What is the value of pH of a NaOH solution that dissociates 2 % in its 0.01 M solution?

Step-by-step solution

The NaOH solution has an initial concentration of 0.01 M with 2 % dissociation. The hydroxide ion concentration is therefore [ OH ^-] = 0.01 0.02 = 2 10⁻⁴ M . Using the logarithmic form for pOH gives pOH = - (2 10⁻⁴) . Breaking down the logarithm: pOH = -( 2 + 10⁻⁴) = - 2 + 4 Substituting 2 0.301 yields pOH -0.301 + 4 = 3.699 . Since pH + pOH = 14 , we find pH = 14 - 3.699 = 10.301 . Final answer: B