Question

The solubility of PbI2 (formula weight = 461) is 0.63 g/L. What is the solubility product?

Can someone explain what solubility product means?

What equation do I use? And what is the step by step process?

The answer is: 4.0x10-7

Answer #1

Given: Solubility = 0.63 g/L

Molar mass of PbI2 = 461 g per mol

Solution:

To calculate ksp, we have to show dissociation.

PbI2 (s) ---- > Pb^{2+} (aq) + 2I- (aq)

We know ksp is the ratio of the product of concentration raised to power = coefficient and product of concentration of reactant at equilibrium.

Ksp = [Pb^{2+}] [ I^{-}]^{2}

Solid phase does not involved in the ksp.

Lets make ICE chart

PbI2 (s) ---- > Pb^{2+} (aq) + 2I- (aq)

I 0 0

C -x (molar +x +2x

Solubility)

E +x +2x

Ksp = x * (2x)^{2}

We calculate value of x (molar solubility)

To calculate value of molar solubility, we divide the g/L (gram solubility ) by molar mass of PbI2

Molar solubility = (0.63 g/L) / 461 g per mol

= 0.001366 M or mol/L

We plug this value in ksp expression

ksp = (0.001366) * 4 * (0.001366)^2 = 1.021 E-8

ksp of **PbI2 = 1.021 E-8**

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