To 100 ml of a solution, which contains 8.32 ×10^-3 g lead ions, 10^-4 moles of H2 SO4 is added

step 1 Dear students, in the given problem, problem number 129 of chapter 5th, we have asked that KSP i

To 100 ml of a solution, which contains 8.32 ×10^-3 g lead...

To $100 \mathrm{ml}$ of a solution, which contains $8.32 \times 10^{-3} \mathrm{~g}$ lead ions, $10^{-4}$ moles of $\mathrm{H}_{2} \mathrm{SO}_{4}$ is added. How much lead remains in the solution unprecipitated? $K_{\text {sp }}$ of $\mathrm{PbSO}_{4}=1.6 \times 10^{-7} \cdot(\mathrm{Pb}=208)$ (a) $4 \times 10^{-4} \mathrm{~g}$ (b) $2.67 \times 10^{-4} \mathrm{~g}$ (c) $2 \times 10^{-4} \mathrm{~g}$ (d) $4.16 \times 10^{-3} \mathrm{~g}$

To $100 \mathrm{ml}$ of a solution, which contains $8.32 \times 10^{-3} \mathrm{~g}$ lead ions, $10^{-4}$ moles of $\mathrm{H}_{2} \mathrm{SO}_{4}$ is added. How much lead remains in the solution unprecipitated? $K_{\text {sp }}$ of $\mathrm{PbSO}_{4}=1.6 \times 10^{-7} \cdot(\mathrm{Pb}=208)$
(a) $4 \times 10^{-4} \mathrm{~g}$
(b) $2.67 \times 10^{-4} \mathrm{~g}$
(c) $2 \times 10^{-4} \mathrm{~g}$
(d) $4.16 \times 10^{-3} \mathrm{~g}$

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Key Concepts

Solubility Product (Ksp)

The solubility product is the equilibrium constant for the dissolution of a sparingly soluble salt. It quantifies the low solubility by expressing the product of the ion concentrations, each raised to the power of its stoichiometric coefficient, and is used to predict whether a precipitate will form under given conditions.

Precipitation Reactions

Precipitation reactions occur when two soluble substances react to form an insoluble salt, which appears as a solid precipitate. The reaction progresses until the ion product reaches or exceeds the solubility product, at which point the system reaches a new equilibrium.

Common Ion Effect

The common ion effect refers to the decrease in solubility of a sparingly soluble salt when an ion common to the salt is added from another source. This effect shifts the dissolution equilibrium, reducing the amount of the salt that can remain in solution.

Stoichiometry and Mass-to-Mole Conversion

Stoichiometry involves calculating the relative quantities of reactants and products in a chemical reaction. Converting mass to moles using the molar mass of a substance is a key step in determining the number of moles present, which is essential for accurately evaluating the reaction’s progress and equilibrium concentrations.

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