Question

Calculate the concentration of all species in a 0.180 M solution of H2CO3. [H2CO3], [HCO−3], [CO2−3], [H3O+], [OH−]

Answer 1

Ans. Step 1: Donation of first proton:

            H₂CO₃(aq) + H₂O ----------> HCO₃^-(aq) + H₃O⁺(aq)

Note: Dissociation of 1 mol H₂CO₃ yields 1 mol each of HCO₃^- and H₃O⁺.

So, at equilibrium, [HCO₃^-] = [H₃O⁺].

Create an ICE table as shown in figure with initial [H₂CO₃] = 0.180 M

Now,

            Ka1 = [HCO₃^-] [H3O⁺] / [H₂CO₃]                             ; [Ka1 = 4.5 x 10^-7]

            Or, 4.5 x 10^-7 = (X) (X) / (0.180 - X)

            Or, 0.00000045 x (0.180 - X) = X²

            Or, 0.0000000810 - 0.00000045X = X²

            Or, X² + 0.00000045X - 0.0000000810 = 0

Solving the quadratic equation, we get following two roots-

            X1 = 0.0002844        ; X2 = -0.0002848

Since concentration can’t be negative, reject X2.   

Therefore, at first equilibrium –

            [H₂CO₃] = 0.180 – X = 0.180 M – 0.0002844 M = 0.1797156 M        

            [HCO₃^-] = X molar = 0.0002844 M         

            [H₃O⁺] = X molar = 0.0002844 M

Step 2: Donation of 2^nd proton:

            HCO₃^-(aq)+ H₂O ----------> CO₃^2-(aq) + H₃O⁺(aq)

Note: Dissociation of 1 mol HCO₃^- yields 1 mol each of CO₃^2- and H₃O⁺.

Initial [HCO₃^-] = 0.0002844 M

Initial [H₃O⁺] = 0.0002844 M

Create an ICE table as shown in figure with initial concentrations (mentioned above) as shown in figure-

Now,

            Ka2 = [CO₃^2-] [H3O⁺] / [HCO₃^-]

            Or, 4.7 x 10^-11 = (X) (0.0002844 + X) / (0.0002844 - X)

[Since X << 0.0002844, its addition of subtraction from the value does not affect it]

            Or, 4.7 x 10^-11 = (X) (0.0002844) / (0.0002844) = X

            Therefore, X = 4.7 x 10^-11  

Therefore, at second equilibrium –

            [HCO₃^-] = 0.0002844 – X = 0.0002844 M (approx.)     

            [H₃O⁺] = 0.0002844 + X = 0.0002844 M (approx.)

            [CO₃^2-] = X = 4.7 x 10^-11 M           

Step 3: The final concentration: The ka2 value is very small compared to ka1.

So, the concentrations at seconds equilibrium is approximately equal to the overall concentrations of the chemical species.

Therefore, the final concentrations are-

            [H₂CO₃] = 0.1797156 M = 1.797 x 10¹ M

[HCO₃^-] = 0.0002844 M (approx.) = 2.844 x 10^-4 M   

            [H₃O⁺] = 0.0002844 M (approx.) = 2.844 x 10^-4 M

            [CO₃^2-] = 4.7 x 10^-11 M      

Now, using the formula, “ [H₃O⁺] [OH^-] = 10^-14”, the [OH^-] is given by-

            [OH^-] = 10^-14 / 0.0002844 = 3.52 x 10^-11 M