The solubility of Al(OH)_3 in a buffer soluti | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The solubility equilibrium for $Al(OH)_3$ is given by: $Al(OH)_3(s) \rightleftharpoons Al^{3+}(aq) + 3OH^-(aq)$.Given $pH = 8.0$,we know $pOH = 14

Vedclass · Accepted Answer

$2.7 	imes 10^{-9} \ M$

Vedclass · Answer

(B) The solubility equilibrium for $Al(OH)_3$ is given by: $Al(OH)_3(s) ightleftharpoons Al^{3+}(aq) + 3OH^-(aq)$.Given $pH = 8.0$,we know $pOH = 14 - pH = 14 - 8 = 6.0$.Therefore,$[OH^-] = 10^{-pOH} = 10^{-6} \ M$.The solubility product expression is $K_{sp} = [Al^{3+}][OH^-]^3$.Let the solubility of $Al(OH)_3$ be $S$. Then $[Al^{3+}] = S$.Substituting the values: $2.7 	imes 10^{-27} = S 	imes (10^{-6})^3$.$2.7 	imes 10^{-27} = S 	imes 10^{-18}$.$S = \frac{2.7 	imes 10^{-27}}{10^{-18}} = 2.7 	imes 10^{-9} \ M$.

The solubility of $Al(OH)_3$ in a buffer solution at $pH = 8.0$ is [$K_{sp}$ of $Al(OH)_3 = 2.7 \times 10^{-27}$]

Similar Questions

The solubility product of $AgBr$ is $4.9 \times 10^{-13}$ at a certain temperature. Calculate the solubility.

The solubility of $AgCl$ is $1 \times 10^{-5} \ mol/L$. Its solubility in $0.1 \ M$ sodium chloride solution is

Solubility of $PbI_2$ is $0.005 \ M$. Then,the solubility product of $PbI_2$ is

The solubility product of $Ni(OH)_2$ at $298 \ K$ is $2 \times 10^{-15} \ mol^3 \ dm^{-9}$. The $pH$ value of its aqueous and saturated solution is

If the solubility of $Mg(OH)_2$ is $x \ mol/L$,then its solubility product $(K_{sp})$ is equal to .....

Vedclass Test Series

Exam Paper Generator

Online Exam Module