This would mean the $K_s_p$ unit would be different for every problem and would be difficult to solve, so in order to make it simpler, chemists generally drop $K_s_p$ units altogether. And what are the $K_s_p$ units? Actually, it doesn’t have a unit! The $K_s_p$ value does not have any units because the molar concentrations of the reactants and products are different for each equation. The more soluble a substance is, the higher its $K_s_p$ chemistry value.

The value of $K_s_p$ varies depending on the solute.

(A solute is insoluble if nothing or nearly nothing of it dissolves in solution.) $K_s_p$ represents how much of the solute will dissolve in solution. $K_s_p$ is used for solutes that are only slightly soluble and don’t completely dissolve in solution. As a reminder, a solute (what is being dissolved) is considered soluble if more than 1 gram of it can be completely dissolved in 100 ml of water. It’s the equilibrium constant used for equations when a solid substance is dissolving in a liquid/aqueous solution. $K_s_p$ is known as the solubility constant or solubility product. At the bottom of this guide, we also have a table with the $K_s_p$ values for a long list of substances to make it easy for you to find solubility constant values. In this $K_s_p$ chemistry guide, we’ll explain the $K_s_p$ chemistry definition, how to solve for it (with examples), which factors affect it, and why it’s important. When you have a solid grasp of $K_s_p$, those questions become much easier to answer! Are you learning chemistry but don’t quite understand the solubility product constant or want to learn more about it? Not sure how to calculate molar solubility from $K_s_p$? The solubility constant, or $K_s_p$, is an important part of chemistry, particularly when you’re working with solubility equations or analyzing the solubility of different solutes.

Ksp mac problem