major buffer systems in the body

Major chemical buffers in the body

Bicarbonate buffer

A normal adult produces about 300 L of CO₂ daily from metabolism. CO₂ from tissues enters the capillary blood, where it reacts with water to form H₂CO₃, which dissociates instantly to yield H⁺ and HCO₃ ^–.

Blood pH would rapidly fall to lethal levels if the H₂CO₃ formed from CO₂ were allowed to accumulate in the body.

Fortunately, H₂CO₃ produced from metabolic CO₂ is only formed transiently in the transport of CO₂ by the blood and does not normally accumulate. Instead, it is converted to CO₂ and water in the pulmonary capillaries and the CO₂ is expired. As long as CO₂ is expired as fast as it is produced, arterial blood CO₂ tension, H₂CO₃ concentration, and pH do not change.

pK of the HCO₃^-/CO₂ is 6.1.

pH of plasma is 7.4.

 According to Henderson-Hasselbalch equation

7.4 = 6.1 + log [HCO₃^-]/ [CO₂]

1.3= log [HCO₃^-]/ [CO₂]

20= [HCO₃^-]/ [CO₂]

i.e. At pH 7.4, concentration of bicarbonate should be 20 times that of Carbon dioxide.

As described in the previous post, it is the salt portion that neutralizes external acids, so a higher concentration of bicarbonate should suffice. This explains the relative importance of bicarbonate/CO2 buffer over other buffers because most metabolic end products are acidic in nature.

Concentration of HCO3 ^-   in plasma or ECF normally averages 24 mmol/L.  CO₂ is measured from PCO₂ which is about 40 mm Hg and that amounts to 1.2mmol/L of CO₂ (0.03* 40) {The solubility coefficient for CO2 in plasma at 37°C is 0.03 mmol CO2/L per mm Hg PCO2 }.  Although the concentration of dissolved CO2 is lower, metabolism provides a nearly limitless supply. Hence [HCO₃-]/ [CO₂] = 24/1.2=20 supports the theoretical calculations performed above using Henderson-Hasselbalch equation.

The Henderson-Hasselbalch Equation for HCO₃^_/CO₂

_{In aqueous solutions:}

CO₂(d)   +    H₂O      give       H₂CO₃

_(dissolved)

At equilibrium, CO₂(d) is greatly favored; at body temperature,  [CO₂(d)] : [H₂CO₃] is about 400:1 [If [CO₂(d)] is 1.2 mmol/L, then [H₂CO₃] equals 3 μmol/L].

H₂CO₃     instantaneously breaks into H⁺ + HCO₃^-

The Henderson-Hasselbalch Equation for the above equation is:

pH= 3.5 +  log [HCO₃^-]/ [H₂CO₃]

H₂CO₃ is a fairly strong acid (pKa = 3.5). Its low concentration in body fluids lessens its impact on acidity. Because [H₂CO₃] is so low and hard to measure so [CO_2(d)] is used instead.

pH= 3.5 +  log [HCO₃^-]/ [CO_2(d)]/400

pH= 3.5+ log 400 + log [HCO₃^-]/ [CO_2(d)]

pH= 6.1+ log [HCO₃^-]/ [CO_2(d)]

pH= 6.1+ log [HCO₃^-]/ 0.03*pCO₂

pH= 6.1  + log [24]/[1.2]

pH=7.4