Q. 0.01 mole of lime (CaO) was dissolved in $100\ cm^{3}$ of water. Assuming the base is completely ionised in the solution, the pH of the solution will be
A) 13.3
B) 8.5
C) 6
D) 8
Explanation
(a)
$\left\lbrack Ca(OH)_{2} \right\rbrack = 0.01\ mol/100\ cc\ = 0.1\ mol\ L^{- 1}$
Assuming complete ionisation,
$\left\lbrack OH^{-} \right\rbrack = 0.2\ M,\left\lbrack H^{+} \right\rbrack = \frac{10^{- 14}}{0.2} = 5 \times 10^{- 14}$
$pH = - \log\left\lbrack H^{+} \right\rbrack$
$= - \log{(5 \times 10^{- 14}})$
$= 14 - 0.669 = 13.3$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.