The value of the equilibrium constant for the following chemical equation is kr 40 at 25°c. calculate the solubility of al(oh)s(s) in an aqueous solution buffered at ph 11.00 at 25°c. al(oh)(s)oh (aq)al(oh)l'(aq) + a) 4.0 x 1010 m b) 6.3 m d) 0.37 m c) 0.040 m

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 =

The concentration of OH⁻ is:

pOH = 14 - pH = 3

pOH = -log [OH⁻]

[OH⁻] = 1x10⁻³

Thus:

40 =

0,04M =  

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!