Answer:
0,040 M
Explanation:
The global reaction of the problem is:
Al(OH) (s) + OH⁻ ⇄ Al(OH)₂⁻(aq) K= 40
The equation of equilibrium is:
K = [tex]\frac{[Al(OH)_{2} ^-]}{[Al(OH)][OH^-]}[/tex]
The concentration of OH⁻ is:
pOH = 14 - pH = 3
pOH = -log [OH⁻]
[OH⁻] = 1x10⁻³
Thus:
40 = [tex]\frac{[Al(OH)_{2} ^-]}{[Al(OH)][1x10^{-3}]}[/tex]
0,04M = [tex]\frac{[Al(OH)_{2} ^-]}{[Al(OH)]}[/tex]
This means that 0,04 M are the number of moles that the solvent can dissolve in 1L, in other words, solubility.
I hope it helps!
