The normal \(\mathrm{pH}\) of blood is \(7.40 \pm 0.05\) and is controlled in part by the \(\mathrm{H}_{2} \mathrm{CO}_{3} / \mathrm{HCO}_{3}^{-}\) buffer system. (a) Assuming that the \(K_{\mathrm{a}}\) value for carbonic acid at \(25^{\circ} \mathrm{C}\) applies to blood, what is the \(\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right] /\left[\mathrm{HCO}_{3}^{-}\right]\) ratio in normal blood? (b) In a condition called acidosis, the blood is too acidic. What is the \(\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right] /\left[\mathrm{HCO}_{3}^{-}\right]\) ratio in a patient whose blood \(\mathrm{pH}\) is \(7.20 ?\)

Short Answer

Expert verified
For normal blood pH 7.4, the ratio is 0.0502, and for pH 7.2 (acidosis), it is 0.0794.

Step by step solution

01

Introduction to Problem

We need to find the \(\frac{[\text{H}_2\text{CO}_3]}{[\text{HCO}_3^-]}\) ratio in normal blood (pH = 7.40) and in a patient with acidosis (pH = 7.20). We'll use the Henderson-Hasselbalch equation to solve this problem.
02

- Understanding the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is given by: \[ \text{pH} = \text{p}K_a + \text{log} \frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]} \] We can rearrange this to solve for the ratio \(\frac{[\text{H}_2\text{CO}_3]}{[\text{HCO}_3^-]}\).
03

- Find the \(\text{p}K_a\) of Carbonic Acid

The \(\text{p}K_a\) of carbonic acid (H_2CO_3) at 25°C is approximately 6.1.
04

- Calculate the Ratio for Normal Blood pH (7.40)

Using the Henderson-Hasselbalch equation: \[ 7.40 = 6.1 + \text{log} \frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]} \] We can rearrange this to solve for the ratio: \[ \frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]} = 10^{7.40 - 6.1} = 10^{1.3} \] Therefore, \[ \frac{[\text{H}_2\text{CO}_3]}{[\text{HCO}_3^-]} = \frac{1}{10^{1.3}} \] Which simplifies to: \[ \frac{[\text{H}_2\text{CO}_3]}{[\text{HCO}_3^-]} \approx 0.0502 \]
05

- Calculate the Ratio for Acidosis (pH 7.20)

Using the Henderson-Hasselbalch equation: \[ 7.20 = 6.1 + \text{log} \frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]} \] Rearranging to solve for the ratio: \[ \frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]} = 10^{7.20 - 6.1} = 10^{1.1} \] Therefore, \[ \frac{[\text{H}_2\text{CO}_3]}{[\text{HCO}_3^-]} = \frac{1}{10^{1.1}} \] Which simplifies to: \[ \frac{[\text{H}_2\text{CO}_3]}{[\text{HCO}_3^-]} \approx 0.0794 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

acid-base equilibrium
Acid-base equilibrium refers to the balance between acidic and basic (or alkaline) components in a solution. In the human body, this balance is essential for maintaining a stable pH, which is critical for various biochemical processes and overall health. When discussing acid-base equilibrium, it is important to understand:

1. **Acids and Bases:** Acids release hydrogen ions (H+), while bases accept hydrogen ions. Carbonic acid (H₂CO₃) is an example of an acid, and bicarbonate (HCO₃) is a base.
2. **Dissociation Constants (Ka):** The acid dissociation constant measures the strength of an acid in a solution. For carbonic acid, this is represented as Ka, which has a pKa value of 6.1 at 25°C.
3. **Henderson-Hasselbalch Equation:** This key equation connects pH, pKa, and the ratio of an acid and its conjugate base. The equation is: \[\text{pH} = \text{p}K_a + \text{log} \frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\text{\text{.}}\]
By rearranging the equation based on known values of pH, we can solve for the concentration ratio of acid and base.
buffer systems
Buffer systems are mixtures of weak acids and their conjugate bases, or weak bases and their conjugate acids. They help maintain a stable pH in a solution despite the addition of acidic or basic components. An important buffer system in the human body is the bicarbonate (HCO₃) and carbonic acid (H₂CO₃) system. Here’s how it works:

