Question

The average \(\mathrm{pH}\) of normal arterial blood is 7.40. At normal body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{w}=2.4 \times 10^{-14}\). Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\), and \(\mathrm{pOH}\) for blood at this temperature.

Short answer

The concentration of hydrogen ions [H+] in the blood is approximately \(4.0 \times 10^{-8} \ \mathrm{M}\), the concentration of hydroxide ions [OH-] is approximately \(6.0 \times 10^{-7} \ \mathrm{M}\), and the pOH is approximately 6.60.

Step-by-step solution

Calculate the concentration of hydrogen ions [H+]

We are given the pH of the blood, and we can use the formula: \(\mathrm{pH} = -\log{\left[\mathrm{H}^{+}\right]}\). We can rearrange this formula to find [H+]: \[\left[\mathrm{H}^{+}\right] = 10^{-\mathrm{pH}}\] Now, we can plug in the given pH value: \[\left[\mathrm{H}^{+}\right] = 10^{-7.40}\] \[\left[\mathrm{H}^{+}\right] \approx 4.0 \times 10^{-8} \ \mathrm{M}\]

Calculate the concentration of hydroxide ions [OH-]

We are given the ion product of water, Kw, and from the previous step, we have calculated the concentration of hydrogen ions [H+]. We can use the relationship between these quantities to find [OH-]: \(K_{\mathrm{w}} = \left[\mathrm{H}^{+}\right] \left[\mathrm{OH}^{-}\right]\) Rearranging the formula to solve for [OH-]: \(\left[\mathrm{OH}^{-}\right] = \frac{K_{\mathrm{w}}}{\left[\mathrm{H}^{+}\right]}\) Plugging in the given values: \(\left[\mathrm{OH}^{-}\right] = \frac{2.4 \times 10^{-14}}{4.0 \times 10^{-8}}\) \(\left[\mathrm{OH}^{-}\right] = 6.0 \times 10^{-7} \ \mathrm{M}\)

Calculate the pOH of blood

We can calculate the pOH using the relationship between pH and pOH: \(\mathrm{pH} + \mathrm{pOH} = 14\) Rearranging the formula to solve for pOH: \(\mathrm{pOH} = 14 - \mathrm{pH}\) Plugging in the given pH value: \(\mathrm{pOH} = 14 - 7.40\) \(\mathrm{pOH} = 6.60\) Now we have the [H+], [OH-], and pOH for blood at this temperature: \(\left[\mathrm{H}^{+}\right] \approx 4.0 \times 10^{-8} \ \mathrm{M}\) \(\left[\mathrm{OH}^{-}\right] \approx 6.0 \times 10^{-7} \ \mathrm{M}\) \(\mathrm{pOH} \approx 6.60\)

Key concepts

pH Scale

The pH scale is a logarithmic scale used to quantify the acidity or basicity of an aqueous solution. It ranges from 0 to 14, where a pH less than 7 indicates an acidic solution, a pH of 7 is neutral, and a pH greater than 7 is basic (or alkaline). The pH value is a direct measure of the hydrogen ion concentration \(\left[\mathrm{H}^{+}\right]\) in a solution, and it is calculated using the formula \(\mathrm{pH} = -\log{\left[\mathrm{H}^{+}\right]}\).

Understanding the pH scale is crucial, since slight changes in pH can have significant physiological effects and can indicate the presence of underlying health conditions. For instance, the normal arterial blood pH is narrowly maintained around 7.40, ensuring optimal functioning for enzymatic and metabolic processes.

Concentration of Hydrogen Ions

The concentration of hydrogen ions \(\left[\mathrm{H}^{+}\right]\) in a solution is a fundamental concept in chemistry and biochemistry as it determines the solution's pH. Hydrogen ion concentration is typically expressed in moles per liter (M), and due to its direct relationship with pH, it can be derived from the pH by using the inverse logarithmic relationship: \(\left[\mathrm{H}^{+}\right] = 10^{-\mathrm{pH}}\).

For example, if a solution has a pH of 7.4, like normal arterial blood, the concentration of hydrogen ions would be \(10^{-7.4}\) or approximately 4.0 x 10^{-8} M. This low concentration reflects the tightly regulated environment necessary for maintaining physiological functions.

Ion Product of Water

The ion product of water \(K_{w}\) represents the equilibrium constant for the self-ionization of water and is an important part of understanding pH and pOH calculations. At 25°C, it has a constant value of approximately \(1.0 \times 10^{-14}\), which may vary with temperature.

It is expressed as the product of the concentrations of hydrogen ions and hydroxide ions: \(K_{w} = [\mathrm{H}^{+}] \times [\mathrm{OH}^{-}]\). In the given problem, at body temperature (37°C), \(K_{w}\) is \(2.4 \times 10^{-14}\), highlighting how temperature affects this equilibrium constant. Understanding \(K_{w}\) allows us to calculate one ionic concentration when we have the other, and it directly contributes to the calculation of pOH.

pOH Calculation

pOH is another scale similar to pH, which measures the basicity of a solution. It is related to the concentration of hydroxide ions \(\left[\mathrm{OH}^{-}\right]\), and is calculated using the formula \(\mathrm{pOH} = -\log{\left[\mathrm{OH}^{-}\right]}\).

Within the framework of the pH scale, pH and pOH are inversely related and always add up to 14, which is derived from the ion product of water at 25°C \(\left(\mathrm{pH} + \mathrm{pOH} = 14\right)\). Therefore, if the pH of a solution is known, the pOH can be easily found, and vice versa. For instance, with a blood pH of 7.40, the pOH would be calculated as \(14 - 7.40 = 6.60\). Knowing the pOH is equally vital, as it can provide insights into the level of basic substances in a solution, which is significant in many chemical and biological processes.