Problem 31 - A Complete Resource Book in Chemistry for JEE Main

Question

At \(90^{\circ} \mathrm{C}\), pure water has \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] 10^{-6}\) mole litre \(^{-1} .\) What is the value of \(\mathrm{K}_{\mathrm{w}}\) at \(90^{\circ} \mathrm{C}\) ? (a) \(10^{-8}\) (b) \(10^{-6}\) (c) \(10^{-12}\) (d) \(10^{-14}\)

Step-by-Step Solution

Answer

The value of \( K_w \) at \(90^{\circ} \mathrm{C}\) is \(10^{-12}\), so the correct answer is (c).

Step 1: Understand the Ion Product of Water

The ion product of water, denoted as \( K_w \), is the product of the concentrations of hydrogen ions \( [\mathrm{H}_3\mathrm{O}^+] \) and hydroxide ions \( [\mathrm{OH}^-] \) in pure water. It is given by the expression \( K_w = [\mathrm{H}_3\mathrm{O}^+] \times [\mathrm{OH}^-] \).

Step 2: Identify the Hydrogen Ion Concentration

From the problem statement, at \(90^{\circ} \mathrm{C}\), the concentration of hydrogen ions \([\mathrm{H}_3\mathrm{O}^+]\) is \(10^{-6}\) mole litre \(^{-1}\).

Step 3: Express the Ion Concentration in Pure Water

In pure water, the concentration of hydroxide ions \([\mathrm{OH}^-]\) is equal to the concentration of hydrogen ions \([\mathrm{H}_3\mathrm{O}^+]\) because the dissociation of water produces one hydroxide ion for each hydrogen ion. Therefore, \([\mathrm{OH}^-] = 10^{-6}\) mole litre \(^{-1}\).

Step 4: Calculate the Ion Product \(K_w\)

Now, substitute the values of \([\mathrm{H}_3\mathrm{O}^+]\) and \([\mathrm{OH}^-]\) into the expression for \( K_w \) to find \( K_w = (10^{-6}) \times (10^{-6}) = 10^{-12} \).

Step 5: Conclusion

The value of \( K_w \) at \(90^{\circ} \mathrm{C}\) is \(10^{-12}\). Therefore, the correct answer is (c) \(10^{-12}\).

Key Concepts

Kw value at 90°C

The ion product of water, represented by the symbol \( K_w \), is essential in understanding the nature of water's behavior at various temperatures. At \(90^{\circ} \mathrm{C}\), the \( K_w \) is especially interesting because temperature affects the ionization of water. The expression for \( K_w \) is the product of the concentrations of the hydrogen ions \([\mathrm{H}_3\mathrm{O}^+]\) and the hydroxide ions \([\mathrm{OH}^-]\). This is expressed as \( K_w = [\mathrm{H}_3\mathrm{O}^+] \times [\mathrm{OH}^-] \). Understanding \( K_w \) at different temperatures helps in predicting how acidic or basic a solution is. Additionally, in pure water, both \([\mathrm{H}_3\mathrm{O}^+]\) and \([\mathrm{OH}^-]\) are equal due to the water dissociation equilibrium. So, at \(90^{\circ} \mathrm{C}\) with \([\mathrm{H}_3\mathrm{O}^+] = 10^{-6}\) mole litre \(^{-1}\), we find \( K_w = 10^{-12} \) by multiplying the hydrogen and hydroxide ion concentrations. This provides valuable insights into the behavior of water under non-standard conditions.

Hydrogen ion concentration

Hydrogen ion concentration in a solution is crucial for determining its acidity level. In scientific terms, it refers to the amount of hydrogen ions present in a liter of solution, denoted as \([\mathrm{H}_3\mathrm{O}^+]\). At \(90^{\circ} \mathrm{C}\), pure water has a hydrogen ion concentration of \(10^{-6} \) mole litre \(^{-1}\). This indicates a stronger presence of hydrogen ions compared to standard conditions (\(25^{\circ} \mathrm{C}\)), where the hydrogen ion concentration would be lower.

• The greater the hydrogen ion concentration, the higher the acidity, leading to a lower pH level.
• The pH can be calculated as \( pH = -\log_{10}[\mathrm{H}_3\mathrm{O}^+]\).

At \(90^{\circ} \mathrm{C}\), the increased \( K_w \) impacts the neutral point of water, meaning the pH for neutrality is lower than \(7\). Recognizing these differences in hydrogen ion concentration aids in understanding chemical behavior and reactions under varying thermal conditions.

Hydroxide ion concentration

Hydroxide ions, represented as \([\mathrm{OH}^-]\), play their role in determining a solution's basicity. At \(90^{\circ} \mathrm{C}\), understanding the hydroxide ion concentration is crucial to compute the ion product of water. Since the disassociation of water yields equal amounts of hydrogen and hydroxide ions, in pure water, the concentration of both ions is identical. Therefore, given \([\mathrm{H}_3\mathrm{O}^+] = 10^{-6} \) mole litre \(^{-1}\), we also have \([\mathrm{OH}^-] = 10^{-6} \) mole litre \(^{-1}\) at \(90^{\circ} \mathrm{C}\).

• The basicity of a solution is inversely related to its hydrogen ion concentration, thus a decrease in \([\mathrm{OH}^-] \) can indicate higher acidity.
• Alkalinity can be expressed with the formula \( pOH = -\log_{10}[\mathrm{OH}^-]\).

At high temperatures, the neutrality condition changes, so both ions increase, leading to changes in the pH and pOH from the standard \(25^{\circ} \mathrm{C}\). This relationship is key when studying reactions and properties of solutions at elevated temperatures.