Question

The \(\mathrm{pH}\) of human blood is steady at a value of approximately \(7.4\) owing to the following equilibrium reactions: \(\mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q)\) Acids formed during normal cellular respiration react with the \(\mathrm{HCO}_{3}^{-}\) to form carbonic acid, which is in equilibrium with \(\mathrm{CO}_{2}(a q)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) During vigorous exercise, a person's \(\mathrm{H}_{2} \mathrm{CO}_{3}\) blood levels were \(26.3 \mathrm{~m} M\), whereas his \(\mathrm{CO}_{2}\) levels were \(1.63 \mathrm{~m} M .\) On resting, the \(\mathrm{H}_{2} \mathrm{CO}_{3}\) levels declined to \(24.9 \mathrm{~m} M\). What was the \(\mathrm{CO}_{2}\) blood level at rest?

Short answer

The CO₂ blood level at rest is \(1.56 \times 10^{-3}\) M or \(1.56 mM\).

Step-by-step solution

Write the equilibrium constant expression for the reaction

Let's write the equilibrium constant expression for the given equilibrium reactions. The primary reaction of interest is: \[H_2CO_3(aq) \rightleftharpoons HCO_3^-(aq) + H^+(aq)\] Since the exercise has provided the pH of the blood, which remains constant throughout, we can directly use this to find the concentration of \({H^+}\) ions. The equilibrium constant expression for the reaction is given by: \[ K_{eq} = \frac{[HCO_3^-][H^+]}{[H_2CO_3]} \]

Calculate the concentration of \({H^+}\) from the pH

We are given the pH of the blood, which is 7.4. We can use the relationship between pH and \({H^+}\) concentration to find the \({H^+}\) concentration: \[ pH = -\log[H^+] \] Now, we can solve for \({H^+}\): \[ [H^+] = 10^{-pH} = 10^{-7.4} = 3.98 \times 10^{-8} M \]

Solve for the equilibrium constant during exercise

Since the \({H^+}\) concentration remains unchanged, we can use the equilibrium constant expression to solve for \(K_{eq}\) during exercise: \[ K_{eq} = \frac{[HCO_3^-][H^+]}{[H_2CO_3]} \] Using the given \(H_2CO_3\) concentration during exercise, which is \(26.3 mM\), and the \(HCO_3^-\) concentration during exercise, which is \(1.63 mM\), we can solve for \(K_{eq}\): \[ K_{eq} = \frac{(1.63 \times 10^{-3})(3.98 \times 10^{-8})}{26.3 \times 10^{-3}} = 2.51 \times 10^{-4} \]

Calculate the \({CO_2}\) blood level at rest

With the equilibrium constant value now known, we can use the equilibrium expression with the \(H_2CO_3\) concentration at rest, which is 24.9 mM, to solve for the \({CO_2}\) blood level at rest. Let the \(CO_2\) blood level at rest be \(x\): \[ K_{eq} = \frac{(x)(3.98 \times 10^{-8})}{24.9 \times 10^{-3}} \] Now, solve for x: \[ x = \frac{K_{eq}(24.9 \times 10^{-3})}{(3.98 \times 10^{-8})} = \frac{(2.51 \times 10^{-4})(24.9 \times 10^{-3})}{(3.98 \times 10^{-8})} = 1.56 \times 10^{-3} M \] Therefore, the \({CO_2}\) blood level at rest is \(1.56 \times 10^{-3}\) M or \(1.56 mM\).

Key concepts

Equilibrium Constant Expression

Understanding the equilibrium constant expression is essential for students studying chemical reactions, especially when exploring the dynamic balance within the human body. The expression quantifies the concentrations of reactants and products in a chemical equilibrium. It's expressed as:ewline ewline
\[K_{eq} = \frac{[\text{products}]}{[\text{reactants}]}\]Where \(K_{eq}\) is the equilibrium constant and the square brackets \([]\) denote concentration in molarity (moles per liter). For reactions with multiple products and reactants, the expression must be adjusted accordingly with each component raised to the power of their respective stoichiometric coefficients.ewline
In the exercise, the equilibrium constant expression is tailored to the human blood's acid-base reactions, focusing on carbonic acid (\(H_2CO_3\)), bicarbonate ions (\(HCO_3^-\)), and hydronium ions (\(H^+\)). It is a crucial component for calculating the changes in concentrations of these species under different physiological conditions, like rest and exercise.

Acid-Base Balance in Human Body

Maintaining the delicate acid-base balance is critical for human survival. Our blood's pH — a measure of acidity or alkalinity — hovers around a narrow range of 7.35 to 7.45. Deviations from this range can result in harmful conditions such as acidosis or alkalosis.To prevent such deviations, the body employs various buffer systems, pathways for excreting or absorbing acids and bases, and even changes in respiratory rates to manage the pH levels. These systems work continuously and in integrated ways to correct pH imbalances that occur from metabolism, exercise, or dietary intake. The exercise given illustrates how blood levels of \(CO_2\) and \(H_2CO_3\) change during physical activity, impacting the blood pH. The ability to calculate these changes through the equilibrium constant expression is an essential tool for understanding the body's compensatory mechanisms in maintaining pH homeostasis.

Carbonic Acid-Bicarbonate Buffer System

The carbonic acid-bicarbonate buffer system is one of the main regulators of blood pH, making it a frequent subject in health and medicine-related studies. It operates on the following reversible reaction:\[CO_2(aq) + H_2O(l) \rightleftharpoons H_2CO_3(aq) \rightleftharpoons HCO_3^-(aq) + H^+(aq)\]Where \(CO_2\) (carbon dioxide) and \(H_2O\) (water) react to form \(H_2CO_3\) (carbonic acid), which quickly dissociates into \(HCO_3^-\) (bicarbonate ions) and \(H^+\) (hydrogen ions). This buffer system is so effective because the concentrations of bicarbonate and carbonic acid can be quickly adjusted by the lungs and kidneys. The lungs modulate carbon dioxide expulsion, and the kidneys regulate bicarbonate ion excretion or retention.In the context of our exercise, understanding this buffer system is crucial to assessing how the body responds to the increased production of \(H_2CO_3\) during exercise and the consequent need to recalibrate the \(CO_2\) levels to maintain a stable pH.