What is Henry’s constant for neon dissolved in water given $C_{N e} = 23.5 m L L^{- 1}$ solution and STP volume $(22, 414 m L m o l e^{- 1}$ gas) and pressure $(1 a t m)$ ?

Answer

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Hint:As we know that Henry’s law states that the amount of the gas that dissolves in a given volume of liquid is directly proportional to the partial pressure of the gas in equilibrium with that solution, at a constant temperature. When this proportionality sign is removed the constant put in its place is the Henry’s constant which is a measure of the solubility of a gas.

Formula used:

m o l e s = \frac{V o l . o f s o l u t e a t S T P}{22, 414}

and

S = K_{H} P

.

Complete answer:
We know that according to Henry’s law which states that “the partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution”. When this proportionality sign is removed the constant put in its place is the Henry’s constant which is a measure of the solubility of a gas. This relation can be given as:

p^{\circ} = K_{H} χ

or in terms of solubility it can be written as

S = K_{H} P

where

K_{H}

is the Henry’s law constant.
Now we are given that

1

litres of solution contains

23.5 m L

of Neon, so we can first calculate the moles of solution at STP. Using the formula:

\Rightarrow m o l e s = \frac{V o l . o f s o l u t e a t S T P}{22, 414}

\Rightarrow m o l e s = \frac{23.5}{22414} = 0.001048 m o l e s

So, we can say that the solubility in moles per litre or concentration of the solution is

0.001048 M

.
Now using the above formula, we can calculate the Henry’s constant as:

\Rightarrow S = K_{H} P

\Rightarrow 0.001048 = K_{H} \times 1

Therefore the Henry’s constant is found to be $K_{H} = 0.001048 L a t m^{- 1}$ .

Note:Different gases have different Henry’s constant values at the same temperature suggesting that Henry’s constant is a function of nature of the gas. At a given pressure, higher the value of Henry’s constant, lower will be solubility and Henry’s constant increases with increase in temperature, so solubility will decrease with increase in temperature.