Question

3. The solubility of lead iodide in water is 0.63 g/litre. Calculate the solubility product of lead iodide. Atof Pb = 207, I = 127)​

Answer 1

Explanation:

PbI

2

⇌Pb

2+

+2I

−

x 2x

given solubility of PbI

2

is 0.7

L

g

also, molar mass of PbI

2

=461.2g/mol

⇒[PbI

2

]=

461.2

0.7

mol/Litre=1.5×10

−3

solubility product, K

sp

=x(2x)

2

K

sp

=4x

3

K

sp

=4(1.5×10

−3

)

3

∴ Solubility product of PbI

2

=135×10

−9

mol

2

/lit

2