Problem 29 - Chemistry: The Central Science in SI Units, Global Edition

Question

(a) What is the ratio of \(\mathrm{HCO}_{3}^{-}\) to \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in blood of \(\mathrm{pH} 7.4\) ? (b) What is the ratio of \(\mathrm{HCO}_{3}^{-}\) to \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in an exhausted marathon runner whose blood \(\mathrm{pH}\) is \(7.1 ?\)

Step-by-Step Solution

Answer

In summary, the ratio of HCO3- to H2CO3 in blood with a pH of 7.4 is approximately 19.95, and the ratio in blood with a pH of 7.1 is 10.

Step 1: Calculate the ratio of [HCO3-] to [H2CO3] for pH 7.4

Given the blood pH is 7.4, we can use the Henderson-Hasselbalch equation to find the ratio of [HCO3-] to [H2CO3]. The pKa value for the H2CO3/HCO3- buffer system is 6.1, so we plug the values into the equation: \[7.4 = 6.1 + \log{\frac{[HCO_{3}^{-}]}{[H_{2}CO_{3}]}}\] (a)

Step 2: Solve for the ratio of [HCO3-] to [H2CO3]

To find the ratio of [HCO3-] to [H2CO3], we first need to isolate the ratio. Rearrange the equation and solve for the ratio: \[7.4 - 6.1 = \log{\frac{[HCO_{3}^{-}]}{[H_{2}CO_{3}]}}\] \[1.3 = \log{\frac{[HCO_{3}^{-}]}{[H_{2}CO_{3}]}}\] Next, take the inverse logarithm of both sides: \[[HCO_{3}^{-}] : [H_{2}CO_{3}] = 10^{1.3}\] \[[HCO_{3}^{-}] : [H_{2}CO_{3}] \approx 19.95\] So, the ratio of [HCO3-] to [H2CO3] is approximately 19.95 when the blood pH is 7.4. (b)

Step 3: Calculate the ratio of [HCO3-] to [H2CO3] for pH 7.1

Now, we repeat the steps for the blood pH of 7.1 using the Henderson-Hasselbalch equation and the pKa value of 6.1: \[7.1 = 6.1 + \log{\frac{[HCO_{3}^{-}]}{[H_{2}CO_{3}]}}\] (b)

Step 4: Solve for the ratio of [HCO3-] to [H2CO3]

Rearrange and solve for the ratio: \[7.1 - 6.1 = \log{\frac{[HCO_{3}^{-}]}{[H_{2}CO_{3}]}}\] \[1 = \log{\frac{[HCO_{3}^{-}]}{[H_{2}CO_{3}]}}\] Take the inverse logarithm of both sides: \[[HCO_{3}^{-}] : [H_{2}CO_{3}] = 10^1\] \[[HCO_{3}^{-}] : [H_{2}CO_{3}] = 10\] So, the ratio of [HCO3-] to [H2CO3] is 10 when the blood pH is 7.1. In summary, the ratio of HCO3- to H2CO3 in blood with a pH of 7.4 is approximately 19.95, and the ratio in blood with a pH of 7.1 is 10.

Key Concepts

Buffer System

A buffer system is a combination of a weak acid and its conjugate base that helps maintain a stable pH level in a solution. In the human body, the bicarbonate buffer system is one of the most important. This system involves the weak acid carbonic acid \( \text{H}_2\text{CO}_3 \) and its conjugate base \( \text{HCO}_3^{-} \). The buffer system's main role is to neutralize pH changes, caused by the addition of an acid or a base. This is vital for processes such as enzyme function and oxygen transport by hemoglobin. When there is an increase in \( \text{H}^+ \) ions, the pH drops, and \( \text{HCO}_3^{-} \) reacts to form \( \text{H}_2\text{CO}_3 \). Conversely, when \( \text{H}^+ \) ions decrease, \( \text{H}_2\text{CO}_3 \) dissociates to form \( \text{HCO}_3^{-} \) and replenish \( \text{H}^+ \) ions.

• This dynamic equilibrium helps keep the blood pH between 7.35 and 7.45.
• Even small deviations from this range can impair bodily functions.
• The buffer systems act as the body's natural defense against extreme pH shifts.

pH Calculation

The pH calculation reveals the acidity or basicity of a solution and is expressed as the negative logarithm of the hydrogen ion concentration \(-\log[\text{H}^+]\). A major tool for calculating pH in buffered solutions is the Henderson-Hasselbalch equation: \[pH = pK_a + \log\left(\frac{[\text{conjugate base}]}{[\text{acid}]}\right)\] For the bicarbonate buffer system specifically, this formula becomes: \[pH = pK_a + \log\left(\frac{[\mathrm{HCO}_3^{-}]}{[\mathrm{H}_2\mathrm{CO}_3]}\right)\] Using this equation:

• The pKa value of carbonic acid, which is 6.1, provides a reference point.
• The pH reflects the balance between \( \text{HCO}_3^{-} \) and \( \text{H}_2\text{CO}_3 \).
• In blood, the typical pH (7.4) is maintained by a favorable ratio of these components.

With this equation, it becomes evident how altering \( \text{HCO}_3^{-} \) or \( \text{H}_2\text{CO}_3 \) concentrations affects pH, displaying the sensitivity of this buffer system.

Acid-Base Equilibria

Acid-base equilibria refer to the balance between acids and bases in a solution, dictated by the reaction of substances that donate protons (acids) and those that accept them (bases). In a buffer system like the bicarbonate buffer, equilibrium is essential to sustaining life by preventing drastic pH changes. In our case: * Carbonic acid \( \text{H}_2\text{CO}_3 \) acts as the weak acid. * Bicarbonate \( \text{HCO}_3^{-} \) serves as its conjugate base. The concept of equilibrium can be understood by the dissociation process: \[\text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-\] This reaction maintains equilibrium between subsequent protons \( \text{H}^+ \) produced and absorbed. In scenarios where blood pH shifts, the equilibrium can adjust swiftly:

• If the pH lowers, indicating too many \( \text{H}^+ \) ions, \( \text{HCO}_3^{-} \) neutralizes them, swinging the equilibrium to more reactants formation.
• Conversely, an increase in pH shifts equilibrium to form more \( \text{H}^+ \) and \( \text{H}_2\text{CO}_3 \).

Such delicate balance ensures that biological systems function optimally despite constant disturbances. Understanding this interplay is key to mastering acid-base chemistry.