
Which solid will precipitate first if an aqueous solution of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{Cr}}{{\text{O}}_{\text{4}}}$ at ${25^{\text{o}}}{\text{C}}$ is slowly added to an aqueous solution containing $0.001{\text{M Pb}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}$ and $0.001{\text{M Ba}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}$ at ${25^{\text{o}}}{\text{C}}$?
A.${\text{BaCr}}{{\text{O}}_{\text{4}}}$
B.${\text{NaN}}{{\text{O}}_{\text{3}}}$
C.${\text{PbCr}}{{\text{O}}_{\text{4}}}$
D.${\text{Pb}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}$
E.None of these
Answer
562.2k+ views
Hint: To answer this question, you must recall the solubility products of the given salts. The salt which has a higher solubility product precipitates first.
Complete step by step answer:
Solubility product is the equilibrium constant for the dissolution of a salt in an aqueous solution. It can be denoted by ${K_{sp}}$. The solubility product generally tends to increase with an increase in temperature as the solubility increases. It is a kind of equilibrium constant.
The solubilities of ionic salts, that dissociate into cations and anions, in water varies a lot. Some salts may be highly soluble and others may be insoluble.
We know that all sodium salts are soluble in water. Thus sodium nitrate is soluble in water. We can rule out option B as it does not precipitate.
Also, we know that all nitrates are soluble in water. Thus, lead nitrate too will not form a precipitate and we can rule out option D.
We have lead chromate and barium chromate. Both the salts sparingly soluble in water and their ${K_{sp}}$ values are $2.0 \times {10^{16}}$ and $2.2 \times {10^{10}}$ respectively. Since lead chromate has a higher value of solubility product, it will precipitate first.
Thus, the correct answer is C.
Note:
For a salt to be soluble in water, the lattice enthalpies of the ions must be overcome by the strength of the interactions between the solvent molecules and the ions. The solvation enthalpy of a salt always has a negative value signifying that energy is always released during dissolution. The nature of the solvent determines the amount of solvation energy. Thus ionic and polar salts dissolve in polar solvents only while non- polar compounds dissolve in non-polar solvents.
Complete step by step answer:
Solubility product is the equilibrium constant for the dissolution of a salt in an aqueous solution. It can be denoted by ${K_{sp}}$. The solubility product generally tends to increase with an increase in temperature as the solubility increases. It is a kind of equilibrium constant.
The solubilities of ionic salts, that dissociate into cations and anions, in water varies a lot. Some salts may be highly soluble and others may be insoluble.
We know that all sodium salts are soluble in water. Thus sodium nitrate is soluble in water. We can rule out option B as it does not precipitate.
Also, we know that all nitrates are soluble in water. Thus, lead nitrate too will not form a precipitate and we can rule out option D.
We have lead chromate and barium chromate. Both the salts sparingly soluble in water and their ${K_{sp}}$ values are $2.0 \times {10^{16}}$ and $2.2 \times {10^{10}}$ respectively. Since lead chromate has a higher value of solubility product, it will precipitate first.
Thus, the correct answer is C.
Note:
For a salt to be soluble in water, the lattice enthalpies of the ions must be overcome by the strength of the interactions between the solvent molecules and the ions. The solvation enthalpy of a salt always has a negative value signifying that energy is always released during dissolution. The nature of the solvent determines the amount of solvation energy. Thus ionic and polar salts dissolve in polar solvents only while non- polar compounds dissolve in non-polar solvents.
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