A 1.0 MN a2 SO4 solution is slowly added to 10.0 mL of a solution that is 0.20 M in Ca^2+ and 0.

step 1 So now we'll work on problem 76 from chapter 17. In this problem, we're told that a one molar so

A 1.0 MN a2 SO4 solution is slowly added to 10.0 mL of a...

A 1.0 $\mathrm{MN} \mathrm{a}_{2} \mathrm{SO}_{4}$ solution is slowly added to 10.0 $\mathrm{mL}$ of a solution that is 0.20 $\mathrm{M}$ in $\mathrm{Ca}^{2+}$ and 0.30 $\mathrm{M}$ in $\mathrm{Ag}^{+} .$ (a) Which compound will precipitate first: $\mathrm{CaSO}_{4}\left(K_{\mathrm{sp}}=2.4 \times 10^{-5}\right)$ or $\mathrm{Ag}_{2} \mathrm{SO}_{4}\left(K_{s p}=1.5 \times 10^{-5}\right) ?(\mathbf{b})$ How much $\mathrm{Na}_{2} mathrm{SO}_{4}$ solution must be added to initiate the precipitation?

A 1.0 $\mathrm{MN} \mathrm{a}_{2} \mathrm{SO}_{4}$ solution is slowly added to 10.0 $\mathrm{mL}$ of a solution that is 0.20 $\mathrm{M}$ in $\mathrm{Ca}^{2+}$ and 0.30 $\mathrm{M}$ in $\mathrm{Ag}^{+} .$ (a) Which compound will precipitate first: $\mathrm{CaSO}_{4}\left(K_{\mathrm{sp}}=2.4 \times 10^{-5}\right)$ or $\mathrm{Ag}_{2} \mathrm{SO}_{4}\left(K_{s p}=1.5 \times 10^{-5}\right) ?(\mathbf{b})$ How much $\mathrm{Na}_{2}
mathrm{SO}_{4}$ solution must be added to initiate the precipitation?

Key Concepts

Solubility Product Constant (Ksp)

The solubility product constant, or Ksp, is an equilibrium constant that quantifies the extent to which a sparingly soluble ionic compound dissolves. It is defined as the product of the equilibrium concentrations of the constituent ions, each raised to the power corresponding to its coefficient in the balanced dissolution equation. When the ionic product of a solution exceeds the Ksp, the solution becomes supersaturated, and precipitation occurs.

Precipitation Reaction

A precipitation reaction involves the formation of an insoluble solid from a solution when ions combine to form a compound whose solubility is exceeded. This process is central in many analytical and synthetic procedures, whereby controlling the ionic concentrations can cause selective formation of precipitates.

Selective Precipitation

Selective precipitation is a method used to separate ions in solution based on differences in their solubility products. By adjusting the concentration of a precipitating agent, it is possible to precipitate one ionic compound while keeping another in solution, as the less soluble salt will reach its precipitation threshold first.

Threshold Concentration for Precipitation

The threshold concentration for precipitation is the point at which the ion product equals or exceeds the respective Ksp of a compound, initiating the formation of a precipitate. Understanding this threshold is key when titrating or gradually adding reagents to induce precipitation in a controlled manner.

Stoichiometry in Precipitation Reactions

Stoichiometry in precipitation reactions involves the precise mole ratios of the ions that form the insoluble compound. Correct accounting is essential to determine the amount of precipitating agent required to reach the critical concentration where precipitation begins, especially when multiple potential precipitates are possible.

Common Ion Effect

The common ion effect refers to the reduction in the solubility of a compound when one of its constituent ions is already present in the solution from another source. This effect can influence the point at which precipitation occurs, as the pre-existing concentration of a common ion shifts the dissolution equilibrium, often leading to earlier or more selective formation of a precipitate.

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