Question

calculate pH of 0.01 molar of NaOH solution.

Answer 1

Hey friend , Harish here.

Here is your answer:

                         NaOH  → Na⁺ + OH⁻ 
                             ↓             ↓         ↓
                        10⁻²M    10⁻²M  10⁻²M

Then [OH⁻] = 10⁻²

$p^{OH} = -\log 10^{-2} = -2\times (-1) =2$

⇒ $p^{H} = 14 - p^{OH}$

⇒  $14 - 2 = 12$

$\bold{Therefore\ p^{H} \ \boxed{12}}}$
_______________________________________________

Hope my answer is helpful to you.

Answer 2

NaOH is a strong base nd we have 0.01 M OH-ph means -log[H+]
so,pOH = -log 0.01M OH-

pOH = 2

pH + 2 = 14

pH = 12
therefore it is strong base