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When a person exercises, muscle contractions produce lactic acid. Moderate increases in lactic acid can be handled by the blood buffers without decreasing the pH of blood. However, excessive amounts of lactic acid can overload the blood buffer system, resulting in a lowering of the blood pH. A condition called acidosis is diagnosed if the blood pH falls to 7.35 or lower. Assume the primary blood buffer system is the carbonate buffer system described in Exercise \(45 .\) Calculate what happens to the \(\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right] /\left[\mathrm{HCO}_{3}^{-}\right]\) ratio in blood when the \(\mathrm{pH}\) decreases from 7.40 to 7.35.

Short Answer

Expert verified
The ratio of bicarbonate ions (\( \text{HCO}_3^-\)) to carbonic acid (\(\text{H}_2\text{CO}_3\)) increases by a factor of approximately 1.12 when the blood pH decreases from 7.40 to 7.35 due to acidosis.

Step by step solution

01

Write the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is given by: \[ \text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \] In our case, the acid is \(\text{H}_2\text{CO}_3\) (carbonic acid) and the conjugate base is \(\text{HCO}_3^-\) (bicarbonate ion). Let's rewrite the equation: \[ \text{pH} = \text{p}K_a + \log\left(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\right) \]
02

Calculate the Change in pH

According to the exercise, the decrease in pH goes from 7.40 to 7.35. The change in pH can be calculated as: \[ \Delta \text{pH} = 7.40 - 7.35 = 0.05 \]
03

Find the \(pK_a\)

The primary blood buffer system is the carbonate buffer system. The \(K_a\) value for carbonic acid is \(4.47 \times 10^{-7}\). In order to use the Henderson-Hasselbalch equation, we need the p\(K_a\) value. To find this, we can use the following equation: \[ \text{p}K_a = -\log_{10}(K_a) \] Plugging in the values, we get: \[ \text{p}K_a = -\log_{10}(4.47 \times 10^{-7}) \approx 6.35\]
04

Calculate the change in the ratio of HCO\(_3^-\) to H\(_2\)CO\(_3\)

Now that we have the change in pH, and the p\(K_a\), we can rearrange the Henderson-Hasselbalch equation to solve for the change in the ratio of bicarbonate ions (\( \text{HCO}_3^-\)) to carbonic acid (\(\text{H}_2\text{CO}_3\)): \[ \Delta \log\left(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\right) = \Delta \text{pH} - \Delta \text{p}K_a \] Since the p\(K_a\) remains constant (it does not change), the change in p\(K_a\) is zero: \[ \Delta \log\left(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\right) = 0.05 \] To calculate the change in ratio, we should find the antilog of this value: \[ 10^{\Delta \log\left(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\right)} = 10^{0.05} \approx 1.12\]
05

Conclusion

The ratio of bicarbonate ions (\( \text{HCO}_3^-\)) to carbonic acid (\(\text{H}_2\text{CO}_3\)) increases by a factor of approximately 1.12 when the blood pH decreases from 7.40 to 7.35 due to acidosis.

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Key Concepts

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

Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is a cornerstone concept in understanding acid-base equilibriums, particularly in biological systems. It's a mathematical formula that relates the pH of a solution to the pKa (the acid dissociation constant) and the ratio of the concentrations of an acid and its conjugate base. This equation is especially useful in calculating the pH of buffer solutions. For a buffer system like the carbonate buffer in blood, the equation is given by:

\[\text{pH} = \text{p}K_a + \log\left(\frac{[\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}\right)\]

This shows that to maintain a certain pH, the relative concentrations of the bicarbonate ion
  • (\(\text{HCO}_3^-\))
  • carbonic acid (\(\text{H}_2\text{CO}_3\))
need to be controlled. If the concentration of carbonic acid increases or the bicarbonate decreases, the pH will drop, and vice versa. In biological terms, this balance is crucial for maintaining a stable pH, particularly in human blood where the pH needs to be tightly regulated around 7.4.
buffer systems
Buffers are solutions that can resist changes in pH when small amounts of acids or bases are added. They are critical in maintaining the pH balance in living organisms. In human blood, the main buffer system is the carbonate buffer system, which involves carbonic acid (\(\text{H}_2\text{CO}_3\)) and bicarbonate ion (\(\text{HCO}_3^-\)). This system works by:
  • Neutralizing added acids or bases to prevent significant pH changes.
  • Balancing the formation and disassociation of carbonic acid.
When there's excess acid, like during intense exercise, bicarbonate ions neutralize this acid, minimizing pH changes. Conversely, when bases are introduced, carbonic acid acts to neutralize them:

\[\text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-\]

In conditions of acidosis, this balance is disrupted, leading to an excess of hydrogen ions which reduces the pH. Buffer systems are crucial as they allow for physiological processes to continue without disruption despite metabolic and environmental changes.
blood pH
Blood pH is a measure of how acidic or basic the blood is and is typically maintained within a narrow range around 7.4. This balance is crucial for normal bodily functions, including enzyme activities and cellular processes. Blood pH is determined by the ratio of bicarbonate ions to carbonic acid, primarily regulated through the carbonate buffer system.

