(d) In 0.001 M $NaOH$, $\left\lbrack OH^{-} \right\rbrack = 0.001 = 10^{- 3}$ $\therefore$ $\left\lbrack H^{+} \right\rbrack = \frac{10^{- 14}}{\lbrack OH^{-}\rbrack} = \frac{10^{- 14}}{10^{- 3}} = 10^{- 11}$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.