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Thursday, July 22, 2021

Finding Concentrations in a Saturated Solution

Let’s solve this problem:

The \(K_{sp}\) of \(PbCl_2\) is \(1.7 \cdot 10^{-5}\), what is the concentration of chloride ion in a saturated solution?

There are several things that are implied, but not stated, that we need to recognize:

  • The reaction involves \(PbCl_2\) breaking up into ions (this comes from the definition of \(K_{sp}\)).
  • The solution is made by adding solid \(PbCl_2\) to pure water. This is implied by the term saturated solution. If the situation was anything OTHER than solid added to water, we would be told.
  • The reaction will run forward until equilibrium is reached. (Again, this is the definition of saturated.)

So, our reaction is:

\(PbCl_{2~(s)} \rightleftarrows Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

We can create an ICE table for this reaction, which would look like this:

Although writing “some” in the midst of an ICE table seems odd, we can get away with it, since the solid is not part of the equilibrium expression.

We can now do the following math:

\(K_{sp}=[Pb^{+2}] \cdot [Cl^{-1}]^2~~~or~~~1.7 \cdot 10^{-5} = [x] \cdot [2x]^2\)

simplified, that is:

\(1.7 \cdot 10^{-5} = 4x^3\)

which gives us x = 0.016 M

Since \([Cl^{-1}] = 2x\) we know that \([Cl^{-1}]=0.032~M\).

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