1. **Mechanisms of Action:** When an excess of H+ (protons) is added, bicarbonate ions neutralize them to form carbonic acid, which then converts to CO₂ and water, thus lowering acidity. Conversely, if the environment becomes too basic, carbonic acid dissociates to release H+ and neutralize excess OH (hydroxide ions).
2. **Importance in pH Regulation:** The bicarbonate-carbonic acid buffer system is crucial for maintaining the blood pH close to 7.40, essential for proper cellular function.
3. **Henderson-Hasselbalch Role:** The Henderson-Hasselbalch equation allows us to determine how shifts in the concentrations of H₂CO₃ and HCO₃ affect blood pH.
In scenarios like acidosis (where blood pH falls below the normal range), the buffer systems work hard to restore balance by adjusting the ratio of bicarbonate and carbonic acid.
blood pH regulation
Blood pH regulation is the process by which the body maintains its pH near the optimal value of 7.40. This is crucial because even slight deviations can severely affect bodily functions. Several systems work together to regulate blood pH:

1. **Buffer Systems:** As mentioned, the bicarbonate-carbonic acid buffer system is pivotal. It handles immediate pH changes by neutralizing excess acids or bases.
2. **Respiratory System:** The lungs help maintain pH by regulating the amount of CO₂ expelled. An increase in CO₂ (which forms carbonic acid in blood) decreases pH, while a decrease in CO₂ increases pH.
3. **Renal System:** The kidneys contribute by excreting excess acids or bases and by reabsorbing bicarbonate from urine. This slower but powerful system ensures long-term pH balance.
Conditions such as acidosis highlight how delicate this balance is. For example, if blood pH drops to 7.20, the body adjusts by altering the H₂CO₃/HCO₃ ratio, as seen from the Henderson-Hasselbalch equation. By knowing pH and pKa, you can calculate the necessary concentration changes to bring pH back to normal. This integrated approach ensures the body quickly adapts to maintain optimal functionality in varying conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Write the ion-product expressions for (a) iron(III) hydroxide; (b) barium phosphate; (c) magnesium fluoride.

Ethylenediaminetetraacetic acid (abbreviated \(\mathrm{H}_{4} \mathrm{EDTA}\) ) is a tetraprotic acid. Its salts are used to treat toxic metal poisoning by forming soluble complex ions that are then excreted. Because EDTA \(^{4-}\) also binds essential calcium ions, it is often administered as the calcium disodium salt. For example, when \(\mathrm{Na}_{2} \mathrm{Ca}\) (EDTA) is given to a patient, the \([\mathrm{Ca}(\mathrm{EDTA})]^{2-}\) ions react with circulating \(\mathrm{Pb}^{2+}\) ions and the metal ions are exchanged: \([\mathrm{Ca}(\mathrm{EDTA})]^{2-}(a q)+\mathrm{Pb}^{2+}(a q) \rightleftharpoons\) $$ [\mathrm{Pb}(\mathrm{EDTA})]^{2-}(a q)+\mathrm{Ca}^{2+}(a q) \quad K_{\mathrm{c}}=2.5 \times 10^{7} $$ A child has a dangerous blood lead level of \(120 \mu \mathrm{g} / 100 \mathrm{~mL}\). If the child is administered \(100 . \mathrm{mL}\) of \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{Ca}(\mathrm{EDTA}),\) what is the final concentration of \(\mathrm{Pb}^{2+}\) in \(\mu \mathrm{g} / 100 \mathrm{~mL}\) blood, assuming that the exchange reaction and excretion process are \(100 \%\) efficient? (Total blood volume is \(1.5 \mathrm{~L} .)\)

Find the \(\mathrm{pH}\) of a buffer that consists of \(1.3 \mathrm{M}\) sodium phenolate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{ONa}\right)\) and \(1.2 M\) phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\left(\mathrm{p} K_{\mathrm{a}}\right.\) of phenol \(\left.=10.00\right)\)

An ecobotanist separates the components of a tropical bark extract by chromatography. She discovers a large proportion of quinidine, a dextrorotatory isomer of quinine used for control of arrhythmic heartbeat. Quinidine has two basic nitrogens \(\left(K_{\mathrm{b} 1}=4.0 \times 10^{-6}\right.\) and \(\left.K_{\mathrm{b} 2}=1.0 \times 10^{-\mathrm{i} 0}\right) .\) To measure the concentration of quinidine, she carries out a titration. Because of the low solubility of quinidine, she first protonates both nitrogens with excess \(\mathrm{HCl}\) and titrates the acidified solution with standardized base. A 33.85-mg sample of quinidine \((\mathscr{M}=324.41 \mathrm{~g} / \mathrm{mol})\) is acidified with \(6.55 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{HCl}\) (a) How many milliliters of \(0.0133 \mathrm{M} \mathrm{NaOH}\) are needed to titrate the excess HCl? (b) How many additional milliliters of titrant are needed to reach the first equivalence point of quinidine dihydrochloride? (c) What is the \(\mathrm{pH}\) at the first equivalence point?

Write the ion-product expressions for (a) calcium chromate; (b) silver cyanide; (c) silver phosphate.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free