What is Henry's Law of Solubility ?

Henry's Law:

(1) The solubility of a gas in a liquid is determined by several factors. In addition to the nature of the gas and the liquid, solubility of the gas depends on the temperature and pressure of the system.

(2) The solubility of a gas in a liquid is governed by Henry's law which states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas.

(3) Dalton, a contemporary of Henry, also concluded independently that the solubility of a gas in a liquid solution is a function of the partial pressure of the gas. If we use the mole fraction of the gas in the solution as a measure of its solubility, then: Mole fraction of the gas in a solution is proportional to the partial pressure of the gas.

Or, partial pressure of the gas in solution = K_H ´ mole fraction of the gas in solution

Here K_H is Henry's law constant

                                  p = K_H X _(Solute)

If we draw a graph between partial pressure of the gas versus mole fraction of the gas in solution, then we should get a plot of the straight line passing through origin.

Experimental result for the solubility of HCl gas in Cyclohexane at 93 K the slope of line is the Henry's law constant

Different gases have different K_H values at the same temperature. This suggests that K_H is a function of the nature of the gas. Table gives K_H values of some common gases at specified temperature

Values of Henry's law constant (K_H) for some selected gases in water:

It is obvious from figure that the higher the value of K_Hat a given pressure, the lower is the solubility of the gas in the liquid. It can be seen from table that K_H value for both N₂ and O₂ increases with increase in temperature indicating that solubility of gases decreases with increase of temperature. It is due to this reason that aquatic species are more comfortable in cold waters rather than warm waters.

If edge fraction unoccupied in ideal antiflourite structure is x . Calculate x/ 0.097

A 1.0 g impure sample containing [Zn(NH3)4CI2] and some inert impurity was treated with 16 ml of 1.0 M NaOH solution, where all the complex was converted into [Na2Zn(OH)4]. Ammonia formed is first boiled off and then the excess of base required 4 ml of 1.0 M HCI solution for complete neutralisation. After the neutralisation the solution is reacted with excess of AgNO3 solution. the mass of AgCI produced (in mg) is x. The value of x10 is.

Give your Answer........

Calculate the millimoles of SeO3(2−​) in solution on the basis of following data: 70 ml of 60/M​ solution of KBrO3​ was added to SeO3(2−​) solution. The bromine evolved was removed by boiling and excess of KBrO3​ was back titrated with 12.5 mL of 25/M​solution of NaAsO2​. The reactions are given below. I. SeO3(2−​) +BrO3(−​) + H+→SeO42−​+Br2​+H2​O II. BrO3(−​) +AsO2(−​) + H2​O→Br(−) +AsO4(3−​)+ H+

SOLUTION:

A six co-ordinate complex of formula CrCl3.6H2O has green colour. One litre of O.1M solution of the complex when treated with excess of AgNO3 gave 28.7 g of white precipitate. The formula of the complex is .... (1) [Cr(H2O)6]CI3 (2) [Cr(H2O)5CI]CI2.H2O (3) [Cr(H2O)4CI2]Cl.2H2O (4) [Cr(H2O)3Cl3].3H2O

SOLUTION: 

Given 28.7 g of white ppt obtained is 28.7/147.5= 0.2 mole Agcl precipitate Since 0.1M complexgives 0.2 mole AgCl means 2Cl− are ionisable or two Cl- ion must be present out of coordination sphere so, complex is [Cr(H2​O)5​Cl]Cl2​.H2​O