B2. A chemist mixes equal volumes of an aqueous solution of CaCl2 (1.34 10–4 M )

Question

B2. A chemist mixes equal volumes of an aqueous solution of CaCl2 (1.34 10–4 M ) and an aqueous solution of Na2CO3 (1.34 10–4 M) to prepare a saturated solution of calcium carbonate (no precipitate is observed).

(a) Calculate the concentration of each of the ions in this saturated solution (assuming 100% dissociation).

[Ca2+] = [Na+] = [Cl–] = [CO32–] =

(b) Assuming that the chemist has calculated the molar solubility for calcium carbonate accurately, write the expression for the solubility product and calculate the value for Ksp for calcium carbonate.

Explanation / Answer

Ans. A. Balanced reaction: CaCl2(aq) + Na2CO3(aq) -------> CaCO3(aq) + 2 NaCl(aq)

Given, [CaCl2] = 1.34 x 10-4 M         ; [Na2CO3] = 1.34 x 10-4 M

Both the solutions are mixed in equal volumes.

So, the final volume of reaction mixture is double the initial volume of each reactant. Doubling the volume halves the concentration as follow-

C1V1 = C2V2                        - equation 1

C1= Concentration, and V1= volume of Initial solution           ; [separate reactant]

C2= Concentration, and V2 = Vol. of final solution                    ; [reaction mixture]

Where, final volume V2 = 2 x V1

Saturated solution means that each chemical species is at equilibrium concertation, i.e. Q = Ksp.

The sole source of mentioned cations are the reactants, CaCl2 and Na2CO3, some of which later form the products. Since all the reactants/products completely dissociate (100% dissociation) into their respective ions, the total concentration (in form of reactant and product) of chemical species can be given as follow-

I. [Ca2+] = (Initial [CaCl2]/ 2)            ; 1 ml CaCl2 yields 1 mol Ca2+ upon dissociation]

                        = (1.34 x 10-4 M)/ 2 = 6.70 x 10-5 M  

II. [Na+] = 2 x (Initial [Na2CO3]/ 2)

                        = 2 x (1.34 x 10-4 M)/ 2 = 1.34 x 10-4 M  

                        [1 mol Na2CO3 yields 2 mol Na+ upon dissociation]

III. [Cl-] = 2 x (Initial [CaCl2]/ 2) = 2 x (1.34 x 10-4 M)/ 2 = 1.34 x 10-4 M  

                        [1 mol CaCl2 yields 2 mol Cl- upon dissociation]

IV. [CO32-] = (Initial [Na2CO3]/ 2) = 6.70 x 10-5 M  

                        [1 mol Na2CO3 yields 1 mol Cl- upon dissociation]

#B. CaCO3 (aq) --------> Ca2+(aq) + CO32-

Ksp of CaCO3 = [Ca2+] [CO32-]

            [where both the concentrations at at equilibrium, Q = Ksp]

Or, Ksp = (6.70 x 10-5) x (6.70 x 10-5) = 4.489 x 10-9

Therefore, Ksp of CaCO3 = 4.49 x 10-9

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