Problem 42 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

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 \(K_{\mathrm{w}}\) at \(90^{\circ} \mathrm{C} ?\) (a) \(10^{-6}\) (b) \(10^{-12}\) (c) \(10^{-14}\) (d) \(10^{-8}\)

Step-by-Step Solution

Answer

The value of \( K_w \) at \( 90^{\circ} \mathrm{C} \) is \( 10^{-12} \).

Step 1: Understand the Relationship Between pH and pOH

In pure water, the concentration of hydroxide ions \([\text{OH}^- ]\) is equal to the concentration of hydronium ions \([ \text{H}_3\text{O}^+ ]\). Therefore, if \([ \text{H}_3\text{O}^+ ] = 10^{-6} \text{ mol/L}\), then \([ \text{OH}^- ] = 10^{-6} \text{ mol/L}\) as well.

Step 2: Calculate the Ion Product of Water

The equilibrium constant for the autoionization of water, denoted as \(K_w\), is calculated by multiplying the concentrations of \([\text{H}_3\text{O}^+]\) and \([\text{OH}^-]\). Thus, \[K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 10^{-6} \times 10^{-6} = 10^{-12}\]

Step 3: Select the Correct Answer from the Options

After calculating \(K_w\) and finding it to be \(10^{-12}\), compare this result to the given options:(a) \(10^{-6}\) (b) \(10^{-12}\) (c) \(10^{-14}\) (d) \(10^{-8}\) The correct choice is option (b) \(10^{-12}\).

Key Concepts

Ion Product of Water

The ion product of water, often denoted as \(K_w\), is a fundamental constant in chemistry that indicates the product of the concentrations of hydrogen ions \([\text{H}^+]\) and hydroxide ions \([\text{OH}^-]\) in pure water at a given temperature. This value is temperature-dependent, meaning it changes with temperature variations. At 25°C, the ion product of water is commonly known to be \(1.0 \times 10^{-14}\), but as the temperature increases, this value can change.- For example, at 90°C, as given in the exercise, the ion concentrations are \(10^{-6}\) moL/L for both \([\text{H}_3\text{O}^+]\) and \([\text{OH}^-]\).- The equation representing the ion product of water is \(K_w = [\text{H}_3\text{O}^+][\text{OH}^-]\).Therefore, in this instance, the ion product of water \(K_w\) equals \(10^{-12}\), which is a higher value compared to the typical \(25^{\circ}C\) condition, indicating increased ionization at higher temperatures.

Autoionization of Water

Autoionization of water is a unique process where water molecules spontaneously ionize into hydrogen ions \(\text{H}^+\) and hydroxide ions \(\text{OH}^-\). This occurs even in pure water without any acid or base added.- It is this self-ionization that gives rise to water's ability to conduct electricity, albeit weakly.- The process can be expressed by the reaction: \[2 \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{OH}^-\]This equilibrium reaction is crucial for understanding the concept of \(K_w\) because the concentrations of the ions produced are used to calculate the ion product of water. Under normal conditions at \(25^{\circ}C\), the concentrations of both ions in neutral water are \(1.0 \times 10^{-7}\) moL/L, leading to the \(K_w\) value commonly cited. However, as the conditions change, like the temperature increase to \(90^{\circ}C\), these concentrations rise, thus increasing \(K_w\).

Equilibrium Constant

The equilibrium constant is a vital concept in chemistry, used to describe the ratio of the concentration of products to reactants at equilibrium in a reversible chemical reaction. Specifically, for the autoionization of water, the equilibrium constant is referred to as \(K_w\).- This constant provides insight into the extent to which water undergoes ionization under certain conditions.- When calculating \(K_w\), it represents the balance achieved in water where \([\text{H}_3\text{O}^+]\) and \([\text{OH}^-]\) reach equilibrium concentrations.The equilibrium constant helps predict whether a reaction will proceed forward or backward during changes, such as temperature shifts. Higher values indicate more extensive ionization, as seen in the exercise where higher temperature at \(90^{\circ}C\) results in a greater \(K_w\) of \(10^{-12}\) compared to the standard \(10^{-14}\) at \(25^{\circ}C\). Understanding this explains why higher temperatures lead to greater concentrations of ions in water, hence a higher \(K_w\).