A solution contains Ba^2+ (1.0 times 10^-3 M) and Ca^2+ (1.0 times 10^-2 M). Dro

Question

A solution contains Ba^2+ (1.0 times 10^-3 M) and Ca^2+ (1.0 times 10^-2 M). Drops of a 5.0 times 10^-1 M KF solution were added using a buret until the [F^-] in the solution mixture reached 1.21 x 10^-2 M. Should a precipitate form? If so, what is the precipitate? K_sp (BaF_2) = 4.0 x 10^-11 [ignore volume changes] (A) a precipitate consisting of BaF_2 only should form (B) a precipitate consisting of NaF should form (C) a precipitate consisting of CaF_2 and BaF_2 should form (D) a precipitate consisting of CaF_2 only should form (E) no precipitate should form

Explanation / Answer

A substance gets precipitated out only when the ionic product (product of concentration of its ions) present in a solution exceeds the solubility product of the substance.

Here, ionic product of [Ba2+][F-] = (1x10-3 )(1.21x 10-2) = 1.21 x 10-5 Ksp of BaF2 = 1.7 x10-6

ionic product of [Ca2+][F-] = (1x10-2 )(1.21x 10-2) = 1.21 x 10-4 Ksp of CaF2 = 4 x10-11

The ionic product of both compounds are exceeded the solubility product, so both can precipitate in the solution. But the KSp of CaF2 is very less than Ksp of BaF2, CaF2 will precipitate first.

We can calculate the amount of F- needed to precipitate each compounds as [F-] = Ksp/ conc of cation

For CaF2 to precipitate, [F-] = 4 x10-11 / (1x10-2 ) = 4 x10-9 M

For BaF2 to precipitate, [F-] = 1.7 x10-6 / (1x10-3 ) = 1.7 x 10-3 M

Since [F-] exceeds the one needed for both the compounds to precipitate, both compounds will precipitate in the solution.

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