Question

A saturated solution was formed when 5.16×10−2 L of argon, at a pressure of 1.0 atm and temperature of 25 ∘C, was dissolved in 1.0 L of water.
Calculate the Henry's law constant for argon. it must be im M/atm

Answer 1

Answer:

The Henry's law constant for argon is $k=2.11*10^{-3}\frac{ M}{atm}$

Explanation:

Henry's Law indicates that the solubility of a gas in a liquid at a certain temperature is proportional to the partial pressure of the gas on the liquid.

C = k*P

where C is the solubility, P the partial pressure and k is the Henry constant.

So, being the concentration $C=\frac{ngas}{V}$  

where ngas is the number of moles of gas and V is the volume of the solution, you must calculate the number of moles ngas. This is determined by the Ideal Gas Law: P*V=n*R*T where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. So $n=\frac{P*V}{R*T}$

In this case:

• P=PAr= 1 atm
• V=VAr= 5.16*10⁻² L
• R=0.082 $\frac{atm*L}{mol*K}$
• T=25 °C=298 °K

Then:

$n=\frac{1 atm*5.16*10^{-2} L}{0.082 \frac{atm*L}{mol*K} *298K}$

Solving:

n= 2.11 *10⁻³ moles

So: $C=\frac{ngas}{V}=\frac{2.11*10^{-3} moles}{1 L} =2.11*10^{-3} \frac{moles}{L}= 2.11*10^{-3} M$

Using Henry's Law and being C=CAr and P =PAr:

2.11*10⁻³ M= k* 1 atm

Solving:

$k=\frac{2.11*10^{-3} M}{1 atm}$

You get:

$k=2.11*10^{-3}\frac{ M}{atm}$

The Henry's law constant for argon is $k=2.11*10^{-3}\frac{ M}{atm}$