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# calculate the concentrations of Pb2+ and Cl - at equilibrium. Ksp for PbCl2(s) is 1.6 10-5.?

A solution is prepared by mixing 50.0 mL of 0.30 M Pb(NO3)2 with 50.0 mL of 2.1 M KCl.

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Lead chloride dissolves and dissociates according to

PbCl₂(s) ⇄ Pb²⁺(ag) + 2 Cl⁻(aq)

So the ionic molarities is a saturated solution satisfy the equation

Ksp = [Pb²⁺]∙[Cl⁻]²

Initially concentrations are

[Pb²⁺]₀ = 50mL ∙ 0.3M / (50mL + 50mL) = 0.15M

[Cl⁻]₀ = 50mL ∙ 2.1M / (50mL + 50mL) = 1.05M

Due to the small solubility of lead chloride it precipitate unless the ionic concentrations satisfy the equilibrium equation. Because we have an excess of chloride ions, the lead ions will almost completely be precipitated in equilibrium while a significant amount of chloride ions remains in solution.

Two chloride ions and one lead ion react with each other forming one solid lead chloride, thus,

∆[Pb²⁺] = ∆[Cl⁻]/2

Using this relation, you can express chloride ion concentration in terms of lead ion concentration:

([Pb²⁺]₀ - [Pb²⁺]) = (1/2)∙([Cl⁻]₀ - [Cl⁻])

<=>

[Cl⁻] = ([Cl⁻]₀ - 2∙([Pb²⁺]₀ - [Pb²⁺])

= (1.05 - 2∙(0.15 - [Pb²⁺])

= 0.75 + 2∙[Pb²⁺]

Due to the excess of chloride and the small solubility of lead chloride [Pb²⁺] will be small compared to 0.75M, i.e. you can approximate:

[Cl⁻] ≈ 0.75

Thus,

Ksp = [Pb²⁺]∙[Cl⁻]²

<=>

1.6×10⁻⁵ = [Pb²⁺] ∙ 0.75²

<=>

[Pb²⁺] = 1.6×10⁻⁵ / 0.75² = 2.8×10⁻⁵M

As you can see the approximation was justified. So the equilibrium concentrations are:

[Pb²⁺] = 2.8×10⁻⁵ M

[Cl⁻] = 0.75 M

• answer to question / Merry Christmas from Jamaica 1st. confirm if PbCl2 will precipitate Pb+2 (aq) + 2 Cl - (aq) --? PbCl2 (s) [Pb+2 ] = ( 50ml) (.25 M) /100ml vol answer = 0.one hundred twenty five M [Cl-a million] = (50 ml) (2.2) /one hundred ml vol answer = a million.a million M Q= (.one hundred twenty five (a million.a million) = a million.512 x 10 -a million hence Q > ok hence a precipitate (pptte) will type. ( word : you are able to continually do this determination first, in the different case you will possibly get tricked by using your professor, only in case the concentrations are no longer sufficient to permit pptte) Pb+2 + 2Cl ----> PbCl2 earlier (50ml) (.25M) (50ml) (2.2M)(2.2M) 0 12.5mmoles 110mmoles After 12.5.12.5 = 0 one hundred ten- 2(12.5) = 85mmoles in extra Conc of [Cl-a million] = 85mmoles/100ml finished vol = 8.80 5 x 10-a million M (yeah!!!!!) (word Pb+2 is the proscribing reagent, and thusly get thoroughly used up) Conc of Pb + 2 could be calculated, on the grounds that at equilibrium there is an infinitismal volume of solubility of the forged PbCl2 So. preliminary [Pb+2 ] = 0 at Eq = X preliminary [Cl-a million] = 8.80 5 x 10-a million M at Eq = (8.80 5 x 10-a million M ) + 2x (word we prefer two times as plenty moles of Cl-a million in step with one mole of Pb + 2 as spoke of interior the stoichiometery) Ksp = a million.6 x 10 -5 = [Pb +2] [Cl-a million]2 = (x) ((8.80 5 x 10-a million M + 2x)2 (it rather is sq.) yet on the grounds that dissociation of PbCl2 is small we ca overlook 2x interior the above expression hence: Ksp = a million.5 x 10 -5 = (x) ((8.80 5 x 10-a million M )2 (it rather is sq.) Maniplate the expression and you may get : X = 2.21 x 10 -5 M = [Pb =2] (yeah !!) This replaced into exciting, have not resolve ay difficulty of this manner in a at an identical time as. working out to get my Christmas tree earlier the grand young babies come over. Cool Runnings !!!!

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