Earlier in the semester, or perhaps in your Chem II class, you learned about acid-base titrations which were used to determine the concentration of an acid or a base. For this, we are using a standard solution (titrant) which is … Read more
When discussing the molar solubility and solubility product constant (Ksp), we mentioned that the values pertain to saturated solutions of the given compound. For example, the molar solubility of BaOS4 is 3.87 x 10-5 mol/L. Therefore, it starts precipitating only … Read more
In this set of practice problems, we will work on calculating the solubility product constant (Ksp) from the molar solubility of the compound, and vice versa, the Ksp of compounds in pure water, in the presence of salts with common … Read more
The solubility of almost any ionic compound is affected by the pH of a solution. The effect is more pronounced when one or both of the ions in the compound are slightly basic or acidic. For example, magnesium hydroxide, an … Read more
In the previous post, we talked about the solubility and solubility product constant (Ksp) of ionic compounds with low solubility. For example, the dissolution equation and the Ksp for CaF2 are: CaF2(s) ⇆ Ca2+(aq) + 2F–(aq) Ksp = [Ca2+][ … Read more
Ksp – The Solubility Product Constant Remember, earlier when discussing precipitation reactions, we used tables with rules for solubility to classify compounds as water-soluble or insoluble. For example, BaCl2, according to the rules, is soluble, while BaSO4 is insoluble in … Read more
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