Fully answer these questions in your laboratory notebook before coming to lab. f
ID: 539422 • Letter: F
Question
Fully answer these questions in your laboratory notebook before coming to lab. for numerical calculations. Review the section on Filtration in the beginning of your lab manual. Compare and contrast: Vacuum Filtration and Gravity Filgration. Define the following terms: a. K_sp b. Molar solubility For each of the following salts: Write the balanced equation for the solution process. Write the Law of Mass Action (Ksp expression). Calculate the value of the molar solubility, s. a. Ca_3(PO_4)_2^3 K_sp = 2.07 times 10^-33 b. AgCN, K_sp = 5.97 times 10^-17 c. Cu_2S, K_sp = 2.26 times 10^-48Explanation / Answer
1. Comparison between gravity filtration and vacuum filtration.
In gravity filtration the suspension of solid present in liquid solution is allowed to flow with the help of gravity through a porous medium ( When performed in laboratory , the porous medium is nothing but the filter paper). The solids present in the solution eventually gets retained by the porous media.
In vacuum filtration , as the name itself suggests, is carried out under vaccum. The filtration process is accomplished by placing filter paper over the Hirsch funnel(Standard funnel used for vacuum operation). When the vaccum process starts the liquid is pulled thorugh this filter paper.
Gravity filtration is generally used for removing Coarse particles.
Vacuum filtration is real advantagous over here as it can filter small particles
Gravity filtration is often of advatnage when the solution to be filtered is hot and has low boiling point. If vacuum filtration is used under such situation it can cause vapourization of the solution and can cause serious problems if the solution is flammable.
Vacuum filtration is beneficial when dry crystals are to be obtained. In gravity filtration the solvent will be in contact with the crystals for a longer period of time. As a result of these, some of the crystals might get dissolved back in the solution and hence will result in low yield os crystals.This problem can be avoided if Vacuum filtration is used.
2.
2.1 Solubility product (Ksp): The solubility product or Ksp is a eqiulibrium constant for a chemical reaction in which an ionic compound dissolves in solution to yield ions in a solution.
This is just one of the definition of solubility product. There are many ways to define it.
2.2 Molar solubility: Molar solubilty is defined as the number of moles dissolved per litre of solution before the solution becomes saturated.
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