Question

What is the molar solubility of PbI2 in a .10M NaI solution?

Pbi2 ksp =7.9 x 10 -9

Answer 1

Answer on the question #48495, Chemistry, Other

Question:

What is the molar solubility of PbI₂ in a .10M NaI solution?

PbI₂ ksp = 7.9 × 10⁻⁹

Solution:

The molar solubility is the molar concentration of saturated solution of PbI₂ at fixed temperature.

$$\mathrm{PbI}_2 = \mathrm{Pb}^{2+} + 2\mathrm{I}^-$$

C / mol L⁻¹ 0 0.1

D +x +x

[ ] x 0.1+x

Solubility product is:

$$K_{sp} = [Pb^{2+}][I^-]^2$$ $$7.9 * 10^{-9} = x(0.1 + x)$$

As x << 0.1, then:

$$x = \frac{7.9 * 10^{-9}}{0.1}$$ $$x = 7.9 * 10^{-8}$$

Then, equilibrium concentration of Pb²⁺ is 7.9 * 10⁻⁸ mol L⁻¹.

$$c(Pb^{2+}) = c(PbI_2) = 7.9 * 10^{-8} \text{ mol L}^{-1}$$

Answer: 7.9 * 10⁻⁸ mol L⁻¹

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