The degree of dissociation of pure water at 25^∘ C is found to be 1.8 ×10^-9. The dissociation c

step 1 So in this question, KA is equal to H positive, H positive, oh, H negative, H2O. So this is equa

The degree of dissociation of pure water at 25^∘ C is found...

The degree of dissociation of pure water at $25^{\circ} \mathrm{C}$ is found to be $1.8 \times 10^{-9}$. The dissociation constant, $K_{\mathrm{d}}$ of water, at $25^{\circ} \mathrm{C}$ is
(a) $10^{-14}$
(b) $1.8 \times 10^{-16}$
(c) $5.56 \times 10^{-13}$
(d) $1.8 \times 10^{-14}$

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Key Concepts

Degree_of_Dissociation

This is the fraction of molecules in a substance that dissociates into ions when a reaction occurs. It quantifies how much of the original compound has split into its constituent ions, and it is an essential parameter in understanding the behavior of weak electrolytes in solution.

Equilibrium_Constant

The equilibrium constant is a measure of the extent to which a reversible reaction proceeds. It is defined by the ratio of the concentrations of the products to reactants, each raised to the power of their stoichiometric coefficients. In the context of weakly dissociating substances, it is often related to the square of the degree of dissociation multiplied by the square of the concentration of the undissociated species.

Water_Autoionization

Water autoionization refers to the process by which water molecules spontaneously dissociate into hydrogen ions and hydroxide ions. This equilibrium process is very limited due to the extremely low degree of dissociation, and the ion product of water, or its dissociation constant, is a fundamental parameter in acid-base chemistry, influencing pH and related calculations.

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