Problem 37 - Chemistry The Central Science In Si Units

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.

Step-by-Step Solution

Answer

\([\mathrm{H}^{+}] \approx 3.98 \times 10^{-8} \, \text{M}\), \([\mathrm{OH}^{-}] \approx 6.03 \times 10^{-7} \, \text{M}\), \( \mathrm{pOH} \approx 6.22 \).

Step 1: Calculate [H⁺] from pH

The concentration of hydrogen ions \([\mathrm{H}^{+}]\) can be calculated using the formula \( \mathrm{pH} = -\log_{10}[\mathrm{H}^{+}] \). Rearranging this formula to solve for \([\mathrm{H}^{+}]\) gives us \([\mathrm{H}^{+}] = 10^{-\mathrm{pH}}\). By substituting the given pH of 7.40, we have \([\mathrm{H}^{+}] = 10^{-7.40}\). Using a calculator, this gives \([\mathrm{H}^{+}] \approx 3.98 \times 10^{-8} \, \text{M}\).

Step 2: Use Kw to Calculate [OH⁻]

The ion product of water \(K_{w}\) at \(37^{\circ} C\) is given as \(2.4 \times 10^{-14}\). The relationship between \([\mathrm{H}^{+}]\) and \([\mathrm{OH}^{-}]\) is \([\mathrm{H}^{+}][\mathrm{OH}^{-}] = K_{w}\). Therefore, \([\mathrm{OH}^{-}] = \frac{K_{w}}{[\mathrm{H}^{+}]}\). Substituting in the values, we get \([\mathrm{OH}^{-}] = \frac{2.4 \times 10^{-14}}{3.98 \times 10^{-8}}\). Calculating this gives \([\mathrm{OH}^{-}] \approx 6.03 \times 10^{-7} \, \text{M}\).

Step 3: Calculate pOH from [OH⁻]

The pOH is calculated using the formula \( \mathrm{pOH} = -\log_{10}[\mathrm{OH}^{-}] \). Substitute \([\mathrm{OH}^{-}] \approx 6.03 \times 10^{-7} \, \text{M}\) into the formula to get \( \mathrm{pOH} = -\log_{10}(6.03 \times 10^{-7}) \). Calculating this using a calculator gives \( \mathrm{pOH} \approx 6.22 \).

Key Concepts

Hydrogen Ion Concentration

Blood pH indicates how acidic or basic it is. A pH of 7.40 tells us blood is slightly basic, as 7.00 is neutral. To find the hydrogen ion concentration \(\text{[H}^{+}\text{]}\), we use the formula for pH, which is \( \text{pH} = -\log_{10}[\text{H}^{+}] \). Rearranging this, we can find the hydrogen ion concentration as follows:

• \( [\text{H}^{+}] = 10^{-\text{pH}} \)

Using a given blood pH of 7.40, we substitute this into the equation: \( [\text{H}^{+}] = 10^{-7.40} \). When calculated, this gives us \( [\text{H}^{+}] \approx 3.98 \times 10^{-8} \, \text{M} \). This concentration signifies a balance necessary for physiological processes.

Ion Product of Water

Understanding blood chemistry requires knowledge of the ion product of water, denoted as \( K_w \). At a body temperature of \(37^{\circ} \text{C}\), \( K_w \) is equal to \( 2.4 \times 10^{-14} \). This value represents the product of the concentrations of hydrogen ions, \( [\text{H}^{+}] \), and hydroxide ions, \( [\text{OH}^{-}] \):

• Equation: \( [\text{H}^{+}][\text{OH}^{-}] = K_w \)

This equation shows the equilibrium state of water's dissociation into ions. At \( 37^{\circ} \text{C}\) for blood, this equilibrium is crucial as it affects \( [\text{H}^{+}]\) and thus blood pH balance.

Hydroxide Ion Concentration

In a balanced system, the concentration of hydroxide ions \(\text{[OH}^{-}\text{]}\) complements hydrogen ions. To find this, apply the ion product of water formula:

• \( [\text{OH}^{-}] = \frac{K_w}{[\text{H}^{+}]} \)

For normal blood at \( 37^{\circ} \text{C}\), with \( K_w = 2.4 \times 10^{-14}\) and \( [\text{H}^{+}] \approx 3.98 \times 10^{-8} \, \text{M} \), we substitute these values to get:

• \( [\text{OH}^{-}] = \frac{2.4 \times 10^{-14}}{3.98 \times 10^{-8}} \)

This results in \( [\text{OH}^{-}] \approx 6.03 \times 10^{-7} \, \text{M} \). Hydroxide ions are higher as the blood is slightly basic.

pOH Calculation

pOH is another measure to express hydroxide ion concentration, complementing pH. It is found using:

• Equation: \( \text{pOH} = -\log_{10} [\text{OH}^{-}] \)

With \( [\text{OH}^{-}] \approx 6.03 \times 10^{-7} \, \text{M} \), plug this into the equation:

• \( \text{pOH} = -\log_{10} (6.03 \times 10^{-7}) \)

This calculation gives us \( \text{pOH} \approx 6.22 \). In the context of blood, knowing both pH and pOH is crucial for medical diagnostics, ensuring that body systems function correctly at their optimal pH range.