Question & Answer: One of the buffers to maintain the pH of blood is the carbonic acid, H_2CO_3, buffer. Carbonic…..

The dissociates of Carbonic acid, H₂CO₃ is occurs in two steps as follows:.

H₂CO₃ (aq) <=====> H⁺ (aq) + HCO₃^– (aq) …..(1)

HCO₃^– (aq) <=====> H⁺ (aq) + CO₃^2- (aq) ….(2)

Given that;

K_a1 = 4.2*10^-7

And;

pK_a1 = -log K_a1

= -log (4.2*10^-7)

= 6.38

K_a2 = 4.8*10^-11;

pK_a2 = -log K_a2

= -log (4.8*10^-11)

= 10.32

The pH of buffer is 7.2 hence we choose H₂CO₃/HCO₃^– ,because the pK_a1 (6.38 )of H₂CO₃ is neatest within unit of the pH (7.2).

Now,

Use Henderson-Hasslebach equation as below:

pH = pK_a1 + log [HCO₃^–]/[H₂CO₃]

7.2 = 6.38 + log [HCO₃^–]/[H₂CO₃]

0.82 = log [HCO₃^–]/[H₂CO₃]

[HCO₃^–]/[H₂CO₃] =10^0.82

[HCO₃^–]/[H₂CO₃] = 6.61

[HCO₃^–] = 6.61*[H₂CO₃] ….(3)

Here given that [H₂CO₃] = 0.250 M;

therefore, [HCO₃^–] = 6.61*0.250 M

= 1.65175 M.

Moles of NaHCO₃ required = molarity * volume in L

=(1.65175 mol/L)*(5 L)

= 8.26 mole (1 M = 1 mol/L).

Mass of NaHCO₃ required = (8.26 mole)*(84.0 g/mol)

= 693.8 g