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

Answer:

0.5 g/L.

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:

$x=\sqrt[3]{\frac{3.3x10^{-8}}{4} }\\\\x=2.02x10^{-3}M$

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

$2.02x10^{-3}\frac{molPbF_2}{L}*\frac{245.2gPbF_2}{1molPbF_2}\\ \\0.5\frac{g}{L}$

Best regards.