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A student found that 0.098 g0.098 \mathrm{~g} of PbF2\mathrm{PbF}_2 was dissolved in 400 . mL\mathrm{mL} of saturated PbF2\mathrm{PbF}_2. What is the KspK_{\mathrm{sp}} for the lead(II) fluoride?

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In this task, we need to calculate KspK_{sp}.

The solubility product constant is the product of the equilibrium concentrations of the precipitated ions raised to the potency of its stoichiometric factor in the equilibrium equation. For any compound of general formula:

AaBb(s)aA+(aq)+bB(aq)\text{A$_a$B$_b$(s)} \rightleftharpoons \text{aA$^+$(aq)} + \text{bB$^-$(aq)}

the constant of the solubility product is:

Ksp=[A+]a[B]bK_{sp}=[\text{A$^+$}]^a \cdot [\text{B$^-$}]^b

First we need to calculate moles using the mass and molar mass:

n=mMn=\dfrac{m}{M}

Then, using the moles and volume, we calculate concentration:

c=nVc=\dfrac{n}{V}

Using the mole ratio, we calculate concentration of ions.

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