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Vinvika [58]
2 years ago
8

The value of Ksp for PbCl2 is 1.6 ×10-5. What is the lowest concentration of Cl-(aq) that would be needed to begin precipitation

of PbCl2(s) in 0.010 M Pb(NO3)2 ?
Chemistry
1 answer:
Dovator [93]2 years ago
3 0

Answer:

0.04 M is the lowest concentration of chloride ions that would be needed to begin precipitation.

Explanation:

Concentration of lead nitarte = [Pb(NO_3)_2]=0.010 M

Pb(NO_3)_2(aq)\rightleftharpoons Pb^{2+}(aq)+2NO_{3}^{-}(aq0

1 Mole of lead nirate gives 1 mole of lead ion.

Concentration of lead ion in the solution = 1\times 0.010 M= 0.010 M

Pb(Cl)_2(aq)\rightleftharpoons  Pb^{2+}(aq)+2Cl^{-}(aq0

Concentration of chloride ions = [Cl^-]

The value of K_{sp} for [tex]PbCl_2= 1.6\times 10^{-5}

K_{sp}=[Pb^{2+}][Cl^{-}]^2

1.6\times 10^{-5}=0.010 M\times [Cl^{-}]^2

[Cl^-]=0.04 M

0.04 M is the lowest concentration of chloride ions that would be needed to begin precipitation.

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