Question

One of the buffers to maintain the pH of blood is the carbonic acid, H_2CO_3, buffer. Carbonic acid is diprotic and it's equilibrium constants are K_a1 = 4.2 times 10^-7 and K_a2= 4.8 times 10^-11. How would you prepare 5.0 L a carbonic acid buffer of a pH of 7.2 from 0.250 M H_2CO_3? You have available NaHCO_3 (MW = 84.0) and Na_2CO_3 (MW = 106).

Answer 1

p^H of sodium bicarbonate is given by p^H = p^k_a 1 + p^k_a 2/2 k_a 1 = 4.2 times 10^- 7 Therefore p^k_a 1 = 7 - log 4.2 = 6.377 k_a 2 = 4.8 times 10^- 11 Therefore p^k_a 2 = 11 - log 4.2 = 10.318 So p^H of sodium bicarbonates = 6.377 + 10.318/2 = 8.34 Since p^H of sodium bicarbonate is greater than 7.2, so carbonic acid sodium bicarbonate buffer will be to make a solution of p^H = 7.2 Acc to Eqn p^H of solution is given by Equation p^H = p^k_1 + log [salt]/[acid] 7.2 = 6.377 + log [salt]/[Acid] or log [Salt]/[Acid] = 7.2 - 6.377 = 0.823 or [Salt]/[Acid] = 10^0.823 = 6.652 So Conc. Of salt (bicarbonate) = 6.652 times conc. Of acid Given conc. of acid = 0.25 M Volume of solution = 5 L Therefore No. of moles of H_2 C O_3 in 5 L solution = 0.25 times 5 = 1.25 L So no. of moles of sodium bicarbonate required = 6.652 times 1.25 = 8.315 = 8.315 times 24 g = 698.5 g