Chemistry Corner

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In which of the following solutions would solid PbCl2 be expected to be the least soluble at 25°C?

A) 0.1 M KNO3 B) 0.1 M HCl C) 0.1 M NaCl D) 0.1 M CaCl2

The answer is D.

I just don't understand why it is D. Is it because like cannot dissolve like? CL2 makes the two similiar so therefore less soluble?

Thanks!

It is due to Le Chatelier's principle, called the "common ion effect" here.

Take the equilibrium equation of solid PbCl₂ in equilibrium with dissolved PbCl₂ in aqueous solution:

PbCl₂ (s)

Pb²⁺ (aq) + 2 Cl^{�?/SUP> (aq)}

If you add more Cl^{�?/SUP> ion, the reverse (<–�? reaction will occur until the reaction is at equilibrium again. Therefore, some solid PbCl₂ would precipitate out, leaving less dissolved in solution. Dissolving the PbCl₂ in a solution that already has the Cl^{�?/SUP> ion in it initially gives the same result in terms of the equilibrium solubility of PbCl₂.}}

Choices (b) and (c) are also solution which have chloride ion in them (0.1 M), but the CaCl₂ solution would have twice the chloride ion, 0.2 M, since each CaCl₂ gives 2 Cl^{�?/SUP> ions.}

You can also calculate the molar solubilities and compare the values.

1. molar solubility of PbCl₂ in pure water:

PbCl₂ (s)

Pb²⁺ (aq) + 2 Cl^{�?/SUP> (aq)}

Initially: �?nbsp; 0 0

At Equilib: �?nbsp; x 2x

K_sp = (x)(2x)²

x is the molar solubility of PbCl₂ in pure water.

2. molar solubility of PbCl₂ in a solution that is 0.1 M in chloride ion:

PbCl₂ (s)

Pb²⁺ (aq) + 2 Cl^{�?/SUP> (aq)}

Initially: �?nbsp; 0 0.1 M

At Equilib: �?nbsp; x 0.1 + 2x

K_sp = (x)(0.1 + 2x)² �?(x)(0.1)²

x is the molar solubility of PbCl₂, but will be less than the molar solubility in pure water.

3. If the chloride concentration is 0.2 M, then we would have,

K_sp = (x)(0.2 + 2x)² �?(x)(0.2)²

x is the molar solubility of PbCl₂, but will be less than the molar solubility in a solution that is 0.1 M chloride ion. As the concentration of Cl^{�?/SUP> ion goes up, the molar solubility of PbCl₂ is seen to go down.}

Steve

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