Question

What is the Ksp expression for the salt PbI2?

Option 1) [Pb2+][I-]2

Option 2) [Pb2+][I2]2

Option 3) [Pb2+][2I-]

Option 4) [Pb2+][2I-]2

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Answer

The correct option is:

Option 1: $$[Pb^{2+}][I^{-}]^{2}$$

To understand why this is the correct expression, we need to consider the dissociation of the salt, lead(II) iodide (PbI$_2$), in water:

$$\text{PbI}_2 \rightleftharpoons \text{Pb}^{2+} + 2\text{I}^{-}$$

From the dissociation, we can observe that one mole of PbI$_2$ produces one mole of Pb$^{2+}$ ions and two moles of I$^{-}$ ions. The solubility product constant (K$_{sp}$) expression for this equilibrium is derived from the concentrations of the ions produced:

$$\text{K}_{sp} = [\text{Pb}^{2+}][\text{I}^{-}]^{2}$$

This expression accounts for the stoichiometry of the dissociation reaction, specifically, the squared term for the I$^{-}$ concentration reflects that there are two iodide ions for every formula unit of PbI$_2$ that dissolves. Hence, option 1 correctly represents the K$_{sp}$ expression.


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