help pls If the molar solubility of aluminum hydroxide, Al(OH)3 is 2.6 x 10-9 M, what is the solubility product, Ksp ?

Question

If the molar solubility of aluminum hydroxide, Al(OH)₃ is 2.6 x 10^-9 M, what is the solubility product, K_sp ?

Answer 1

Hello, Ina,

The solubility product of a substance is defined for a solute dissolving in water and having the general dissociation equation of

aA(s) Ξ cC(aq) + dD(aq)

The concentrations of the dissolved species (C and D, here) are raised to a power equal to their coefficients in the balanced equation and then multiplies to produce the solubility product:

K_sp=[C]^c[D]^d

In this dissolution we have:

Al(OH)₃ Ξ Al³⁺ + 3OH^-

If the solubility of the aluminum hydroxide is 2.6x10^-9, we can say since that much dissolved, we should produce the following concentrations of Al³⁺ and (OH)^-:

Al³⁺ is 2.6x10^-9M

(OH)^- is 2 times 2.6x10^-9M, or 5.2x10^-9M (since 2 OH groups are produced for every 1 Al(OH)₃)

Now we can calculate K_sp

K_sp=[C]^c[D]^d

K_sp=[2.6x10^-9M]¹[5.2x10^-9M]³

K_sp = 3.66x10^-34

Wow. That's why it has the nickname, "Insoluble."

Bob

Answer 2

Al(OH)₃(s) <==> Al³⁺(aq) + 3OH^-(aq)

Ksp = [Al³⁺][OH-]³ = (2.6x10^-9)(7.8x10-9)³ = answer