Calculate the pH value of the following
0.01 M NaOH.

Explanation

NaOH is a strong base and may be supposed to be completely ionised
` therefore " " [OH^(-) ] =[NaOH ]=0.01 M= 1.0 xx 10 ^(-2) " mol " L^(-1)`
Hence , `" " [H_3O^(+) ] =(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_3O^(+) ]`
` " " = - log _(10 ) (1.0 xx 10 ^(-12) )= 12`
Alternately : `" " pOH =-log _(10 ) [OH^(-) ] `
` " " = -log _(10) (1.0 xx 10 ^(-2) )=2`
` " " therefore pH= 14- pOH =14-2 =12`