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Question: If an equal volume of \[BaC{l_2}\] and \[NaF\] solutions are mixed which of these combinations will ...

If an equal volume of BaCl2BaC{l_2} and NaFNaF solutions are mixed which of these combinations will not give a precipitate. (Kspof BaF2=1.7×10−7)\left( {{K_{sp}}{\text{of}}\,Ba{F_2} = 1.7 \times {{10}^{ - 7}}} \right)
A.10−3M BaCl2{10^{ - 3}}{\text{M }}BaC{l_2} and 2×10−2M NaF2 \times {10^{ - 2}}{\text{M }}NaF
B.10−3M BaCl2{10^{ - 3}}{\text{M }}BaC{l_2} and 1.5×10−2M NaF1.5 \times {10^{ - 2}}{\text{M }}NaF
C.1.5×10−2M BaCl21.5 \times {10^{ - 2}}{\text{M }}BaC{l_2} and 10−3M NaF{10^{ - 3}}{\text{M }}NaF
D.2×10−2M BaCl22 \times {10^{ - 2}}{\text{M }}BaC{l_2} and 2×10−2M NaF2 \times {10^{ - 2}}{\text{M }}NaF

Explanation

Solution

To answer this question, you should recall the concept of solubility product. The solubility product constant refers to the equilibrium constant for the dissolution of a solid substance into an aqueous solution. The solution will precipitate if the ionic product is greater than the solubility product.

Complete step by step answer:
Since for precipitate to start the product of [Ba2+]  [B{a^{2 + }}]\; and square [F−]  [{F^ - }]\; should be greater than Ksp{K_{sp}} ​ of BaF2.Ba{F_2}.
Let us analyse the options systematically:
A) [Ba2+][F−]2=0.04×(0.02)2=1.6×10−5>Ksp  of  BaF2[B{a^{2 + }}]{[{F^ - }]^2} = 0.04 \times {\left( {0.02} \right)^2} = 1.6 \times {10^{ - 5}} > {K_{sp}}\;of\;Ba{F_2}
∴\therefore We can conclude that BaF2Ba{F_2} ​ will precipitate.
B) [Ba2+][F−]2=0.01×(0.015)2=2.25×10−6>Ksp  of  BaF2[B{a^{2 + }}]{[{F^ - }]^2} = 0.01 \times {\left( {0.015} \right)^2} = 2.25 \times {10^{ - 6}} > {K_{sp}}\;of\;Ba{F_2}.
∴\therefore We can conclude that BaF2Ba{F_2}will precipitate.
C) [Ba2+][F−]2=[0.015]×[0.01]2=1.5×10−6>Ksp  of  BaF2[B{a^{2 + }}]{[{F^ - }]^2} = \left[ {0.015} \right] \times {\left[ {0.01} \right]^2} = 1.5 \times {10^{ - 6}} > {K_{sp}}\;of\;Ba{F_2} .
∴\therefore We can conclude that BaF2Ba{F_2}​ will precipitate.
D) [Ba2+][F−]2=0.02×(0.002)2=8×10−8<Ksp  of  BaF2[B{a^{2 + }}]{[{F^ - }]^2} = 0.02 \times {\left( {0.002} \right)^2} = 8 \times {10^{ - 8}} < {K_{sp}}\;of\;Ba{F_2} ​.
∴\therefore We can conclude that BaF2Ba{F_2}will not precipitate.
Since for precipitate to start the product of [Ba2+]  [B{a^{2 + }}]\; and square of [F−]  [{F^ - }]\; should be greater than Ksp{K_{sp}} ​ of BaF2.Ba{F_2}.

Hence, the correct answer for this question is option D.

Note:
You should know about the difference between solubility and solubility product constant. The solubility of a substance in a solvent is the total amount of the solute that can be dissolved in the solvent at equilibrium. The solubility product constant is an equilibrium constant that provides insight into the equilibrium between the solid solute and its constituent ions that are dissociated across the solution.
Solubility is a property of a substance being dissolved called solute to get dissolved in a solvent to form a solution. Solubility depends on various factors such as lattice enthalpy of salt and solvation enthalpy of ions. Upon dissolution of any salt, the strong forces of attraction of solute must be overcome by the interactions between ions and the solvent.
This solvation enthalpy of ions is always negative. This means that there is a loss of energy during this process. Unlike polar solvents the non-polar solvents have a small value of solvation enthalpy, meaning that this energy is not sufficient to overcome the lattice enthalpy.