Solveeit Logo
Login

Question

Question: If equal volumes of \(BaC{l_2}(aq)\) and \(NaF(aq)\) solutions are mixed together, which will give a...

If equal volumes of BaCl2(aq)BaC{l_2}(aq) and NaF(aq)NaF(aq) solutions are mixed together, which will give a precipitate?
[Ksp{K_{sp}} of BaF2=1.7×10−7Ba{F_2} = 1.7 \times {10^{ - 7}} ]
A. 0.004M0.004M BaCl2BaC{l_2} and 0.002M0.002M NaFNaF
B. 0.010M0.010M BaCl2BaC{l_2} and 0.015M0.015M NaFNaF
C. 0.015M0.015M BaCl2BaC{l_2} and 0.002M0.002M NaFNaF
D. 0.020M0.020M BaCl2BaC{l_2} and 0.002M0.002M NaFNaF

Explanation

Solution

The solubility product constant of a compound is the equilibrium constant for the dissolution of a solid substance/ compound into an aqueous solution. It is denoted by the symbol Ksp{K_{sp}} . The solubility product is a type of equilibrium constant and its value depends on the temperature. Ksp{K_{sp}} usually increases with an increase in temperature due to increased rate of dissolution or solubility.

Complete step by step answer:
As per the question, equal volumes of BaCl2(aq)BaC{l_2}(aq) and NaF(aq)NaF(aq) solutions are mixed together. The equation of the chemical reaction that takes place here is as follows:
BaCl2(aq)+NaF(aq)→BaF2(aq)+NaCl(aq)BaC{l_{2(aq)}} + Na{F_{(aq)}} \to Ba{F_{2(aq)}} + NaC{l_{(aq)}}
Ka{K_a} , the acid ionization constant, is the equilibrium constant for chemical reactions involving weak acids in aqueous solution. The numerical value of Ka{K_a} is used to predict the extent of acid dissociation.
In this reaction, the value of Ka{K_a} is equal to:
Ka=[Ba2+][F−]2{K_a} = [B{a^{2 + }}]{[{F^ - }]^2} ….(i)
In the question, various concentrations of barium and fluoride ions are given. We have to check for various concentration pairs in order to determine in which case is the value of Ka>Ksp{K_a} > {K_{sp}} .

Let us take the concentration of [Ba2+]=0.010M[B{a^{2 + }}] = 0.010M and the concentration of [F−]=0.015M[{F^ - }] = 0.015M
Substituting the values in equation (i), we have:
Ka=[Ba2+][F−]2=(0.010)(0.015)2=2.25×10−6{K_a} = [B{a^{2 + }}]{[{F^ - }]^2} = (0.010){(0.015)^2} = 2.25 \times {10^{ - 6}}
Thus, the value of the acid ionization constant (Ka{K_a}) = 2.25×10−62.25 \times {10^{ - 6}}
As we can clearly see that, Ka>Ksp{K_a} > {K_{sp}} . Thus, this combination of concentrations will give a precipitate.

So, the correct answer is Option B.

Note:
Two compounds in their aqueous states react with each other and give a precipitate only when the solubility product constant has a value lesser than the acid ionization constant. If the value is high, the new products formed will remain in their aqueous state in the form of ions and will not give a precipitate.