Regarding the Ksp for aluminum hydroxide AlOH3 which is 13 x

Regarding the Ksp for aluminum hydroxide Al(OH)₃ which is 1.3 x 10^-33, what would be the hydroxide ion concentration were the aluminum ion concentration to be adjusted to .05 M?

Al(OH)3 (s) --------------------> Al+3 (aq) + 3 OH- (aq)

is the solubility equilibrium expression.

Thus Ksp = [Al+3][OH-]³

Given [Al+3] = 0.5M

Hence Ksp = 1.3 x10^-33 = 0.5 x[OH-]³

thus [OH-] =1.375x10^-11 M

Regarding the Ksp for aluminum hydroxide Al(OH)3 which is 1.3 x 10-33, what would be the hydroxide ion concentration were the aluminum ion concentration to be a

Regarding the Ksp for aluminum hydroxide AlOH3 which is 13 x

Solution

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