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(a) What is the ratio of \(\mathrm{HCO}_{3}^{-}\)to \(\mathrm{H}_{2} \mathrm{CO}_{3}\) in blood of pH 7.4? (b) What is the ratio of \(\mathrm{HCO}_{3}^{-}\)to \(\mathrm{H}_{3} \mathrm{CO}_{3}\) in an exhausted marathon runner whose blood \(\mathrm{pH}\) is \(7.1\) ?

Short Answer

Expert verified
(a) The ratio of \(\mathrm{HCO}_{3}^-\) to \(\mathrm{H}_{2}\mathrm{CO}_{3}\) in blood at pH 7.4 is approximately 19.95. (b) The ratio of \(\mathrm{HCO}_{3}^-\) to \(\mathrm{H}_{2}\mathrm{CO}_{3}\) in blood at pH 7.1 is 10.

Step by step solution

01

(a) Ratio in blood at pH 7.4

First, let's find the logarithmic expression: \(\mathrm{pH} - \mathrm{pK_a} = \log \frac{\mathrm{[HCO_3^-]}}{\mathrm{[H_2CO_3]}}\) \(7.4 - 6.1 = \log \frac{\mathrm{[HCO_3^-]}}{\mathrm{[H_2CO_3]}}\) \(1.3 = \log \frac{\mathrm{[HCO_3^-]}}{\mathrm{[H_2CO_3]}}\) Now we find the ratio: \(\frac{\mathrm{[HCO_3^-]}}{\mathrm{[H_2CO_3]}} = 10^{1.3} \approx 19.95\) The ratio of \(\mathrm{HCO}_{3}^-\) to \(\mathrm{H}_{2}\mathrm{CO}_{3}\) in blood at pH 7.4 is approximately 19.95.
02

(b) Ratio in blood at pH 7.1

Now let's find the logarithmic expression for blood at pH 7.1: \(\mathrm{pH} - \mathrm{pK_a} = \log \frac{\mathrm{[HCO_3^-]}}{\mathrm{[H_2CO_3]}}\) \(7.1 - 6.1 = \log \frac{\mathrm{[HCO_3^-]}}{\mathrm{[H_2CO_3]}}\) \(1.0 = \log \frac{\mathrm{[HCO_3^-]}}{\mathrm{[H_2CO_3]}}\) Now we find the ratio: \(\frac{\mathrm{[HCO_3^-]}}{\mathrm{[H_2CO_3]}} = 10^{1.0} = 10\) The ratio of \(\mathrm{HCO}_{3}^-\) to \(\mathrm{H}_{2}\mathrm{CO}_{3}\) in blood at pH 7.1 is 10.

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Key Concepts

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

Henderson-Hasselbalch Equation
Understanding the Henderson-Hasselbalch equation is pivotal for students diving into the complexities of chemical equilibrium in blood pH. This powerful equation connects pH, the measure of acidity or alkalinity of a solution, with the concentrations of an acid and its conjugate base.

In essence, the equation is expressed as \[\text{pH} = \text{p}K_a + \log \frac{\text{[conjugate base]}}{\text{[acid]}}\] where \(\text{p}K_a\) is the acid dissociation constant, and the bracketed terms represent the molar concentrations of the conjugate base and the acid, respectively.

This equation is derived from the acid dissociation constant expression and the relationship between pH and the concentration of hydrogen ions in a solution. By rearranging this relationship, we can better understand how the proportions of acid and base forms impact blood pH. Answers from the original exercise reveal how shifts in these ratios reflect changes in the blood's pH, a fine balance essential to our physiology.
Bicarbonate Buffer System
The bicarbonate buffer system is the main system that maintains the acid-base balance in our blood. It involves bicarbonate (\(\mathrm{HCO}_3^-\)) acting as a base, and carbonic acid (\(\mathrm{H}_2\mathrm{CO}_3\)) as the acid.

When pH levels in the blood drop, indicating increased acidity, the bicarbonate ions can accept hydrogen ions, thereby neutralizing the excess acid and raising the pH. Conversely, if the blood becomes too basic, carbonic acid can release hydrogen ions to lower the pH.

You can see the system in action in the exercise results. The ratios of \(\mathrm{HCO}_3^-\) to \(\mathrm{H}_2\mathrm{CO}_3\) show how the buffer system balances the pH in different states — resting and post-physical activity, where metabolism produces more acid. Thus, it's a system in constant dynamic equilibrium, finely tuned to sustain life.
Acid-Base Balance in Blood
The acid-base balance in blood is critical for proper physiological function, and our bodies possess impressive regulatory mechanisms to ensure this balance is maintained.

