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Chapter 15: Problem 45

Carbonate buffers are important in regulating the \(\mathrm{pH}\) of blood at \(7.40\). If the carbonic acid concentration in a sample of blood is \(0.0012 M\), determine the bicarbonate ion concentration required to buffer the \(\mathrm{pH}\) of blood at \(\mathrm{pH}=7.40\). \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=4.3 \times 10^{-7}\)

Short Answer

Expert verified
The bicarbonate ion concentration required to buffer the \(\mathrm{pH}\) of blood at \(\mathrm{pH}=7.40\) is \(0.0013 M\).

Step by step solution

01

Equilibrium constant expression

Write the \(K_{a}\) expression using the given equilibrium constant: \[K_{a} = \frac{[\mathrm{HCO}_{3}^{-}][\mathrm{H}^{+}]}{[\mathrm{H}_{2}\mathrm{CO}_{3}]}\]
02

Calculate the concentration of \(\mathrm{H}^{+}\) from the given \(\mathrm{pH}\)

To calculate the concentration of \(\mathrm{H}^{+}\) ions, use the \(\mathrm{pH}\) equation: \[\mathrm{pH} = -\log[\mathrm{H}^{+}]\] We are given \(\mathrm{pH}=7.40\), so solve for \([\mathrm{H}^{+}]\): \[7.40 = -\log[\mathrm{H}^{+}]\] \[[\mathrm{H}^{+}] = 10^{-7.40} = 3.98\times 10^{-8} M\]
03

Plug in to \(K_{a}\) expression and solve for \([\mathrm{HCO}_{3}^{-}]\)

Substitute the values from the given equilibrium constant, carbonic acid concentration, and the calculated \(\mathrm{H}^{+}\) concentration into the \(K_{a}\) expression: \[\begin{aligned} K_{a} &= \frac{[\mathrm{HCO}_{3}^{-}][\mathrm{H}^{+}]}{[\mathrm{H}_{2}\mathrm{CO}_{3}]} \\ 4.3 \times 10^{-7} &= \frac{[\mathrm{HCO}_{3}^{-}](3.98\times 10^{-8})}{0.0012} \end{aligned}\] Now solve for \([\mathrm{HCO}_{3}^{-}]\): \[[\mathrm{HCO}_{3}^{-}] = \frac{4.3 \times 10^{-7} \times 0.0012}{3.98\times 10^{-8}} = 0.0013 M\]
04

Conclusion

The bicarbonate ion concentration required to buffer the \(\mathrm{pH}\) of blood at \(\mathrm{pH}=7.40\) is \(0.0013 M\).

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Key Concepts

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

Bicarbonate Ion
The bicarbonate ion, denoted as \(\mathrm{HCO}_3^-\), plays a crucial role in maintaining the acid-base balance in biological systems, particularly in human blood. It acts as a base that can accept hydrogen ions to neutralize acidity. By forming carbonic acid when bonded with a hydrogen ion, the bicarbonate ion contributes to the dynamic equilibrium necessary for regulating the body's \(\mathrm{pH}\).

When bicarbonate ions are present in sufficient quantities, they effectively buffer changes in \(\mathrm{pH}\), helping maintain a stable environment crucial for biological processes. This stability is essential for enzyme function and overall cellular health.
Carbonic Acid
Carbonic acid, represented as \(\mathrm{H}_2\mathrm{CO}_3\), is a weak acid formed in aqueous solution. It's created when carbon dioxide (\(\mathrm{CO}_2\)) dissolves in water, and in biological systems, it鈥檚 a key player in the carbonate buffer system.

In the blood, carbonic acid can dissociate into bicarbonate ions and hydrogen ions, driving the reaction between carbon dioxide and water. This reversible reaction allows the body to buffer changes in \(\mathrm{pH}\) effectively. The concentration of carbonic acid tends to establish the baseline for buffering since it is part of the equilibrium with bicarbonate ions.
pH Regulation
pH regulation is critical for biological processes, as enzymes and biochemical reactions are sensitive to \(\mathrm{pH}\) changes. The body uses buffer systems, such as the carbonate buffer system, to manage \(\mathrm{pH}\) precisely.

