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To 1.0 L of water, 2.8 × 10–6 mol of Pb(NO3)2, 6.6 × 10–5 mol of K2CrO4,...

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Chemistry

To 1.0 L of water, 2.8 × 10–6 mol ofPb(NO3)2, 6.6 × 10–5 mol ofK2CrO4, and 1.0 mol of NaCl are added. Whatwill happen?

SaltKsp
PbCrO41.8× 10–14
PbCl21.6× 10–5

A.No precipitate will form.

B.A precipitate of PbCrO4 will form.

C.Both a precipitate of PbCl2 and a precipitate ofPbCrO4 will form.

D.A precipitate of PbCl2 will form.

E.A precipitate of KCl will form.

Answer & Explanation Solved by verified expert
3.7 Ratings (679 Votes)

A is flase because of PbCrO4 form pricipitate

B. Ksp = [Pb+2][CrO42-]

      1.8*10-14 = [Pb+2][CrO42-]

[Pb+2][CrO42-] = 2.8*10-6 *6.6*10-5

                                 = 1.8*10-10

So will be pricipitate form PbCrO4 because of concentration exeed the KsP .

B.A precipitate of PbCrO4 will form.

Ksp = [Pb+2][cl-]2

1.6*10-5 = [Pb+2][cl-]2

[Pb+2][cl-]2    = 2.8*10-6 *(1.0)2

                  = 2.8*10-6

will not form PbCl2 pricipitate because of concentration not exeed the Ksp

E.A precipitate of KCl will form. flase Kcl is soluble

C.Both a precipitate of PbCl2 and a precipitate of PbCrO4 will form. Flase becuse of only pbcro4 pricipitate.



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