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Question: The solubility of \(Pb{{(OH)}_{2}}\) in water is \(6.7\text{ x 1}{{\text{0}}^{-6}}\) M. Calculate th...

The solubility of Pb(OH)2Pb{{(OH)}_{2}} in water is 6.7 x 10−66.7\text{ x 1}{{\text{0}}^{-6}} M. Calculate the solubility of Pb(OH)2Pb{{(OH)}_{2}} in a buffer solution pH = 8.
(a)- 3.2 x 10−33.2\text{ x 1}{{\text{0}}^{-3}}
(b)- 12 x 10−312\text{ x 1}{{\text{0}}^{-3}}
(c)- 1.2 x 10−31.2\text{ x 1}{{\text{0}}^{-3}}
(d)- 6 x 10−46\text{ x 1}{{\text{0}}^{-4}}

Explanation

Solution

When Pb(OH)2Pb{{(OH)}_{2}} dissociates, then there will the formation of three ions, so the solubility product for the compound will be:
Ksp=S x 2S{{K}_{sp}}=S\text{ x }2S
We can also use another formula to solve this question:
pH + pOH = 14.0

Complete step by step solution: When we take the product of the concentration of the ions in a saturated solution, in which all the concentrations are raised to the power equal to the number of ions produced, it is known as solubility product.
The given compound is Pb(OH)2Pb{{(OH)}_{2}}, so the ions produced in this compound will be:
Pb(OH)2⇌Pb2++2OH−Pb{{(OH)}_{2}}\rightleftharpoons P{{b}^{2+}}+2O{{H}^{-}}
Therefore, the solubility of Lead ions will be S and the solubility of hydroxyl ions will be 2S. Hence we can write the solubility product of the compound will be:
Ksp=S x (2S)2{{K}_{sp}}=S\text{ x (}2S{{)}^{2}}
We are given the value of solubility of Pb(OH)2Pb{{(OH)}_{2}} in water as 6.7 x 10−66.7\text{ x 1}{{\text{0}}^{-6}} M. Putting the value, we get:
Ksp=(6.7 x 10−6) x (2 x 6.7 x 10−6)2=1.2 x 10−15{{K}_{sp}}=(6.7\text{ x 1}{{\text{0}}^{-6}})\text{ x (2 x 6}\text{.7 x 1}{{\text{0}}^{-6}}{{)}^{2}}=1.2\text{ x 1}{{\text{0}}^{-15}}
So, the value of solubility product is 1.2 x 10−151.2\text{ x 1}{{\text{0}}^{-15}}
We are given the pH of the buffer solution as 8. From this we can calculate the pOH, by using the formula:
pH + pOH = 14.0
the pOH will be:
pOH = 14.0 – 8.0 = 6.0
so, we can write:
[OH−]=10−pOH=10−6[O{{H}^{-}}]={{10}^{-pOH}}={{10}^{-6}}
Now, we have the value of the solubility product and the concentration of hydroxyl ions in the buffer solution, therefore, can calculate the solubility of the lead ions by:
Ksp=[Pb2+][OH−]2{{K}_{sp}}=[P{{b}^{2+}}]{{[O{{H}^{-}}]}^{2}}
Putting the values, in this, we get:
1.2 x 10−15=[Pb2+] x (10−6)21.2\text{ x 1}{{\text{0}}^{-15}}=[P{{b}^{2+}}]\text{ x (1}{{\text{0}}^{-6}}{{)}^{2}}
[Pb2+]=1.2 x 10−3[P{{b}^{2+}}]=1.2\text{ x 1}{{\text{0}}^{-3}} M
So, the solubility of Pb(OH)2Pb{{(OH)}_{2}} in the buffer solution is 1.2 x 10−31.2\text{ x 1}{{\text{0}}^{-3}}M

Therefore, the correct answer is an option (c).

Note: When you write the dissociation of any product, then write the balanced reaction, because if you don't balance the reaction, then the value of the solubility product will be wrong for the compound.