The ionic product of water is 1.0 ×10^-14 at 25^∘ C. Assuming the density of water independent f

step 1 So in this question, LN K2 by K1 is equal to DLH by R, 1 by T1 minus 1 by T2. So LN K2 by 10x2 p

The ionic product of water is 1.0 ×10^-14 at 25^∘ C....

The ionic product of water is $1.0 \times 10^{-14}$ at $25^{\circ} \mathrm{C}$. Assuming the density of water independent from change in temperature, the ionic product of water at $50^{\circ} \mathrm{C}$ will be (a) $2.0 \times 10^{-14}$ (b) $5.0 \times 10^{-15}$ (c) $5.9 \times 10^{-14}$ (d) $1.0 \times 10^{-14}$

The ionic product of water is $1.0 \times 10^{-14}$ at $25^{\circ} \mathrm{C}$. Assuming the density of water independent from change in temperature, the ionic product of water at $50^{\circ} \mathrm{C}$ will be
(a) $2.0 \times 10^{-14}$
(b) $5.0 \times 10^{-15}$
(c) $5.9 \times 10^{-14}$
(d) $1.0 \times 10^{-14}$

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Key Concepts

Ionic Product of Water

This is the equilibrium constant (denoted as Kw) for the self-ionization of water into hydrogen ions (H+) and hydroxide ions (OH-). It is typically expressed as the product of the concentrations of these ions and is fundamental in understanding the acid-base properties of water and aqueous solutions.

Autoionization of Water

Autoionization of water refers to the process by which water molecules spontaneously dissociate into hydrogen ions (or hydronium ions) and hydroxide ions. This equilibrium process underlies the concept of water's ionic product and is essential for defining pH in aqueous systems.

Temperature Dependence of Equilibrium Constants

The equilibrium constant for a chemical reaction, including the autoionization of water, is temperature dependent. Changes in temperature alter the equilibrium position because reactions either absorb or release heat, thus affecting the concentration of products and reactants as predicted by thermodynamic principles such as Le Chatelier's principle.

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