Question

In biological and medical applications, it is often necessary to study the autoionization of water at \(37^{\circ} \mathrm{C}\) instead of \(25^{\circ} \mathrm{C}\). Given that \(K_{\mathrm{w}}\) for water is \(2.5 \times 10^{-14}\) at \(37^{\circ} \mathrm{C},\) calculate the \(\mathrm{pH}\) of pure water at this temperature.

Short answer

The pH of pure water at 37°C is approximately 6.80.

Step-by-step solution

Define the Relationship Between Kw and pH

The ion product of water, denoted as \(K_{\mathrm{w}}\), is the equilibrium constant for the autoionization of water. At any temperature, the relationship between \(K_{\mathrm{w}}\) and the concentration of hydrogen ions \([\mathrm{H}^+]\) in pure water can be expressed as: \[ K_{\mathrm{w}} = [\mathrm{H}^+][\mathrm{OH}^-] \] In pure water, the concentration of hydrogen ions is equal to the concentration of hydroxide ions, i.e., \([\mathrm{H}^+] = [\mathrm{OH}^-]\). Thus, \(K_{\mathrm{w}} = [\mathrm{H}^+]^2\).

Solve for Hydrogen Ion Concentration

To find \([\mathrm{H}^+]\), take the square root of \(K_{\mathrm{w}}\): \[ [\mathrm{H}^+] = \sqrt{K_{\mathrm{w}}} = \sqrt{2.5 \times 10^{-14}} \] Calculating this gives: \([\mathrm{H}^+] \approx 1.58 \times 10^{-7}\).

Calculate the pH Value

The \(\mathrm{pH}\) of a solution is defined as the negative logarithm (base 10) of the hydrogen ion concentration: \[ \mathrm{pH} = -\log_{10}([\mathrm{H}^+]) \] Substituting the value of \([\mathrm{H}^+]\) from the previous step, \( \mathrm{pH} = -\log_{10}(1.58 \times 10^{-7}) \), which approximately equals \(6.80\).

Key concepts

Kw (ion product of water)

The autoionization of water is a fundamental concept where water molecules spontaneously form hydrogen ions ([H⁺]) and hydroxide ions ([OH^-]). This process is in equilibrium, and the equilibrium constant for this reaction is known as the ion product of water, or \( K_{\mathrm{w}} \).

At any given temperature, the expression for \( K_{\mathrm{w}} \) is \( K_{\mathrm{w}} = [\mathrm{H}^+][\mathrm{OH}^-] \). This represents the concentrations of hydrogen and hydroxide ions in pure water.

• At room temperature (25°C), \( K_{\mathrm{w}} \) is typically \( 1.0 \times 10^{-14} \).
• At 37°C, as in biological contexts, \( K_{\mathrm{w}} \) increases slightly to \( 2.5 \times 10^{-14} \).

In pure water, the concentrations of [H⁺] and [OH^-] are equal, leading to \( K_{\mathrm{w}} = [\mathrm{H}^+]^2 \). This relationship is crucial for understanding pH calculations.

pH calculation

Understanding how to calculate pH involves knowing that the pH is determined by the concentration of hydrogen ions in a solution. The formula for pH is given by:

\[ \text{pH} = -\log_{10}([\mathrm{H}^+]) \]
This means you're taking the base-10 logarithm of the hydrogen ion concentration and then multiplying by -1.

For example, if you have calculated \([\mathrm{H}^+] \approx 1.58 \times 10^{-7}\) at 37°C, you would insert this value into the formula:
\[ \text{pH} = -\log_{10}(1.58 \times 10^{-7}) \]
The result of this calculation gives a pH of approximately 6.80.

• This value indicates that pure water is slightly less neutral at 37°C due to the increased ion product.
• Pure water is neutral when \([\mathrm{H}^+]=[\mathrm{OH}^-]\), aligning with \( \text{pH} = 7 \) at room temperature.

Hydrogen ion concentration

Calculating the hydrogen ion concentration is an essential step in understanding water's autoionization and its effect on pH. Given the equilibrium condition, in pure water, \([\mathrm{H}^+] = [\mathrm{OH}^-]\). This simplifies the expression for \( K_{\mathrm{w}} \) to:

\[ K_{\mathrm{w}} = [\mathrm{H}^+]^2 \]
To find \([\mathrm{H}^+] \), you take the square root of \( K_{\mathrm{w}} \):
\[ [\mathrm{H}^+] = \sqrt{K_{\mathrm{w}}} \]
At 37°C, with \( K_{\mathrm{w}} = 2.5 \times 10^{-14} \), the calculation is:
\[ [\mathrm{H}^+] = \sqrt{2.5 \times 10^{-14}} \approx 1.58 \times 10^{-7} \]
This concentration influences the pH, showing how variations in temperature cause changes in water's ionization and subsequent pH.