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antoniya [11.8K]
3 years ago
5

The solubility product of PbI2 is 7.9x10^-9. What is the molar solubility of PbI2 in distilled water?

Chemistry
1 answer:
Agata [3.3K]3 years ago
3 0

<u>Answer:</u> The molar solubility of PbI_2 is 1.25\times 10^{-3}mol/L

<u>Explanation:</u>

Solubility is defined as the maximum amount of solute that can be dissolved in a solvent at equilibrium.

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The balanced equilibrium reaction for the ionization of calcium fluoride follows:

PbI_2\rightleftharpoons Pb^{2+}+2I^-

                s       2s

The expression for solubility constant for this reaction will be:

K_{sp}=[Pb^{2+}][I^-]^2

We are given:

K_{sp}=7.9\times 10^{-9}

Putting values in above equation, we get:

7.9\times 10^{-9}=(s)\times (2s)^2\\\\7.9\times 10^{-9}=4s^3\\\\s=1.25\times 10^{-3}mol/L

Hence, the molar solubility of PbI_2 is 1.25\times 10^{-3}mol/L

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