Predicting a precipitate

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Intros
Lessons
  1. Predicting if a solution will form a precipitate.
  2. Recalling Ksp.
  3. The ionic product Q (with worked example).
  4. Predicting precipitates using Ksp and Q.
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Examples
Lessons
  1. Use the solubility product expression to predict a precipitate.1
    1. A solution was made by combining 75 mL of 0.2M Mg2+(aq) and 110 mL of 0.08M OH-(aq). Use these values and the solubility product expression to predict whether a precipitate of Mg(OH)2 will form in the resultant mixture.
    2. A solution was made by combining 20 mL of 0.01M Fe3+(aq) and 25 mL of 0.01M OH-(aq). Use these values and the solubility product expression to predict whether a precipitate of Fe(OH)3 will form in the resultant mixture.
    3. A solution was made combining 50 mL of 0.003M Ca2+(aq) and 40 mL of 0.008M SO42-(aq). Use these values and the solubility product expression to predict whether a precipitate of CaSO4 will form in the resultant mixture.

      1 Source for Ksp solubility constant data at 250C: http://www4.ncsu.edu/~franzen/public_html/CH201/data/Solubility_Product_Constants.pdf
Topic Notes
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In this lesson, we will learn:

  • To recall the ionic product Q and how it relates to the solubility product Ksp.
  • How to calculate the ionic product and use it to predict whether a precipitate will form in solution.

Notes:

  • As mentioned in the lesson Solubility product; the value of Ksp for a saturated solution of a given compound is constant. It doesnโ€™t matter how the ion concentrations are made up (whether by one solution or multiple combined); if the solution is saturated then the product of the individual ion concentrations will be equal to Ksp.
    • Using this fact, we can predict if a precipitate forms in the solution or not before making it by using the Ksp value. This Ksp product value is the product of the ion concentrations required to make a saturated solution (for the equilibrium to be established).
    • If this product of ion concentration is smaller than Ksp, the solution will not be saturated and the equilibrium will not be established.
  • The โ€˜product of ion concentrationโ€™ is known simply as the ionic product with label Q. As a product, Q is calculated by multiplying the concentrations of ions together. For aqueous ions Mx+ and Xm- forming precipitate MmXx:

    Q = [M+]m [X-]x

  • Compare this to Ksp which is the ion concentration needed to form a saturated solution. Using these two terms, three situations can be described when two solutions are mixed to create a new ionic species.
    • If Q is smaller than Ksp, then there are less ions in solution than necessary to form a saturated solution. Without a saturated solution, a precipitate will not form.
    • If Q is equal to Ksp, then there are just enough ions in solution necessary for a saturated solution to form. In this scenario, a solution that has just reached saturation will form. This means that any further addition of aqueous ions will form a precipitate, as the equilibrium shifts to the left to favor the undissolved state.
    • If Q is greater to Ksp, then there are more ions in solution than are necessary to form a saturated solution. In this scenario, an equilibrium will be established to form a precipitate with the excess ions; a precipitate will form.

  • For example: a solution made by combining 100 mL of 0.1 M Ca2+(aq) and 75mL of 0.2 M F-(aq) :

    Q = [Ca2+][F-]2\qquad To Form CaF2

    These solutions dilute each other. Find the new concentration by dividing original volume by combined volume.

    [Ca2+]sol = 0.1 M * 0.1โ€‰L0.175โ€‰L\large \frac{0.1 \, L}{0.175 \, L} = 0.057 M

    [F-]sol = 0.2 M * 0.075โ€‰L0.175โ€‰L\large \frac{0.075 \, L}{0.175 \, L} = 0.086 M

    Q = [0.057] x [0.086]2โ€‰\, = 4.22*10-4

    Ksp (CaF2) = 3.45*10-11 M


    Q > Ksp therefore a precipitate will form.