Question

Calculate the solubility of PbF2 in water at . You'll find data in the ALEKS Data tab. Round your answer to significant digits.

Answer 1

0.5 g/L.

Step-by-step explanation:

Hello,

In this case, for this solubility problem, we can write for the lead (II) fluoride:

`PbF_2(s)\rightleftharpoons Pb^(2+)(aq)+2F^-(aq)`

And the equilibrium expression is:

`Ksp=[Pb^(2+)][F^-]^2`

Whereas Ksp of lead (II) fluoride is 3.3x10⁻⁸. In such a way, we can write the equilibrium expression in terms of the molar solubility
`x` as follows:

`Ksp=(x)(2x)^2=3.3x10^(-8)`

Hence, solving for
`x` we find:

`x=\sqrt[3]{(3.3x10^(-8))/(4) }\\\\x=2.02x10^(-3)M`

Moreover, since the molar mass of lead (II) fluoride is 245.2 g/mol, the solubility turns out:

`2.02x10^(-3)(molPbF_2)/(L)*(245.2gPbF_2)/(1molPbF_2)\\ \\0.5(g)/(L)`

Best regards.