Chapter 17: Problem 59
Calculate the molar solubility of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) at \(25^{\circ} \mathrm{C}\) in (a) \(0.200 \mathrm{M} \mathrm{AgNO}_{3}\) and (b) \(0.200 \mathrm{M} \mathrm{Na}_{2} \mathrm{CrO}_{4}\). For \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) at \(25^{\circ} \mathrm{C}, K_{\mathrm{cn}}=1.1 \times 10^{-12}\)
Short Answer
Step by step solution
- Write down the dissolution equation
- Write the expression for the solubility product constant (Ksp)
- Solve for molar solubility in \(0.200 \mathrm{M} \mathrm{AgNO}_3\) solution
- Calculate the molar solubility in \(0.200 \mathrm{M} \mathrm{Na}_2\mathrm{CrO}_4\) solution
- Neglect \(s\) and \(s'\) if small and solve for Ksp
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solubility Product Constant
For example, consider the compound \(Ag_2CrO_4\), which dissociates into two silver ions (\(Ag^+\)) and one chromate ion (\(CrO_4^{2-}\)). The equilibrium expression for its Ksp would be \(K_{sp} = [Ag^+]^2[CrO_4^{2-}]\), where the brackets indicate the concentration of ions in mol/L. The Ksp value is unique for every salt and depends on the temperature, illustrated by the given Ksp of \(1.1 \times 10^{-12}\) for \(Ag_2CrO_4\). This information enables us to calculate the extent to which the salt can dissolve in a particular solvent at a given temperature.
Common Ion Effect
Taking our earlier example of \(Ag_2CrO_4\), if we introduce it into a solution containing \(Ag^+\) ions from another source like \(AgNO_3\), the increased concentration of \(Ag^+\) decreases the solubility of \(Ag_2CrO_4\). In the exercise, we saw this effect in action when calculating the molar solubility in a \(0.200 M AgNO_3\) solution. As the silver ion concentration is already significant, less \(Ag_2CrO_4\) can dissolve to reach equilibrium, consequently decreasing its molar solubility. This concept is critical when predicting the solubility of ionic compounds in solutions with shared ions and has broad applications in fields such as pharmacology, environmental science, and chemical manufacturing.
Equilibrium Concentration
In our dissolution equation for \(Ag_2CrO_4\), the molar solubility is directly linked to the equilibrium concentrations of the resultant ions. This concept not only helps in calculating the extent of solubility but also in understanding how various external factors, such as the common ion effect, temperature, and pressure, can shift the equilibrium. Particularly, it is important for carrying out precise calculations in the lab, for example, when a given concentration of \(Na_2CrO_4\) in the solution influences the dissociation of \(Ag_2CrO_4\), as seen in part (b) of the exercise. Understanding equilibrium concentration is vital in predicting the behavior of systems in dynamic equilibrium, allowing chemists to manipulate conditions to favor desired outcomes in synthesis and to explore reaction mechanisms.