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dalvyx [7]
3 years ago
10

Suppose 231.8 mgmg of PbCl2PbCl2 was added to 15.0 mLmL of water in a flask, and the solution was allowed to reach equilibrium a

t 20.0 oCoC. Some solute remained at the bottom of the flask after equilibrium, and the solution was filtered to collect the remaining PbCl2PbCl2, which had a mass of 74 mgmg . What is the solubility of PbCl2PbCl2 (in g/Lg/L)?
Chemistry
1 answer:
vovangra [49]3 years ago
8 0

Answer:

ksp = 2.2 x ⁻⁴

Explanation:

The equilibrium here is:

PbCl₂ (s)     ⇄ Pb²⁺ + 2 Cl⁻

we can recognize it as a product solubilty equilibrium once we are told that some undissolved PbCl₂ remained.

The equilibrium constant, Ksp is given by the equation

Ksp = [Pb²⁺][Cl⁻]²

where [Pb²⁺] and [Cl⁻]² are the concentrations (M) of Pb²⁺ and Cl⁻ in solution.

we have the mass of solid PbCl₂ placed in solution, so we can determine the number of moles it represents, and if  we  substract the moles of undissolved PbCl₂ we will know the moles of Pb²⁺ and Cl⁻ which went into solution.

From there we can calculate the molarity (M= moles/L solution) and finally plug the values into our expression for Ksp to answer this question.

molar mas PbCl₂ = 278.1 g/mol

1 milligram = 1 x 10⁻³ g

mol PbCl₂ initially = 231.8 x 10⁻³ g / 278.1 mol = 8.3 x 10⁻⁴ mol

Volume solution = 15 mL x 1L / 1000 mL = 0.015 L

mol undissolved PbCl₂ = 74 x 10⁻³ g / 278.1 g/mol = 2.7 x 10⁻⁴ mol

mol PbCl dissolved =   8.3 x 10⁻⁴ mol -  2.7 x 10⁻⁴ mol = 5.7 x 10⁻⁴ mol

Concentration of Pb²⁺ in solution = 5.7 x 10⁻⁴ mol / 0.015 L = 3.8 x 10⁻² M

Concentration of Cl⁻ in solution = 2 x 3.8 x 10⁻² M = 7.6 x 10⁻² M

(Note from the formula we we get 2 mol Cl⁻ per mol PbCl₂)

Plugging these values into the expression for Ksp we have

Ksp = 3.8 x 10⁻² x (7.6 x 10⁻²)² = 2.2 x 10⁻⁴

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