Question

A 1.100−L flask at 25∘C and 1.00 atm pressure contains CO2(g) in contact with 100.0 mL of a saturated aqueous solution in which [CO2(aq)] = 3.29×10−2 M.

a) What is the value of Kc at 25∘C for the equilibrium CO2(g)⇌CO2(aq)

Answer: Kc = 0.804

b) If 0.01400 M of radioactive 14CO2 is added to the flask, how many moles of the 14CO2 will be found in the gas phase and in the aqueous solution when equilibrium is reestablished?
[Hint: The radioactive 14CO2 distributes itself between the two phases in exactly the same manner as the nonradioactive 12CO2.]

14CO2(g), 14CO2(aq) = ? mol

Answer 1

Ans :- a) The equilibrium reaction given is

CO₂(g)CO₂(aq)

we know,

equilibrium constant,Kc = [product] / [reactant]

                                   =[CO₂(aq)] /[CO₂(g)]

Here given [CO₂(aq) ] = 3.29x10^-2 = 0.0329M

Given ,temperature = 25⁰C = 298K

pressure ,P= 1atm

R=0.082057Latm/mol/K

we know from ideal gas equation

PV=nRT

n=PV/RT

Also we know ,concentration = n/V

[CO₂(g)]=n/V

   = PV / VRT

   =P/RT

   = 1/ 0.082057x298

= 0.04089

K_c= [CO₂(aq)] /[CO₂(g)]

=0.0329/0.04089

   = 0.804

Ans b) Here 0.0140M of ¹⁴CO₂ is added to the flask

The given reaction is CO₂(g) CO₂(aq)

	[CO2(g)]	[CO2(aq)]
initial	0.04089	0.0329
added	+0.0140	0
change	-x	+x
equilibrium	0.05498-x	0.0329+x

Given Kc=0.804

Now

0.804=[0.0329+x] / [0.05498-x]

x=0.0062

Now,

[CO₂(g)]=0.05498-0.0062 =0.04878

[CO₂(aq)]=0.0329+x = 0.0391

Volume for CO₂(g) = 1L

Volume for CO₂(aq) = 0.1L

moles of CO₂(g) = [CO₂(g) ] x volume

                         = 0.0487 x1

                          = 0.0487 mol

moles of CO₂(aq) =[CO₂(aq)] x volume

                          = 0.1 x0.0391

                            = 0.00391

For radioactive component ¹⁴CO₂

number of moles of component = 1

total number of moles both gaseous and aqueous CO₂ after adding 0.0140M of ¹⁴CO₂ = 0.0487+0.00391

                                                                                                                             = 0.05261

mole fraction of ¹⁴CO₂ = 1/ 0.05261

                                   =0.19

moles of ¹⁴CO₂(g) = moles of CO₂(g) x mole fraction of ¹⁴CO₂

                             = 0.0487 x 0.19

                              = 0.0092

moles of ¹⁴CO₂ (aq) = moles of CO₂(aq) x mole faction of ¹⁴CO₂

                        = 0.00391 x 0.19

                          = 0.00074