Question

The ionic product of water at $25^{\circ} C$ is $10^{-14}$. Its ionic product at $90^{\circ} C$ will be

• A. $1 \times 10^{-14}$
• B. $1 \times 10^{-12}$
• C. $1 \times 10^{-20}$
• D. $1 \times 10^{-16}$

Answer 1

The Correct Option is B

To determine the ionic product of water at a higher temperature, we need to understand how temperature affects the equilibrium constant of water dissociation.

The ionic product of water, denoted as \( K_w \), is given by:

K_w = [H^+][OH^-]

At 25^{\circ} C, the ionic product of water (\( K_w \)) is 10^{-14}. This value changes with temperature because the dissociation of water is an endothermic process. According to Le Chatelier's Principle, an increase in temperature shifts the equilibrium towards the right, increasing the concentrations of H^+ and OH^− ions, thus increasing \( K_w \).

Let's examine the options:

1. 1 \times 10^{-14} - This would imply no change in the ionic product at higher temperature, which is unlikely as the equilibrium shifts with temperature change.
2. 1 \times 10^{-12} - Reflects an increase in the ionic product, which is expected at 90^{\circ} C.
3. 1 \times 10^{-20} - This value is too low and does not align with the process of increasing temperatures causing increased dissociation.
4. 1 \times 10^{-16} - This indicates a smaller increase than option 2, which might not be significant enough given the extent of temperature increase.

Thus, the correct answer is 1 \times 10^{-12}, indicating a significant increase in the ionic product of water at 90^{\circ} C compared to 25^{\circ} C.

This reflects the expected behavior of increased ionization of water molecules at elevated temperatures.