Question

How is the Henderson-Hasselbalch equation used to calculate the ratio of ${H_2}C{O_3}$ to $HCO_3^ -$ in blood having a pH of 7.40?

Answer 1

The Henderson-Hasselbalch equation is a mathematical equation which gives relation between the pH of the solution and the $p{K_a}$ which is equal to the $- \log {K_a}$. The ${K_a}$ is the acid dissociation constant of the weak base. We need to determine the ratio of weak acid ${H_2}C{O_3}$to its conjugate base $HCO_3^ -$.

Complete step by step answer:
The equation which relates the pH of an aqueous solution of an acid to the acid dissociation constant of the acid is described as the Henderson-Hasselbalch equation.
The equation is given as shown below.
$pH = p{K_a} + \log \left( {\dfrac{{[Conjugate\;base]}}{{[weak\;acid]}}} \right)$
In this question it is given that the weak acid is ${H_2}C{O_3}$and its conjugate base is $HCO_3^ -$.
Substitute it in the given equation.
$\Rightarrow pH = p{K_a} + \log \left( {\dfrac{{[HCO_3^ - ]}}{{[{H_2}C{O_3}]}}} \right)$
$K{a_1}({H_2}C{O_3}) = 4.5\times{10^{ - 7}}$
The $p{K_a}$ value is equal to the negative logarithm of acid dissociation constant of the weak acid.
It is given as shown below.
$p{K_a} = - \log \left[ {{K_a}} \right]$
Where,
${K_a}$ is the acid dissociation constant of the weak acid.
Substitute the value in the given equation.
$\Rightarrow p{K_a} = - \log \left[ {4.5\times{{10}^{ - 7}}} \right]$
$\Rightarrow p{K_a} = 6.4$
It is given that the pH is 7.40
Now we need to determine the ratio which exists between the concentration of the conjugate base, $HCO_3^ -$ and the concentration of the weak acid ${H_2}C{O_3}$.
Substitute the value in the equation.
$\Rightarrow 7.40 = 6.4 + {\log _{10}}\left( {\dfrac{{[HCO_3^ - ]}}{{[{H_2}C{O_3}]}}} \right)$
$\Rightarrow {\log _{10}}\left( {\dfrac{{\left[ {HCO_3^ - } \right]}}{{{H_2}C{O_3}}}} \right) = 7.4 - 6.4$
$\Rightarrow {\log _{10}}\left( {\dfrac{{\left[ {HCO_3^ - } \right]}}{{{H_2}C{O_3}}}} \right) = 1.0$
$\Rightarrow \left( {\dfrac{{\left[ {HCO_3^ - } \right]}}{{{H_2}C{O_3}}}} \right) = {10^{1.0}}$
$\Rightarrow \left( {\dfrac{{\left[ {HCO_3^ - } \right]}}{{{H_2}C{O_3}}}} \right) = 10$
$\Rightarrow ([HCO_3^ - ]:[{H_2}C{O_3}]) = 10:1$

Therefore, the ratio of ${H_2}C{O_3}$ to $HCO_3^ -$ in blood having a pH of 7.40 is 10:1.

Note: The $p{K_a}$ value measures the strength of the acid is solution. The weak acid has $p{K_a}$ value ranging from 2-12 in water. The Henderson-Hasselbalch equation is also used to determine the pH of the buffer solution and the equilibrium pH in an acid-base reaction.