The degree of dissociation of water at 25^∘ C is 1.8 ×10^-7 % and density is 1.0 g cm^-3. The

step 1 In this question for the equilibrium for the equilibrium H2O gives H positive plus OH negative C

The degree of dissociation of water at 25^∘ C is 1.8 ×10^-7...

The degree of dissociation of water at $25^{\circ} \mathrm{C}$ is $1.8 \times 10^{-7} \%$ and density is $1.0 \mathrm{~g} \mathrm{~cm}^{-3}$. The ionic constant for water is
(a) $1.0 \times 10^{-14}$
(b) $2.0 \times 10^{-16}$
(c) $1.0 \times 10^{-16}$
(d) $1.0 \times 10^{-8}$

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Key Concepts

Degree of Dissociation

This concept refers to the fraction or percentage of molecules in a substance that dissociate into ions in a given process or reaction. In the context of water, although its degree of dissociation is extremely small, it plays a critical role in defining the ionic concentration and thereby influences the ionic product, impacting the overall equilibrium of the system.

Autoionization of Water

This is the process by which water molecules dissociate into hydrogen ions (H+) and hydroxide ions (OH?). Even in pure water, a very small fraction of molecules undergo this transformation, establishing an equilibrium between undissociated water molecules and the ions, which is fundamental in understanding aqueous chemistry and pH levels.

Ionic Product of Water (Kw)

The ionic product of water is the equilibrium constant for the autoionization of water, defined as the product of the molar concentrations of H+ and OH? ions. At 25°C, Kw typically has a value of 1.0 × 10?¹?, making it a crucial parameter in acid-base chemistry for determining the pH and understanding the behavior of aqueous solutions.

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