Th pH of blood is steady at a value of approximately 7.4 as a result of the following equilibrium reactions: $$ \mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \leftrightharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \leftrightharpoons \mathrm{HCO}_{3}-(a q)+\mathrm{H}^{+}(a q) $$ The actual buffer system in blood is made up of $\mathrm{H}_{2} \mathrm{CO}_{3}\( and \)\mathrm{HCO}_{3}$ - One way the body keeps the pH of blood at 7.4 is by regulating breathing. Under what blood ph conditions will the body increase breathing and under what blood pH conditions will the body decrease breathing? Explain.

Short Answer

Expert verified
In conclusion, the body increases breathing under low blood pH (acidic) conditions to decrease the concentration of hydrogen ions (H+) and restore the pH balance. On the other hand, the body decreases breathing under high blood pH (alkaline) conditions to increase the concentration of hydrogen ions (H+) and restore the pH balance. Breathing plays a vital role in regulating the blood pH by controlling the concentrations of CO2 and H2CO3 in the blood, through the blood buffering system.

Step by step solution

01

Understanding the blood buffering system

Blood buffering system consists of carbon dioxide (CO2) dissolved in water (H2O), carbonic acid (H2CO3), bicarbonate ion (HCO3-), and a hydrogen ion (H+). These substances are in equilibrium with each other, as shown in the provided reactions: $$ \mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \leftrightharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \leftrightharpoons \mathrm{HCO}_{3}-(a q)+\mathrm{H}^{+}(a q) $$ The pH of the blood is mainly determined by the concentration of hydrogen ions (H+) in the blood. The more hydrogen ions present, the lower the pH value will be, and the more acidic the blood becomes.
02

Identify the role of Breathing in the Blood Buffering System

Breathing plays a crucial role in maintaining the appropriate pH in the blood. When we breathe in, we take in oxygen (O2) and expel carbon dioxide (CO2). The process of inhaling and exhaling air helps in regulating the concentrations of CO2 and H2CO3 in the blood, thereby controlling the blood pH.
03

Determine the conditions for increased Breathing

If the blood pH becomes too low (acidic) due to a higher concentration of hydrogen ions (H+), the body tries to restore the pH balance by eliminating more carbon dioxide (CO2) through increased breathing. Increased breathing causes the blood buffering system to shift the equilibrium to the left, according to the reactions: $$ \mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \leftrightharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \leftrightharpoons \mathrm{HCO}_{3}-(a q)+\mathrm{H}^{+}(a q) $$ As a result, the concentration of hydrogen ions (H+) decreases, and the pH of the blood increases.
04

Determine the conditions for decreased Breathing

If the blood pH becomes too high (alkaline) due to a lower concentration of hydrogen ions (H+), the body tries to restore the pH balance by retaining more carbon dioxide (CO2) through decreased breathing. Decreased breathing causes the blood buffering system to shift the equilibrium to the right, according to the reactions: $$ \mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \leftrightharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \leftrightharpoons \mathrm{HCO}_{3}-(a q)+\mathrm{H}^{+}(a q) $$ As a result, the concentration of hydrogen ions (H+) increases, and the pH of the blood decreases.
05

Conclusion

In conclusion, the body increases breathing under low blood pH (acidic) conditions to decrease the concentration of hydrogen ions (H+) and restore the pH balance. On the other hand, the body decreases breathing under high blood pH (alkaline) conditions to increase the concentration of hydrogen ions (H+) and restore the pH balance. Breathing plays a vital role in regulating the blood pH by controlling the concentrations of CO2 and H2CO3 in the blood, through the blood buffering system.

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!

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

A buffer is prepared by dissolving HONH_ and $\mathrm{HONH}_{3} \mathrm{NO}_{3}$ in some water. Write equations to show how this buffer neutralizes added \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\)

Which of the following mixtures would result in a buffered solution when 1.0 \(\mathrm{L}\) of each of the two solutions are mixed? $$ \begin{array}{l}{\text { a. } 0.2 M \mathrm{HNO}_{3} \text { and } 0.4 \mathrm{M} \mathrm{NaNO}_{3}} \\ {\text { b. } 0.2 \mathrm{M} \mathrm{HNO}_{3} \text { and } 0.4 \mathrm{M} \mathrm{HF}}\end{array} $$ $$ \begin{array}{l}{\text { c. } 0.2 M \mathrm{HNO}_{3} \text { and } 0.4 \mathrm{M} \mathrm{NaF}} \\ {\text { d. } 0.2 \mathrm{M} \mathrm{HNO}_{3} \text { and } 0.4 \mathrm{M} \mathrm{NaOH}}\end{array} $$

One method for determining the purity of aspirin $\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\right)$ is to hydrolyze it with NaOH solution and then to titrate the remaining NaOH. The reaction of aspirin with NaOH is as follows: \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}(s)+2 \mathrm{OH}^{-}(a q)\) $$ \mathrm{C}_{7} \mathrm{H}_{3} \mathrm{O}_{3}^{-}(a q)+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ A sample of aspirin with a mass of 1.427 g was boiled in 50.00 \(\mathrm{mL}\) of 0.500\(M \mathrm{NaOH}\) . After the solution was cooled, it took 31.92 \(\mathrm{mL}\) of 0.289 \(\mathrm{M} \mathrm{HCl}\) to titrate the excess NaOH. Calculate the purity of the aspirin. What indicator should be used for this titration? Why?

Repeat the procedure in Exercise \(67,\) but for the titration of 25.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{HNO}_{3}\) with 0.100 $\mathrm{M} \mathrm{NaOH} .$

A \(0.400-M\) solution of ammonia was titrated with hydrochloric acid to the equivalence point, where the total volume was 1.50 times the original volume. At what pH does the equivalence point occur?

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