7. The pH of blood plasma from a patient was found to be 7.6, a life-threatening situation. Calculate the ratio [HCO3-]/[H2CO3] in this sample of blood plasma.

Question

7. The pH of blood plasma from a patient was found to be 7.6, a life-threatening situation. Calculate the ratio [HCO3-]/[H2CO3] in this sample of blood plasma. (Ka of H2CO3, carbonic acid, is 4.5×10-7.)

Answer 1

The ratio would be 18.1/. The Henderson-Hasselbach equation is : pH=pKa + log [base]/[acid]. So 7.6=pKa carbonic acid (-log 4.5 x10^-7)=6.34, so 7.6-6.34=log x, x= 10^1.26=18/1.

Answer 2

H₂CO₃ <==> H⁺ + HCO₃^-

This creates a buffer because you have a weak acid (H₂CO₃) and the conjugate base (HCO₃^-).

Using the Henderson Hasselbalch equation, we can find the ratio of conjugate base/weak acid:

pH = pKa + log [conj.base]/[weak acid]

The pKa is obtained from the given Ka: pKa = -log Ka = -log 4.5x10^-7 = 6.35

Substituting the values into the HH equation we have...

7.6 = 6.35 + log [HCO₃^-/H₂CO₃]

1.25 = log x

x = 17.8 so ratio is 17.8 : 1 or ~18 to 1