Question

Calculate the pH of 6.66 x ${10}^{-3}$ M solution of $AI(OH{)}_3$. its first dissociation is 100% where as second dissociation is 50% and third dissociation is negligible.

• A. 2
• B. 12
• C. 11
• D. 3

Answer 1

Answer: (B)

12

$Al(OH{)}_3⟶[Al(OH{)}_2{]}^{+}{+}^{-}OH$ $\alpha =1$

$\left[Al\right(OH{)}_2{]}^{+}⟶\left[Al\right(OH){]}^{2+}{+}^{-}OH$ ${\alpha }^{{}^{\prime }}=0.5$

$[AlOH{]}^{2+}⟶A{l}^{+3}{+}^{-}OH$ ${\alpha }^{{}^{\prime \prime }}=0$

As, $\alpha =\frac{x}{c}$ $\alpha$ = degree of dissociation, $x$ = concentration of ${}^{-}OH$, $c$ = concentration of $Al(OH{)}_3$

$\alpha =\frac{x}{6.66\times {10}^{-3}}=1$ $x=6.66\times {10}^{-3}M$

${\alpha }^{{}^{\prime }}=0.5=\frac{x^{{}^{\prime }}}{6.66\times {10}^{-3}}\Rightarrow x^{{}^{\prime }}=3.33\times {10}^{-3}M$

${\alpha }^{{}^{\prime \prime }}=0\Rightarrow \frac{x^{{}^{\prime }}}{6.66\times {10}^{-3}}=0$ $x^{{}^{\prime \prime }}=0$

So, Total concentration of ${[}^{-}OH]=x+x^{{}^{\prime }}=9.99\times {10}^{-3}M$

$pOH=-\log \left[O{H}^{-}\right]=-\log [9.99\times {10}^{-3}]=2.0$

as $pH-14-pOH=14-2=12.0$

$pH=12.0$