Question

If the concentration of [OH-] in the saturated
solution of Al(OH)3 is 3 x 10-5 mol L-1 then
solubility product of Al(OH)3 will be
(1) 2.7 x 10-18
(2) 2.7 x10-29
(3) 2.7 x 10-19
(4) 2.7 x 10-9

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Answer 1

Al(OH)₃ dissociates as follows,

Al(OH)₃ ⇔ Al³⁺ + 3 OH⁻

So, x moles of Al(OH)₃ dissociates to give x moles of Al and 3x moles of OH.

Given that [OH⁻] = 3 x 10⁻⁵ = 3x

x = 1 x 10⁻⁵ mol/lit

Solubility product, Ksp is given by product of the concentrations raised to their stoichiometric power.

i.e., Ksp = (x).(3x)³

= (1 x 10⁻⁵).(3 x 10⁻⁵)³

=27 x 10⁻²⁰

= 2.7 x 10⁻¹⁹ mol⁴ lit⁻⁴

So, we get the solubility product for the above dissociation as 2.7 x 10⁻¹⁹ mol⁴ lit⁻⁴

Hope it helps!!

Answer 2

Answer:

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