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

In summary, at a body temperature of 37°C and an average arterial blood pH of 7.40, the concentration of H+ ions in blood is \(1.00 \times 10^{-7.40}\) mol/L, the concentration of OH- ions is \(2.41 \times 10^{-6.60}\) mol/L, and the pOH value is approximately 6.60.

Step-by-step solution

Finding the concentration of H+ ions [H+] using pH

We are given the pH of blood, which is 7.40. We can use the equation pH = -log[H+] to find the concentration of H+ ions: 7.40 = -log[H+] To get [H+], we need to take the antilog of both sides of the equation: [H+] = 10^(-7.40) Calculate the value: [H+] = \(1.00 \times 10^{-7.40}\) mol/L So, the concentration of H+ ions in the blood is \(1.00 \times 10^{-7.40}\) mol/L.

Finding the concentration of OH- ions [OH-] using Kw

Now that we have the concentration of H+ ions, we can find the concentration of OH- ions using the ion product constant of water, Kw: Kw = [H+] * [OH-] We are given Kw = \(2.4 \times 10^{-14}\), and [H+] = \(1.00 \times 10^{-7.40}\) mol/L. Plugging these values into the equation: \(2.4 \times 10^{-14}\) = \(1.00 \times 10^{-7.40}\) * [OH-] Solve for [OH-]: [OH-] = \(\frac{2.4 \times 10^{-14}}{1.00 \times 10^{-7.40}}\) mol/L Calculate the value: [OH-] = \(2.41 \times 10^{-6.60}\) mol/L So, the concentration of OH- ions in the blood is \(2.41 \times 10^{-6.60}\) mol/L.

Finding the pOH value

Now that we have found the concentration of OH- ions, we can calculate the pOH value for blood using the equation pOH = -log[OH-]: pOH = -log[\(2.41 \times 10^{-6.60}\)] Calculate the value: pOH ≈ 6.60 So, the pOH of blood at this temperature is approximately 6.60. In summary, at a body temperature of 37°C and an average arterial blood pH of 7.40, the concentration of H+ ions in blood is \(1.00 \times 10^{-7.40}\) mol/L, the concentration of OH- ions is \(2.41 \times 10^{-6.60}\) mol/L, and the pOH value is approximately 6.60.

Key concepts

Ion Product Constant

The ion product constant of water, often denoted as \( K_w \), is a crucial concept in understanding the self-ionization of water. At any given temperature, water can dissociate very slightly into hydrogen ions \([H^+]\) and hydroxide ions \([OH^-]\). This dissociation can be represented by the equation:\[H_2O \rightleftharpoons H^+ + OH^-\]The ion product constant \( K_w \) is thus the product of the concentrations of these ions:\[K_w = [H^+] \times [OH^-]\]At 25°C, the value of \( K_w \) is \(1.0 \times 10^{-14}\), but it varies with temperature. In the context of this problem, at 37°C, the given \( K_w \) is \(2.4 \times 10^{-14}\). This value accounts for the increased kinetic energy and ionization capacity of water molecules at higher temperatures.The concept of \( K_w \) is foundational in determining the balance between \([H^+]\) and \([OH^-]\) in aqueous solutions. Recognizing its temperature dependence is essential when performing pH and pOH calculations.

H+ Ion Concentration

The concentration of hydrogen ions \([H^+]\) directly relates to the acidity of a solution. It is often calculated from the \(\text{pH}\) value using the formula:\[\text{pH} = -\log[H^+]\]To find \([H^+]\), you can take the antilog (inverse log) of the given pH:\[[H^+] = 10^{-\text{pH}}\]For example, in this problem, the pH of blood is 7.40. So, to find \([H^+]\), calculate:\[[H^+] = 10^{-7.40}\]This gives:\[[H^+] = 1.00 \times 10^{-7.40}\, \text{mol/L}\]The \([H^+]\) concentration is a measure of how acidic or basic a solution is, with lower values corresponding to more acidic solutions. This relationship is important for many biological and chemical reactions, especially in maintaining proper body function.

OH- Ion Concentration

Hydroxide ions \([OH^-]\) are pivotal in determining the basicity of a solution. Given the ion product constant \( K_w \) and the hydrogen ion concentration \([H^+]\), you can easily calculate \([OH^-]\):\[K_w = [H^+] \times [OH^-]\]Solving for \([OH^-]\) involves rearranging the formula:\[[OH^-] = \frac{K_w}{[H^+]}\]In the exercise, we are given that \( K_w = 2.4 \times 10^{-14} \) and \([H^+] = 1.00 \times 10^{-7.40}\, \text{mol/L}\). Plugging these values into the equation:\[[OH^-] = \frac{2.4 \times 10^{-14}}{1.00 \times 10^{-7.40}}\]This calculation reveals \([OH^-] = 2.41 \times 10^{-6.60}\, \text{mol/L}\).Understanding \([OH^-]\) helps in gauging how a solution will interact with acids or bases. The relative concentrations of \([H^+]\) and \([OH^-]\) ions allow one to classify solutions as acidic, neutral, or basic.

pOH Value

The pOH value of a solution complements the pH and offers insight into the basicity of the solution. It is defined similarly to pH, using the hydroxide ion concentration:\[\text{pOH} = -\log[OH^-]\]In this scenario, with \([OH^-] = 2.41 \times 10^{-6.60}\, \text{mol/L}\), the pOH is calculated as:\[\text{pOH} = -\log(2.41 \times 10^{-6.60})\]This results in a pOH value of approximately 6.60.The relationship between pH and pOH is also significant; they add up to 14 at 25°C in pure water. However, since temperature can affect ion concentrations, the sum can be slightly different at other temperatures, but the principle remains useful. Together, pH and pOH provide a comprehensive picture of the ion balance in the solution, important for understanding chemical equilibria and biological systems.