Question

The solubility of lead (ii) chloride (pbcl2) is 1.6 ⋅ 10-2 m. what is the ksp of pbcl2? the solubility of lead (ii) chloride (pbcl2) is 1.6 10-2 m. what is the ksp of pbcl2? 3.1 ⋅ 10-7 5.0 ⋅ 10-4 1.6 ⋅ 10-5 1.6 ⋅ 10-2 4.1 ⋅ 10-6

Answer 1

Answer:The Solubility product of
`PbCl_2` is
`1.63* 10^(-5)`.

Explanation;

Solubility of
`PbCl_2,S=1.6* 10^(-2) M`

`PbCl_2\rightleftharpoons Pb^(2+)+2Cl^-`

S 2S

The expression of
`K_(sp)` is given as:

`K_(sp)=S* (2S)^2=4S^3`

`K_(sp)=4* (1.6* 10^(-2))^3=1.63* 10^(-5)`

The Solubility product of
`PbCl_2` is
`1.63* 10^(-5)`.

Answer 2

According to the equilibrium equation for this reaction:

PbCl2(s) ↔ Pb2+(aq) + 2Cl-(aq)

So when Ksp is the solubility product constant for a solid substance when it

dissolved in the solution. and measure how a solute dissolves in the solution

So, Ksp expression = [Pb2+] [Cl-]^2

and when the solubility is the maximum quantity of solute which can dissolve in a certain solute.

So, we assume the solubility = X

∴[Pb2+] = X = 1.6 x 10^-2 M

[Cl-] = 2X = 2 * 1.6 x 10^-2 = 0.032 M

by substitution:

∴ Ksp = (1.6 x 10^-2) * (0.032)^2

= 1.64 x 10^-5