Q. What mole of $Ca(OH)_{2}$ is dissolved in 250 mL aqueous solution to given a solution of $pH$ 10.65, assuming full dissociation?
A) $0.47 \times 10^{- 4}$
B) $0.48 \times 10^{- 4}$
C) $0.56 \times 10^{- 4}$
D) $0.58 \times 10^{- 4}$
Explanation
(c)
$pOH = 14 - pH = 14 - 10.65 = 3.35$
$pOH = - \log\left\lbrack OH^{-} \right\rbrack$
$- 3.35 = \log{\lbrack OH^{-}\rbrack}$
$\therefore\left\lbrack OH^{-} \right\rbrack = 4.447 \times 10^{- 4}$ M
$Ca(OH)_{2} = Ca^{2 +} + \ 2OH^{-}$
$4.447 \times 10^{- 4}\ M$
$\frac{\left\lbrack OH^{-} \right\rbrack}{2} = \left\lbrack Ca(OH)_{2} \right\rbrack = 2.2235 \times 10^{- 4}$ M
Hence, in 250 mL, moles of
$Ca(OH)_{2} = \frac{2.2235 \times 10^{- 4} \times 250}{1000}$
$= 0.56 \times 10^{- 4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.