Question

The bicarbonate buffering system can be described as shown below:

CO2(d) + H2O  H2CO3 Keq= 0.003 at 37C (Hint: this is a simple equilibrium) H2CO3  H+ + HCO3- Ka= 0.000269 at 37C (Hint: this is an ionization)

A. (2 pts) If the plasma pH is 7.4 and the plasma concentration of of HCO3- is 15 mM, what is the plasma concentration of H2CO3?

B. (1 pts) What is the concentration of CO2 (dissolved) ?

C. (1.) If the metabolic activity changes the concentration of CO2 (dissolved) to 3 mM and the [HCO3-] remains at 15 mM, what is the pH of the plasma?

Answer 1

The two equilibrium reactions are

CO₂ (g) + H₂O (l) <=====> H₂CO₃ (aq)

K_eq = 0.003 = [H₂CO₃]/[CO₂] (concentration of solids and liquids are not included) ……(1)

H₂CO₃ (aq) -------> H⁺ (aq) + HCO₃^- (aq)

K_a = 0.000269 = [H⁺][HCO₃^-]/[H₂CO₃] …….(2)

A) K_eq and K_a are equilibrium constants and hence must remain constant at a particular temperature.

Given pH = 7.4, [H⁺] = antilog (-pH) = 10^-7.4 = 3.98*10^-8 M and [HCO₃^-] = 15 mM = (15 mM)*(1 M/1000 mM) = 1.5*10^-3 M, plug in values in expression (2) to get

0.000269 = (3.98*10^-8).(1.5*10^-3)/[H₂CO₃]

====> [H₂CO₃] = (3.98*10^-8).(1.5*10^-3)/(0.000269) = 2.219*10^-7 ≈ 2.22*10^-7

The equilibrium concentration of H₂CO₃ is 2.22*10^-7 M (ans).

B) Put the value of [H₂CO₃] in expression (1) and get

0.003 = (2.22*10^-7 M)/[CO₂]

===> [CO₂] = (2.22*10^-7 M)/(0.003) = 7.4*10^-5 M (ans).

C) Put [CO₂] = 3 mM = (3 mM)*(1 M/1000 mM) = 0.003 M. Plug in the value in expression (1) and get

[H₂CO₃] = 0.003*[CO₂] = 0.003*(0.003 M) = 9.0*10^-6 M.

Use the value of [H₂CO₃] in expression (2) to find [H⁺].

[H⁺] = 0.000269*[H₂CO₃]/[HCO₃^-] = 0.000269*(9.0*10^-6 M)/(1.5*10^-3 M) = 1.614*10^-6 M.

Therefore, pH = -log [H⁺] = -log (1.614*10^-6) = 5.792 ≈ 5.79 (ans).