Under normal circumstances:

  • The bicarbonate ion acts as a base.
  • Carbonic acid acts as an acid.
This balance can be represented through the Henderson-Hasselbalch equation, where a stable pH reflects a balanced ratio. If blood pH shifts significantly, physiological processes can be adversely affected, leading to conditions such as acidosis or alkalosis. Regular monitoring and quick response to deviations in blood pH are crucial in medical settings to prevent critical complications. Deviations are quickly managed by:
  • Adjusting respiration (to release or retain \(\text{CO}_2\)).
  • Through renal mechanisms (to excrete or retain bicarbonate).
acidosis
Acidosis is a condition where the pH of the blood falls below 7.35, indicating an increase in acidity. This condition can arise from multiple origins, such as excessive production of lactic acid during vigorous exercise, or metabolic issues affecting normal acid balance.
When acidosis occurs, the body's ability to function optimally is compromised. For instance, enzyme activity can be affected, and essential body processes might slow down. The symptoms of acidosis can range from mild fatigue and confusion to severe outcomes like coma, depending on the severity and cause.
To counteract acidosis:
  • The body relies on buffer systems to restore pH balance.
  • The respiratory system can increase breathing rate to expel \(\text{CO}_2\), thus reducing acidity.
  • The kidneys can also help by excreting more hydrogen ions and retaining bicarbonate.
Accordingly, understanding and managing acidosis involves not only recognizing changes in blood pH but also the mechanisms through which the body attempts to reestablish equilibrium. Effective management is crucial for patient recovery and maintaining physiological stability.

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Most popular questions from this chapter

Consider the titration of \(100.0 \mathrm{mL}\) of \(0.100 \mathrm{M}\) HCN by \(0.100 M \mathrm{KOH}\) at \(25^{\circ} \mathrm{C} .\left(K_{\mathrm{a}} \text { for } \mathrm{HCN}=6.2 \times 10^{-10} .\right)\) a. Calculate the \(\mathrm{pH}\) after \(0.0 \mathrm{mL}\) of \(\mathrm{KOH}\) has been added. b. Calculate the \(\mathrm{pH}\) after \(50.0 \mathrm{mL}\) of \(\mathrm{KOH}\) has been added. c. Calculate the \(\mathrm{pH}\) after \(75.0 \mathrm{mL}\) of \(\mathrm{KOH}\) has been added. d. Calculate the \(\mathrm{pH}\) at the equivalence point. e. Calculate the pH after 125 mL of KOH has been added.

Derive an equation analogous to the Henderson-Hasselbalch equation but relating \(\mathrm{pOH}\) and \(\mathrm{p} K_{\mathrm{b}}\) of a buffered solution composed of a weak base and its conjugate acid, such as \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4}^{+}\).

A best buffer has about equal quantities of weak acid and conjugate base present as well as having a large concentration of each species present. Explain.

A buffered solution is made by adding \(50.0 \mathrm{g} \mathrm{NH}_{4} \mathrm{Cl}\) to 1.00 \(\mathrm{L}\) of a 0.75-M solution of \(\mathrm{NH}_{3}\). Calculate the pH of the final solution. (Assume no volume change.)

a. Calculate the pH of a buffered solution that is 0.100 \(M\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\) (benzoic acid, \(K_{\mathrm{a}}=6.4 \times 10^{-5}\) ) and \(0.100 M\) in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{Na}\) b. Calculate the \(\mathrm{pH}\) after \(20.0 \%\) (by moles) of the benzoic acid is converted to benzoate anion by addition of a strong base. Use the dissociation equilibrium $$ \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}^{+}(a q)$$ to calculate the pH. c. Do the same as in part b, but use the following equilibrium to calculate the pH: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}(a q)+\mathrm{OH}^{-}(a q)\) d. Do your answers in parts \(b\) and c agree? Explain.

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