   Chapter 14, Problem 146AE

Chapter
Section
Textbook Problem

# The pH of human blood is steady at a value of approximately 7.4 owing to the following equilibrium reactions: CO 2 ( a q ) + H 2 O ( l ) ⇌ H 2 CO 3 ( a q ) ⇌ H 2 CO 3 − ( a q ) + H + ( a q ) Acids formed during normal cellular respiration react with the HCO3− to form carbonic acid, which is in equilibrium with CO2(aq) and H2O(l). During vigorous exercise, a person’s H2CO3 blood levels were 26.3 mM whereas his CO2 levels were 1.63 mM. On resting, the H2CO3 levels declined to 24.9 mM. What was the CO2 blood level at rest?

Interpretation Introduction

Interpretation: The pH value of human body, the reaction of normal cellular respiration, concentration of H2CO3 and CO2 during exercise and the concentration of H2CO3 is given. By using these values, the concentration of CO2 is to be calculated.

Concept introduction: The equilibrium constant Ka is the ratio of product of equilibrium concentrations of product to the equilibrium concentration of reactant acid. A reaction of dissociation of ion is,

HA(aq)+H2O(l)H3O+(aq)+A(aq)

Ka is the acid dissociation constant for the above reaction where proton is removed from HA to form conjugate base A . The Ka for above reaction is,

Ka=[H3O+][A][HA]

Explanation

Explanation

To determine: The CO2 blood level at rest.

The acid dissociation constant of given reaction is 16.1_ .

The stated reaction is,

CO2(g)+H2O(l)H2CO3(aq)HCO3(aq)+H+(aq)

Given

The concentration of H2CO3 during exercise is 26.3mM .

The concentration of CO2 is during exercise is 1.63mM .

The concentration of H2CO3 is at rest is 24.9mM .

The reaction is,

CO2(g)+H2O(l)H2CO3(aq)

Formula

The acid dissociation constant is calculated using the formula,

Ka=[H2CO3][CO2]

Where,

• Ka is acid dissociation constant.
• [H2CO3] is the concentration of H2CO3 during exercise.
• [CO2] is the concentration of CO2 during exercise.

Substitute the values of [H2CO3] and [CO2] in the above equation.

Ka=[H2CO3][CO2]=26.3mM1.63mM=16

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