(b) $\left\lbrack OH^{-} \right\rbrack = 1 \times 10^{- 5}$ $pOH = - \log{\lbrack OH^{-}\rbrack}$ $= - \log{\lbrack 1 \times 10^{- 5}\rbrack}$ $pOH$ = 5 $pH + pOH = 14$ $pH = 14 - pOH$ $= 14 - 5$ $pH = 9$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.