The carbonate buffer system in the blood uses a balance between bicarbonate ions and carbonic acid to keep \(\mathrm{pH}\) around \(7.40\), a value compatible with normal physiological functions. If \(\mathrm{pH}\) falls, bicarbonate ions react with excess \(\mathrm{H}^{+}\) ions to form carbonic acid, reducing acidity. Conversely, when \(\mathrm{pH}\) rises, carbonic acid dissociates to release \(\mathrm{H}^{+}\) ions, neutralizing excess base.
Equilibrium Constant
The equilibrium constant, symbolized as \(K_a\) for acids, describes the ratio of the concentrations of the products to the reactants at equilibrium. For the carbonate buffer system, the relevant expression is:

\[K_a = \frac{[\mathrm{HCO}_3^-][\mathrm{H}^+]}{[\mathrm{H}_2\mathrm{CO}_3]} \]

This expression helps to quantify the extent of dissociation of carbonic acid into bicarbonate ions and hydrogen ions. A given value of \(K_a\) helps in calculating the required concentrations of ions to achieve desired \(\mathrm{pH}\) levels in buffering processes.
Hydrogen Ion Concentration
The concentration of hydrogen ions \([\mathrm{H}^+]\) is directly related to \(\mathrm{pH}\) by the formula: \( \mathrm{pH} = -\log[\mathrm{H}^+] \). A higher hydrogen ion concentration means a lower \(\mathrm{pH}\), indicating more acidic conditions, whereas lower concentrations indicate higher \(\mathrm{pH}\), or more basic conditions.

In the context of the carbonate buffer system, a finely tuned balance of \([\mathrm{H}^+]\), \([\mathrm{HCO}_3^-]\), and \([\mathrm{H}_2\mathrm{CO}_3]\) ensures that the blood \(\mathrm{pH}\) remains around \(7.40\). This precise maintenance of \([\mathrm{H}^+]\) is critical for normal metabolic functions.

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Most popular questions from this chapter

A student titrates an unknown weak acid, HA, to a pale pink phenolphthalein end point with \(25.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\). The student then adds \(13.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}\). The \(\mathrm{pH}\) of the resulting solution is \(4.70\). How is the value of \(\mathrm{p} K_{\mathrm{a}}\) for the unknown acid related to 4.70?

A student intends to titrate a solution of a weak monoprotic acid with a sodium hydroxide solution but reverses the two solutions and places the weak acid solution in the buret. After \(23.75 \mathrm{~mL}\) of the weak acid solution has been added to \(50.0 \mathrm{~mL}\) of the \(0.100 \mathrm{M} \mathrm{NaOH}\) solution, the \(\mathrm{pH}\) of the resulting solution is \(10.50 .\) Calculate the original concentration of the solution of weak acid.

A friend asks the following: "Consider a buffered solution made up of the weak acid HA and its salt NaA. If a strong base like \(\mathrm{NaOH}\) is added, the HA reacts with the \(\mathrm{OH}^{-}\) to form \(\mathrm{A}^{-}\). Thus the amount of acid (HA) is decreased, and the amount of base \(\left(\mathrm{A}^{-}\right)\) is increased. Analogously, adding \(\mathrm{HCl}\) to the buffered solution forms more of the acid (HA) by reacting with the base \(\left(\mathrm{A}^{-}\right) .\) Thus how can we claim that a buffered solution resists changes in the \(\mathrm{pH}\) of the solution?" How would you explain buffering to this friend?

Calculate the volume of \(1.50 \times 10^{-2} M \mathrm{NaOH}\) that must be added to \(500.0 \mathrm{~mL}\) of \(0.200 M \mathrm{HCl}\) to give a solution that has \(\mathrm{pH}=2.15\)

Consider a buffered solution containing \(\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) and \(\mathrm{CH}_{3} \mathrm{NH}_{2}\). Which of the following statements concerning this solution is(are) true? \(\left(K_{\mathrm{a}}\right.\) for \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}=2.3 \times 10^{-11}\).) a. A solution consisting of \(0.10 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) and \(0.10 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) would have a higher buffering capacity than one containing \(1.0 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\) and \(1.0 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{2}\). b. If \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]>\left[\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}\right]\), then the \(\mathrm{pH}\) is larger than the \(\mathrm{p} K_{\mathrm{a}}\) value. c. Adding more \(\left[\mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{Cl}\right]\) to the initial buffer solution will decrease the \(\mathrm{pH}\). d. If \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]<\left[\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}\right]\), then \(\mathrm{pH}<3.36\). e. If \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=\left[\mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}\right]\), then \(\mathrm{pH}=10.64\).
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