Question

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

Answer 1

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 = $\frac{[Al(OH)_{2} ^-]}{[Al(OH)][OH^-]}$

The concentration of OH⁻ is:

pOH = 14 - pH = 3

pOH = -log [OH⁻]

[OH⁻] = 1x10⁻³

Thus:

40 = $\frac{[Al(OH)_{2} ^-]}{[Al(OH)][1x10^{-3}]}$

0,04M =  $\frac{[Al(OH)_{2} ^-]}{[Al(OH)]}$

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!