 # Enter the solubility-product expression for Al(OH)3(s). [Al^{3+}][O Ernest Ryland 2021-12-16 Answered
Enter the solubility-product expression for $Al\left(OH\right)3\left(s\right)$.
$\left[A{l}^{3+}\right]{\left[O{H}^{-}\right]}^{3}$
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it Daniel Cormack
Step 1
S be the solubility of $Al{\left(OH\right)}_{3}$
${K}_{sp}=\left[A{l}^{3+}\right]{\left[O{H}^{-}\right]}^{3}=\left(S\right){\left(3S\right)}^{3}=27{S}^{4}$
${S}^{4}=\frac{{K}_{sp}}{27}=\frac{27×{10}^{-11}}{27×10}=1×{10}^{-12}$

Step 2
Solution of $Al{\left(OH\right)}_{3}$: Molar mass of $Al{\left(OH\right)}_{3}$ 78 g. Therefore, solubility of $Al{\left(OH\right)}_{3}$ in
$g{L}^{-1}=1×{10}^{3}×78g{L}^{-1}=78×{10}^{-3}g{L}^{-1}=7.8×{10}^{-2}g{L}^{-1}$
Step 3
pH of the solution:
$\left[OH\right]=3S=3×1×{10}^{-3}=3×{10}^{-3}$
$p\left[OH\right]=3-\mathrm{log}3$
$pH=14-pOH=11+\mathrm{log}3=11.4771$

We have step-by-step solutions for your answer! vrangett

Step 1
Suppose the solubility is S mol ${L}^{-1}$. Then
$Al\left(OH{\right)}_{3}⇒\begin{array}{cc}A{l}^{3+}& +3O{H}^{-}\\ S& 3S\end{array}$
${K}_{sp}=S×{\left(3s\right)}^{3}=27{S}^{4}$
$\therefore 27{S}^{4}=2.7×{10}^{-11}$ or ${S}^{4}={10}^{-12}$ or
Molar mass of $Al{\left(OH\right)}_{3}$ in $g{L}^{-1}={10}^{-3}×78=7.8×{10}^{-2}g{L}^{-1}$

$\therefore pOH=-\mathrm{log}\left(3×{10}^{-3}=3-0.4771=3.5229$
$pH=14-2.5229=11.4771$

We have step-by-step solutions for your answer! Don Sumner

Step 1
The equation is as follows:
$Al\left(OH{\right)}_{3}\left(s\right)$
$A{L}^{3+}\left(aq\right)+3O{H}^{-}\left(aq\right)$
The solubility equation of $Al\left(OH{\right)}_{3}\left(s\right)$ is given above. The ionization of $AL\left(OH{\right)}_{3}\left(s\right)$ gives in water. The coefficient of $A{l}^{3+}$ ion is one whereas the coefficient of $O{H}^{-}$ ion is three.
Step 2
The solubility product expression ${K}_{sp}$ is as follows:
${K}_{sp}=\left[A{l}^{3+}\right]\left[O{H}^{-}{\right]}^{3}$
Solubility-product expression for $Al\left(OH{\right)}_{3}\left(s\right)$ is ${K}_{sp}=\left[A{l}^{3+}\right]\left[O{H}^{-}{\right]}^{3}$
NSK
${K}_{sp}$ of $Al\left(OH{\right)}_{3}\left(s\right)$ is the product of molar solubility of ions $A{l}^{3+}$ and $O{H}^{-}$ formed in the ionization of $Al\left(OH{\right)}_{3}\left(s\right)$. Therefore, the ${K}_{sp}$ expression for $Al\left(OH{\right)}_{3}\left(s\right)$ is ${K}_{sp}=\left[A{l}^{3+}\right]\left[O{H}^{-}{\right]}^{3}$
NSK
Solubility-product expression for $Al\left(OH{\right)}_{3}\left(s\right)$ is ${K}_{sp}=\left[A{l}^{3+}\right]\left[O{H}^{-}{\right]}^{3}$

We have step-by-step solutions for your answer!