Normally, the blood pH is maintained between 7.35 and 7.45, which is slightly basic. This delicate balance is achieved mainly through the bicarbonate buffer system, as well as other mechanisms such as respiratory and renal control.

The exercise given showcases this balance in two different scenarios: at a normal resting pH of 7.4, and a slightly acidotic state in an exhausted runner with a pH of 7.1. The differential in bicarbonate to carbonic acid ratio between these two scenarios reflects the body’s ability to buffer acids and the inherent dynamic nature of the blood's buffering capacity.
Logarithmic Expressions in Chemistry
Logarithmic expressions are indispensable in chemistry, especially when dealing with pH and equilibrium expressions.

The pH scale itself is a logarithmic measure of the hydrogen ion concentration in a solution. Since the concentration of H+ ions in typical solutions can range across many orders of magnitude, the logarithmic scale provides a more manageable format for expressing these values. In mathematical terms, pH is the negative logarithm of the hydrogen ion concentration: \[\text{pH} = -\log[\text{H}^+]\].

In the context of the exercise given, logarithmic expressions are used to calculate the ratio of bicarbonate to carbonic acid concentration, providing a clear understanding of the blood's buffering capacity. By understanding these logarithmic relationships, students can easily analyze and predict how changes in these ratios affect the overall pH of a system.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) at the equivalence point in titrating \(0.100 \mathrm{M}\) solutions of each of the following with \(0.080 \mathrm{M} \mathrm{NaOH}\) : (a) hydrobromic acid (HBr), (b) chlorous acid \(\left(\mathrm{HClO}_{2}\right)\), (c) benzoic acid ( \(\left.\mathrm{C}_{4} \mathrm{H}_{3} \mathrm{COOH}\right)\).

The value of \(K_{a p}\) for \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) is \(2.1 \times 10^{-20}\). The \(\mathrm{AsO}_{4}{ }^{3-}\) ion is derived from the weak acid \(\mathrm{H}_{3} \mathrm{AsO}_{4}\left(\mathrm{pK}_{\mathrm{a} 1}=\right.\) \(\left.2.22 ; \mathrm{pK}_{u 2}=6.98 ; \mathrm{p} K_{u 3}=11.50\right)\). When asked to calculate the molar solubility of \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) in water, a student used the \(K_{p p}\) expression and assumed that \(\left[\mathrm{Mg}^{2+}\right]=1.5\left[\mathrm{AsO}_{4}{ }^{3-}\right]\). Why was this a mistake?

(a) Will \(\mathrm{Co}(\mathrm{OH})_{2}\) precipitate from solution if the pH of a \(0.020 \mathrm{M}\) solution of \(\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}\) is adjusted to \(8.5\) ? (b) Will \(\mathrm{AgIO}_{3}\) precipitate when \(20 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{AgIO}\) is mixed with \(10 \mathrm{~mL}\) of \(0.015 \mathrm{M} \mathrm{NalO}_{3}\) ? \(\left(K_{4}\right.\) of \(\mathrm{AglO}_{3}\) is \(\left.3.1 \times 10^{-t}\right)\).

Fluoridation of drinking water is employed in many places to aid in the prevention of tooth decay. Typically the F ion concentration is adjusted to about \(1 \mathrm{ppb}\). Some water supplies are also "hard"; that is, they contain certain cations such as \(\mathrm{Ca}^{2+}\) that interfere with the action of soap. Consider a case where the concentration of \(\mathrm{Ca}^{2+}\) is \(8 \mathrm{ppb}\). Could a precipitate of \(\mathrm{CaF}_{2}\) form under these conditions? (Make any necessary approximations.)

Excess \(\mathrm{Ca}(\mathrm{OH})_{2}\) is shaken with water to produce a saturated solution. The solution is filtered, and a \(50.00-\mathrm{mL}\) sample titrated with \(\mathrm{HCl}\) requires \(11.23 \mathrm{~mL}\) of \(0.0983 \mathrm{M} \mathrm{HQ}\) to reach the end point. Calculate \(K_{\mathrm{wp}}\) for \(\mathrm{Ca}(\mathrm{OH})_{2}\). Compare your result with that in Appendix D. \(25^{\circ} \mathrm{C}\). Suggest a reason for any differences you find between your value and the one in Appendix D.

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