Question

Like any equilibrium constant, \(K_{w}\) changes with temperature. (a) Given that autoionization is endothermic, how does \(K_{w}\) change with rising \(T ?\) Explain with a reaction that includes heat as reactant or product. (b) In many medical applications, the value of \(K_{w}\) at \(37^{\circ} \mathrm{C}\) (body temperature) may be more appropriate than the value at \(25^{\circ} \mathrm{C}, 1.0 \times 10^{-14} .\) The \(\mathrm{pH}\) of pure water at \(37^{\circ} \mathrm{C}\) is \(6.80 .\) Calculate \(K_{\mathrm{w}}, \mathrm{pOH},\) and \(\left[\mathrm{OH}^{-}\right]\) at this temperature.

Short answer

At 37°C, pOH is 7.20, [OH-] is approximately 6.31 × 10^{-8} M, and Kw is approximately 9.98 × 10^{-15}.

Step-by-step solution

- Understanding the effect of temperature on endothermic reactions

Given that autoionization is endothermic, according to Le Chatelier's Principle, increasing the temperature will favor the forward reaction where heat is absorbed. Thus, the equilibrium constant, \(K_{w}\), will increase with rising temperature.

- Writing the reaction showing heat as a reactant

The autoionization of water can be expressed with heat included as: \[ \text{H}_2\text{O} (l) + \text{heat} \rightleftharpoons \text{H}^+ (aq) + \text{OH}^- (aq) \]

- Determining the relationship between pH, pOH, and Kw at 37°C

Given \( \text{pH} = 6.80 \) at \( 37^{\text{C}} \), using the equation \[ \text{pH} + \text{pOH} = 14 \], we find: \[ 6.80 + \text{pOH} = 14 \] Re-arranging, \[ \text{pOH} = 14 - 6.80 = 7.20 \]

- Calculating hydroxide ion concentration, [OH-]

Using the relationship \[ \text{pOH} = -\text{log} [\text{OH}^-] \], we can solve for \[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-7.20} \approx 6.31 \times 10^{-8} \]

- Calculating Kw

Using \[ K_{w} = [\text{H}^+] [\text{OH}^-] \] and \[ \text{pH} = 6.80 \Rightarrow [\text{H}^+] = 10^{-6.80} = 1.58 \times 10^{-7} \], we find: \[ K_{w} = (1.58 \times 10^{-7})(6.31 \times 10^{-8}) \approx 9.98 \times 10^{-15} \]

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as K, is a crucial concept in chemical equilibrium. It quantifies the ratio of concentrations of products to reactants at equilibrium. For the autoionization of water, the equilibrium constant is specifically termed as the ion-product constant, Kw.

In water at 25°C, Kw is equal to 1.0 × 10^{-14}. This constant is crucial for understanding the balance between hydrogen ions (H+) and hydroxide ions (OH-) in water. Essentially, Kw = [H+][OH-]. Changes in temperature can shift this equilibrium, affecting the values of [H+] and [OH-].

Le Chatelier's Principle

Le Chatelier's Principle helps predict how a change in conditions affects a chemical equilibrium. It states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

For the autoionization of water, which is endothermic, introducing heat will shift the equilibrium to favor the products (H+ and OH-), thus increasing Kw. This principle can be summarized as:

• Adding heat to an endothermic reaction increases the products.
• Removing heat will favor the reactants and reduce the product formation.

Endothermic Reactions

Endothermic reactions absorb heat from their surroundings. For instance, the autoionization of water can be represented as:
\[ \text{H}_2\text{O} (l) + \text{heat} \rightleftharpoons \text{H}^+ (aq) + \text{OH}^- (aq) \]
In this reaction, heat acts as a reactant. As the temperature increases, the equilibrium shifts to produce more H+ and OH- ions, indicating the system absorbs heat.

This behavior aligns with Le Chatelier's Principle, confirming that endothermic reactions result in a higher equilibrium constant as temperature rises.

pH Calculation

pH is a measure of the acidity or basicity of a solution. It is calculated using the formula: \[ \text{pH} = -\text{log} [\text{H}^+] \]
The value of pH can also be used to determine pOH, which is the measure of hydroxide ion concentration: \[ \text{pH} + \text{pOH} = 14 \]

In the case of pure water at 37°C, given that pH is 6.80, we calculate pOH as:

• pOH = 14 - pH = 14 - 6.80 = 7.20

The concentration of OH- ions can then be determined using: \[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-7.20} \]

Temperature Dependency of Kw

The ion-product constant of water, Kw, is temperature-dependent. This means its value changes with temperature shifts. At higher temperatures, the value of Kw increases.

For example, at 37°C, pH of water is 6.80, different from the neutral pH of 7 at 25°C.
To find Kw at 37°C:

• Calculate [H+] using pH: \[ [\text{H}^+] = 10^{-6.80} \]
• Calculate [OH-] using pOH: \[ [\text{OH}^-] = 10^{-7.20} \]

Finally, multiply the two concentrations to find Kw: \[ K_{w} = [\text{H}^+][\text{OH}^-] = (1.58 \times 10^{-7})(6.31 \times 10^{-8}) \]