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Chapter 16: Problem 76

Use the Henderson-Hasselbalch equation to calculate the ratio of \(\mathrm{HCO}_{3}^{-}\) to \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in blood having a \(\mathrm{pH}\) of \(7.40\). The value of \(\mathrm{K}_{\mathrm{a}}\) for carbonic acid at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) is \(7.9 \times 10^{-7}\).

Short Answer

Expert verified
The ratio of \( \mathrm{HCO}_3^- \) to \( \mathrm{H}_2 \mathrm{CO}_3 \) is approximately 20.

Step by step solution

01

Write down the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is used to relate the pH of a solution to the pKa and the ratio of the concentrations of the anion (base form) to the protonated form (acid form). The equation is: \[pH = pK_a + \log \left( \frac{\text{[Base]}}{\text{[Acid]}} \right)\]For this problem, \( \text{[Base]} = \text{[HCO}_3^-\text{]} \) and \( \text{[Acid]} = \text{[H}_2 \text{CO}_3\text{]} \).
02

Calculate the pKa

The pKa is calculated from the given K_a value using the formula:\[ pK_a = -\log(K_a) \]Given \( K_a = 7.9 \times 10^{-7} \), we calculate:\[ pK_a = -\log(7.9 \times 10^{-7}) = 6.10 \]
03

Substitute values into the Henderson-Hasselbalch equation

Substitute the known values into the Henderson-Hasselbalch equation:\[ 7.40 = 6.10 + \log \left( \frac{\text{[HCO}_3^-\text{]}}{\text{[H}_2 \text{CO}_3\text{]}} \right) \]
04

Solve for the ratio of base to acid

Rearrange the equation to solve for the log ratio:\[ \log \left( \frac{\text{[HCO}_3^-\text{]}}{\text{[H}_2 \text{CO}_3\text{]}} \right) = 7.40 - 6.10 \]This simplifies to:\[ \log \left( \frac{\text{[HCO}_3^-\text{]}}{\text{[H}_2 \text{CO}_3\text{]}} \right) = 1.30 \]
05

Calculate the actual ratio

To find the ratio \( \frac{\text{[HCO}_3^-\text{]}}{\text{[H}_2 \text{CO}_3\text{]}} \), use the inverse log operation:\[ \frac{\text{[HCO}_3^-\text{]}}{\text{[H}_2 \text{CO}_3\text{]}} = 10^{1.30} \]Calculating this gives approximately:\[ \frac{\text{[HCO}_3^-\text{]}}{\text{[H}_2 \text{CO}_3\text{]}} \approx 20 \]

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Key Concepts

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

Buffer Systems in Blood
Buffer systems are vital for maintaining the pH balance in blood, ensuring proper physiological functioning. The primary buffer system in blood consists of bicarbonate ions ( [HCO_3^-] ) and carbonic acid ( [H_2CO_3] ). This system is crucial because:
  • It maintains blood pH around 7.4, essential for enzyme and cellular functions.
  • The system responds to acid-base changes by modulating the concentration of [HCO_3^-] and [H_2CO_3] , thus stabilizing pH.
Bicarbonate acts as a base, neutralizing excess acids; carbonic acid acts as an acid, neutralizing excess bases. Together, they buffer pH changes, allowing the body to function correctly even under stress.
Understanding how these components work together helps to explain the body's resilience to pH fluctuations and highlights the importance of the bicarbonate-carbonic acid buffer in medical contexts.
Acid-Base Equilibrium
Acid-base equilibrium refers to the balance between acid and base concentrations in a solution, critical in maintaining the body's physiological environment. In the context of blood:
  • The equilibrium ensures that the body's pH stays within a narrow range, preventing cellular damage and ensuring metabolic processes continue smoothly.
  • It involves complex interactions between weak acids (like carbonic acid) and their conjugate bases (bicarbonate ions).
When an increase in acid or base occurs, the buffer system absorbs the excess ions, minimizing free hydrogen ion changes and thus limiting pH shifts.
In biological systems, acid-base equilibrium allows the body to adapt to varying conditions, such as changes in diet, physical activity, or environmental factors, by using mechanisms like breathing adjustments or kidney function changes to keep pH levels stable.
pKa Calculation
The pKa value is a numerical representation of an acid's strength, indicating how easily an acid gives up protons. It is especially useful in buffer chemistry because:
  • pKa helps to determine the position of equilibrium in acid-base reactions, which is essential for effective buffering.
  • In the given exercise, the pKa is used in the Henderson-Hasselbalch equation to understand the relationship between pH, pKa, and the concentration ratio of bicarbonate to carbonic acid.
The formula used is:\[pKa = -\log(K_a)\]This conversion from K_a to pKa gives a more intuitive measure of an acid's behavior in solution, particularly in physiological conditions.
In real-world applications, a calculated pKa provides insight into how an acid will act under specific pH conditions, helping in designing buffer solutions that closely mimic natural environments, such as blood, enhancing biological system stability.

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Most popular questions from this chapter

A saturated solution of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) has \(\left[\mathrm{Ca}^{2+}\right]=2.01 \times 10^{-8} \mathrm{M}\) and \(\left[\mathrm{PO}_{4}^{3-}\right]=1.6 \times 10^{-5} \mathrm{M}\). Calculate \(K_{\mathrm{sp}}\) for \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

Prior to having an X-ray exam of the upper gastrointestinal tract, a patient drinks an aqueous suspension of solid \(\mathrm{BaSO}_{4}\). (Scattering of \(\mathrm{X}\) rays by barium greatly enhances the quality of the photograph.) Although \(\mathrm{Ba}^{2+}\) is toxic, ingestion of \(\mathrm{BaSO}_{4}\) is safe because it is quite insoluble. If a saturated solution prepared by dissolving solid \(\mathrm{BaSO}_{4}\) in water has \(\left[\mathrm{Ba}^{2+}\right]=1.05 \times 10^{-5} \mathrm{M}\), what is the value of \(K_{\mathrm{sp}}\) for \(\mathrm{BaSO}_{4} ?\)

A \(0.0100 \mathrm{~mol}\) sample of solid \(\mathrm{Cd}(\mathrm{OH})_{2}\left(K_{\mathrm{sp}}=5.3 \times 10^{-15}\right)\) in \(100.0 \mathrm{~mL}\) of water is titrated with \(0.100 \mathrm{M} \mathrm{HNO}_{3}\). (a) What is the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}\) in pure water? What is the \(\mathrm{pH}\) of the solution before the addition of any \(\mathrm{HNO}_{3} ?\) (b) What is the \(\mathrm{pH}\) of the solution after the addition of \(90.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HNO}_{3} ?\) (c) How many milliliters of \(0.100 \mathrm{M} \mathrm{HNO}_{3}\) must be added to completely neutralize the \(\mathrm{Cd}(\mathrm{OH})_{2} ?\)

Consider the titration of \(25.0 \mathrm{~mL}\) of \(0.125 \mathrm{M} \mathrm{HCl}\) with \(0.100 \mathrm{M}\) KOH. Calculate the pH after the addition of each of the following volumes of base: (a) \(3.0 \mathrm{~mL}\) (b) \(20 \mathrm{~mL}\) (c) \(65 \mathrm{~mL}\)

Explain why the uptake of \(\mathrm{CO}_{3}{ }^{2-}\) by calcifying marine organisms causes more \(\mathrm{CO}_{2}\) to dissolve in the ocean than in pure water.
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