# How do you write the Ksp expression for lead chromate (PbCrO_4) and calculate its solubility in mol/L? Ksp= 2.3 *10^-13

Jul 3, 2016

You can do it like this:

#### Explanation:

In a saturated solution of lead chromate the solid is in equilibrium with its ions:

$\textsf{P b C r {O}_{4 \left(s\right)} r i g h t \le f t h a r p \infty n s P {b}_{\left(a q\right)}^{2 +} + C r {O}_{4 \left(a q\right)}^{2 -}}$

For which:

$\textsf{{K}_{\left(s p\right)} = \left[P {b}_{\left(a q\right)}^{2 +}\right] \left[C r {O}_{4 \left(a q\right)}^{2 -}\right] = 2.3 \times {10}^{- 13} {\text{ ""mol"^2."l}}^{- 2}}$

This is temperature dependent and the temperature should have been specified in the question.

From the equilibrium you can see that $\textsf{\left[P {b}_{\left(a q\right)}^{2 +}\right] = \left[C r {O}_{4 \left(a q\right)}^{2 -}\right]}$

$\therefore$$\textsf{{\left[P {b}_{\left(a q\right)}^{2 +}\right]}^{2} = 2.3 \times {10}^{- 13} {\text{ ""mol"^2."l}}^{- 2}}$

$\therefore$$\textsf{\left[P {b}_{\left(a q\right)}^{2 +}\right] = \sqrt{2.3 \times {10}^{- 13}} = 4.8 \times {10}^{- 7} \text{ ""mol/l}}$

Since lead chromate is 1 molar with respect to lead(II) ions, then this is also equal to the solubility at that particular temperature.