Question

Calculate the pH of $6.66\times {10}^{-3}M$ solution $Al(OH{)}_3$ its first dissociation is 100% whereas second dissociation is 50% and third dissociation is negligible.

• A. 7
• B. 12
• C. 14
• D. less than 7

Answer 1

The correct option is B 12

$Al(OH{)}_3\rightleftharpoons Al(OH{)}_2^{+}+O{H}^{-}\left[O{H}^{-}\right]=6.66\times {10}^{-3}(\alpha =100\%)Al(OH{)}_2^{+}\rightleftharpoons Al(OH{)}^{+2}+O{H}^{-}\left[O{H}^{-}\right]=3.33\times {10}^{-3}(\alpha =50\%)$
Total $[OH{]}^{-}=6.66\times {10}^{-3}+3.33\times {10}^{-3}=10\times {10}^{-3}$
Total $[OH{]}^{-}={10}^{-2}$
$pOH=-log\left[O{H}^{-}\right]$
$pOH=-log{10}^{-2}$
$pOH=2$
also, $pH+pOH=14$
$pH+2=14$
$pH=12$
$\therefore pH=12$