# Using the following data, how do you calculate the Ksp value given that the solubility of "Pb"_3("PO"_4)_2 is 6.2xx10^(-12) "mol/L" ?

Jun 26, 2016

${K}_{s p} = 9.9 \cdot {10}^{- 55}$

#### Explanation:

You know that you're dealing with a solution of lead(II) phosphate, an insoluble ionic compound that does not dissociate completely when dissolved in water.

The first thing to do here is write the equilibrium reaction that describes the partial dissociation of the salt.

${\text{Pb"_ 3("PO"_ 4)_ (2(s)) rightleftharpoons color(blue)(3)"Pb"_ ((aq))^(2+) + color(red)(2)"PO}}_{4 \left(a q\right)}^{3 -}$

Notice that for every mole of lead(II) phosphate that dissociates in solution you get $3$ moles of lead(II) cations, ${\text{Pb}}^{2 +}$, and $3$ moles of phosphate anions, ${\text{PO}}_{4}^{3 -}$.

Now, the molar solubility of lead(II) phosphate in water is said to be equal to $6.2 \cdot {10}^{- 12} {\text{mol L}}^{- 1}$.

This means that in one liter of water, presumably at room temperature, you can only hope to dissolve $6.2 \cdot {10}^{- 12}$ moles of lead(II) phosphate.

The concentration of the resulting ions will be

["Pb"^(2+)] = color(blue)(3) xx 6.2 * 10^(-12)"mol L"^(-1) = 1.86 * 10^(-11)"mol L"^(-1)

["PO"_4^(3-)] = color(red)(2) xx 6.2 * 10^(-12)"mol L"^(-1) = 1.24 * 10^(-11)"mol L"^(-1)

The solubility product constant, ${K}_{s p}$, is defined as

${K}_{s p} = {\left[{\text{Pb"^(2+)]^color(blue)(3) * ["PO}}_{4}^{3 -}\right]}^{\textcolor{red}{2}}$

Plug in the values you have for the concentrations of the two ions to get

${K}_{s p} = {\left(1.86 \cdot {10}^{- 11}\right)}^{\textcolor{b l u e}{3}} \cdot {\left(1.24 \cdot {10}^{- 11}\right)}^{\textcolor{red}{2}}$

${K}_{s p} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{9.9 \cdot {10}^{- 55}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to tow sig figs, the number of sig figs you have for the molar solubility of lead(II) phosphate.