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Why is high relative supersaturation undesirable in a gravimetric precipitation?

Short Answer

Expert verified

The high relative supersaturation undesirable in a gravimetric precipitation due to the fast nucleation- suspension of colloids would be higher would isn’t desirable.

Step by step solution

01

Definition

It is an analytical technique that uses a precipitation reaction to separate ions from a solution. The chemical that is added to cause the precipitation is called the precipitant or precipitating agent

02

Formation of Aggregrates

During the process of nucleation in gravimetric precipitation process, molecules are closer to one another in a solution which leads to the formation of aggregrates.

03

Relative supersaturation is undesirable

This process appear to be faster in supersaturated solutions, and due to fast nucleation, suspension of colloids would be higher would isn’t desirable. Therefore, in gravimetric precipitation high relative supersaturation is undesirable.

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Most popular questions from this chapter

Consider a mixture of the two solids,

BaCl22H2O(FM244.26)andKCI(FM74.551), in an unknown ratio. (The notation BaCl22H2O means that a crystal is formed with two water molecules for each BaCl2) When the unknown is heated to 160°C for 1h, the water of crystallization is driven off:

BaCl22H2O(s)160°BaCI(s)+2H2O(g)

A sample originally weighing 1.7839 g weighed 1.5623 g after heating. Calculate the weight percent Ba,K and Cl ofin the original sample.

Write a balanced equation for the combustion of benzoic acid,C6H5CO2H, to giveCO2 and H2O. How many milligrams of and of will be produced by the combustion of 4.635mgof benzoic acid?

State four desirable properties of a gravimetric precipitate.

Combustion analysis of a compound known to contain justC,H,N, and Odemonstrated that it is 46.21wt%C,9.02wt% H, 13.74wt N and, by difference, 100 - 46.21 - 9.02 -13.74 = 31.03% O. This means that of unknown would contain46.21gof C,9.02g ofH, and so on. Find the atomic ratio C : H : N : Oand express it as the lowest reasonable integer ratio.

Man in the vat problem.15 Long ago, a workman at a dye factory fell into a vat containing hot, concentrated sulfuric and nitric acids. He dissolved completely! Because nobody witnessed the accident, it was necessary to prove that he fell in so that the man’s wife could collect his insurance money. The man weighed 70 kg, and a human body contains |6.3 parts per thousand (mg/g) phosphorus. The acid in the vat was analyzed for phosphorus to see whether it contained a dissolved human.

(a) The vat contained8.00×103Lof liquid, and a 100.0-mL sample was analyzed. If the man did fall into the vat, what is the expected quantity of phosphorus in 100.0 mL?

(b) The 100.0-mL sample was treated with a molybdate reagent that precipitated ammonium phosphomolybdate,(NH4)3[P(Mo12O40)]12H2OThis substance was dried at110°Cto remove waters of hydration and heated to400°Cuntil it reached the constant compositionP2O5×24MoO3, which weighed 0.371 8 g. When a fresh mixture of the same acids (not from the vat) was treated in the same manner, 0.033 1 g ofP2O5×24MoO3(FM3596.46)was produced. This blank determination gives the amount of phosphorus in the starting reagents. TheP2O5×24MoO3that could have come from the dissolved man is therefore0.3718-0.0331=0.3387g.How much phosphorus was present in the 100.0-mL sample? Is this quantity consistent with a dissolved man?

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