Question

A 2.50 L container is filled with gaseous H2S at 13.0ºC and a pressure of 169.1 torr. Then 200.0 mL of water is added to the container (without allowing any of the H2S to escape). A small amount of H2S dissolves in the water. When the system has reached equilibrium, the gas pressure in the container is 155.9 torr and the temperature is still 13.0ºC. The vapor pressure of water at this temperature is 11.2 torr. What is the molar concentration of H2S in the water?

Answer 1

first Use the ideal gas equation to calculate the number of moles of H2S :

PV = n RT

P = Pressure 169.1 torr or 0.2225 atm   , T= temperature 13 or 286 K , R = gas constant; 0.082057 L atm mol^-1K^-1 V= Volume, 2.50 L

OR n =PV/ RT

= 0.2225 * 2.50/ 0.082057 L atm mol^-1K^-1 *286

= 0.0237 Moles H2S

Now again calculate the moles of H2S:

Calculate the moles of H2S remaining in the vapor space. Note the volume has been reduced by 200 mL.

Thus new volume = 2500 ml-200 ml = 2300 ml

= 2.3 L

Here we use the partial pressure of H2S= total pressure - pressure of water at this temperature

= 155.9 torr- 11.2 torr

= 144.7 torr or 0.190 atm

PV = n RT

OR n =PV/ RT

= 0.190 * 2.30/ 0.082057 L atm mol^-1K^-1 *286

= 0.0186 Moles H2S

The difference is the moles of H2S dissolved in the water=
0.0237 Moles H2S -0.0186 Moles H2S

= 0.0051 Moles H2S
Molar concentration = moles of H2S dissolved / volume of water in L.

= 0.0051 Moles H2S /0.200 L

= 0.0255 M H2S

Thus the molar concentration of H2S in the water is 0.0255 M H2S