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

To 1.0 L of water, 2.8 × 10^–6 mol of Pb(NO₃)₂, 6.6 × 10^–5 mol of K₂CrO₄, and 1.0 mol of NaCl are added. What will happen?

Salt	K_sp
PbCrO₄	1.8 × 10^–14
PbCl₂	1.6 × 10^–5

A.No precipitate will form.

B.A precipitate of PbCrO₄ will form.

C.Both a precipitate of PbCl₂ and a precipitate of PbCrO₄ will form.

D.A precipitate of PbCl₂ will form.

E.A precipitate of KCl will form.

Solutions

Expert Solution

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.

queen_honey_blossom answered 1 year ago

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