_(10) (1.0 xx 10^(-12))`= 12 or `pOH= - log_(10) [OH^(-)]`
`= -log_(10)(1.0 xx 10^(-2))`
=2
Therefore, `pH= 14-pOH`
`=14-2`
=12