Question

The $pH$ of $0.001MBa(OH{)}_2$ solution will be

• A. $2$
• B. $8.4$
• C. $11.3$
• D. $2.7$

Answer 1

Answer: (C)

$Ba(OH{)}_2\rightleftharpoons B{a}^{2+}+2O{H}^{-}$
$\left[O{H}^{-}\right]=2\times 1\times 10^{-3}M$
$pOH=-log\left[O{H}^{-}\right]$
$=-log\left(2\times 10^{-3}\right)=2.7$
$pOH+pH=14$
$pH=14-2.7=11.3$