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.
- As mentioned in the lesson Solubility product; the value of Ksp for a saturated solution of a given compound is constant. It doesn’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 ‘product of ion concentration’ 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 To Form CaF2
These solutions dilute each other. Find the new concentration by dividing original volume by combined volume.
[Ca2+]sol = 0.1 M * = 0.057 M
[F-]sol = 0.2 M * = 0.086 M
Q = [0.057] x [0.086]2 = 4.22*10-4
Ksp (CaF2) = 3.45*10-11 M
Q > Ksp therefore a precipitate will form.