(a) How much aluminium hydroxide will dissolve in 500ml of water at #25^@C# given that #K_(sp)=3xx10^(-34)#?
(b) How much will dissolve in 500ml of a solution of 0.04M #Ba(OH)_2# ?

Question

Ellie Andrews · Accepted Answer

(a). #7.12xx10^(-8)"g"#

(b). #2.28xx10^(-28)"g"#

Section A:

When aluminum hydroxide separates:

#Al(OH)_(3(s))rightleftharpoons#

#Al_((aq))^(3+)+3OH_((aq))^(-)# #color(red)((1))#

Thus:

#K_(sp)=[Al_((aq))^(3+)][OH_((aq))^(-)]^3=3xx10^(-34)mol^3.l^(-3)# #color(red)((2))#

We can let#[Al_((aq))^(3+)]# #="s"#

We can see that #[OH_((aq))^(-)]="3s"#

Thus:

#sxx(3s)^(3)=3xx10^(-34)#

Thus:

#27s^(4)=3xx10^(-34)#

whereby:

#s=1.826xx10^(-9)"mol/l"#

#M_r=78#

#s=1.826xx10^(-9)xx78=142.3xx10^(-9)"g/l"#

To obtain the solubility in 500 milliliters, then

#=142.3/2xx10^(-9)=7.12xx10^(-8)"g"#

Section B

Now we are trying to dissolve the compound in a solution that already has a lot of #OH^-# ions.

From #color(red)((1))# we can see that Le Chatelier's Principle predicts that increasing #[OH_((aq))^-]# like this will shift the position of equilibrium to the left thus reducing the solubility of the compound.

This is known as "The Common Ion Effect" as the #OH^-# ions are common to both solutions.

We can now make an assumption that will make things a lot easier for ourselves. Because #K_(sp)# is so small we can assume that the concentration of #OH^(-)# from the #Al(OH)_3# is tiny compared with that from the #Ba(OH)_2#.

So we can set #[OH_((aq))^(-)]# as equal to #0.04xx2=0.08"mol/l"#

From #color(red)((2))rArr#

#3xx10^(-34)="s"xx(0.08)^(3)#

whereby:

#s=5.85xx10^(-30)"mol/l"#

#s=5.85xx10^(-30)xx78 = 4.563xx10^(-28)"g/l"#

Here, you can observe the significant reduction.

Thus, half of that will be soluble in 500 milliliters:

#=2.28xx10^(-28)"g"#

Wyatt Adkins · Answer

undefined (a) To find out how much aluminum hydroxide will dissolve in 500 ml of water at 25°C, we can use the solubility product constant (Ksp) and the molar solubility (s) of aluminum hydroxide. At equilibrium, the product of the concentrations of aluminum ions (Al3+) and hydroxide ions (OH-) should be equal to Ksp. Given that Ksp = 3 × 10^(-34), we can calculate the molar solubility (s) of aluminum hydroxide and then determine the mass of aluminum hydroxide that will dissolve.

(b) To find out how much aluminum hydroxide will dissolve in 500 ml of a solution of 0.04 M Ba(OH)2, we first need to calculate the concentration of hydroxide ions (OH-) using the given concentration of Ba(OH)2. Then, we can use this concentration to calculate the solubility product constant (Ksp) of aluminum hydroxide and determine the molar solubility (s). Finally, we can use the molar solubility to find the mass of aluminum hydroxide that will dissolve in the solution.

(a) How much aluminium hydroxide will dissolve in 500ml of water at #25^@C# given that #K_(sp)=3xx10^(-34)#? (b) How much will dissolve in 500ml of a solution of 0.04M #Ba(OH)_2# ?