Question

At what minimum $pH$, will $1.0\times {10}^{-3}$ mol of $Al{\left(OH\right)}_3$ go into $1L$ solution as $\left[Al{\left(OH\right)}_4^{-}\right]$?
Given : $K_{sp}\left[Al{\left(OH\right)}_3\right]=5.0\times {10}^{-33}$ and for $\left[Al{\left(OH\right)}_4^{-}\right]\leftrightharpoons {Al}^{3+}+4{OH}^{-}$

$K=1.3\times {10}^{-34}$ (write the value to the nearest integer).

Answer 1

$Al{\left(OH\right)}_3\rightleftharpoons {Al}^{+3}+3{OH}^{-};$ $K_{sp}$

$Al{\left(OH\right)}_4^{-}\rightleftharpoons {Al}^{+3}+4{OH}^{-};$ $K$

$Al{\left(OH\right)}_3+{OH}^{-}\rightleftharpoons Al{\left(OH\right)}_4^{-}$

$\frac{K_{sp}}{K}=38.46$ $=\left[\frac{Al{\left({OH}^{-}\right)}_4}{\left[{OH}^{-}\right]}\right]=\frac{{10}^{-3}}{\left[{OH}^{-}\right]}$

$\left[{OH}^{-}\right]=2.6\times {10}^{-5}$

$pOH=4.585$

$pH=9.415$

Hence, the answer to the nearest integer is $9$.