Solubility products (Ksp) are equilibrium constants for saturated solution and solid formation. where precipitations are formed. It is an equilibrium constant for solution which contains sparingly soluble salt. It is linked to molar solubility, as the solubility product of salts are indication (indirect) of their solubilities expressed in mol/L, which is called molar solubility. When common ions are present in the solution, molar solubilities can be affected. So, this Part 3 of Berry Berry Easy notes on Heterogeneous Ionic Equilibrium for STPM Chemistry is focused on solubility product and molar solubility.
[Tips: Note the expression for ion products, Qsp which has the same expression as Ksp is used when we have unsaturated and supersaturated solutions, where the systems are not at equilibrium. In the second case, no precipitation is formed and the product is the ion product.]
STPM Chemistry Form 6 Notes – Heterogeneous Ionic Equilibrium (Part 3)
Solubility Product and Molar Solubility
- Molar solubility and solubility product are very different terms. If one salt has a lower Ksp value than another salt, it does not mean that it is less soluble (has a lower molar solubility).
- Solubility is determined by the molar solubility of a salt, not the solubility product
- Ksp values are calculated as products over reactants, but because solid are ignored, there is no denominator.
- Remember to multiply the ions by the coefficient from the balanced equation!
|Solubility Reaction||Ksp Expression||Ksp Calculation|
|MX(s) <—-> M+(aq) + X-(aq)||Ksp = [M+][X-]||Ksp = (x)(x) = x2|
|MX2(s) <—-> M2+(aq) + 2X-(aq)||Ksp = [M2+][X-]2||Ksp = (x)(2x)2 = 4x3|
|MX3(s) <—-> M2+(aq) + 3X-(aq)||Ksp = [M2+][X-]3||Ksp = (x)(3x)3 = 27x4|
|M2X(s) <—-> 2M+(aq) + X2-(aq)||Ksp = [M+]2[X2-]||Ksp = (2x)2(x) = 4x3|
|M3X(s) <—-> 3M+(aq) + X3-(aq)||Ksp = [M+]3[X3-]||Ksp = (3x)3(x) = 27x4|
- It is difficult to predict relative solubility and the only way to compare is through experiment or data analysis.
Example: Comparing the solubility of calcium carbonate (CaCO3) and calcium fluoride (CaF2). Which of the two salts is more soluble? Given the solubility product of CaCO3 is 8.7 x 10-9 mol2 dm-6 and the solubility product for CaF2 is 4.0 x 10-11 mol3 dm-9.
Solution: the question asks for the highest solubility, it refers to the compound that produces the greatest amount of dissociated salt (the greatest molar solubility) not solubility product. Never compare numbers with unlike units. Solubility products are used to determine the molar solubility.
- CaCO3 is an MX salt, Ksp = x2
- CaF2 is an MX2 salt, Ksp = 4x3
Therefore, molar solubility for CaCO3 is 9.3 x 10-5 mol dm-3 and CaF2 is 2.2 x 10-4 mol dm-3. CaF2 is more soluble than CaCO3 even though it has a smaller solubility product. Relative solubility questions are asked as:
- Which salt exhibits greater salvation?
- Which salt precipitate first?
- Use molar solubility to answer these questions.
This signals the end of Part 3 notes on the topic of Heterogeneous Ionic Equilibrium for STPM Form 6 Chemistry students from Berry Berry Easy. In Part 4, Berry Readers will be shown examples of heterogeneous equilibria and be given the tabulated list of solubility product constants. So make sure you understand the first 3 posts in the series before you read the next post.