Question:easy

At $25^\circ\text{C}$, ionic product ($K_w$) of $0.01\text{M HCl}$ solution is

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Don't let the concentration of the acid ($0.01\text{M}$) distract you! $K_w$ is an equilibrium constant. Like all equilibrium constants, its value only changes with temperature. As long as the temperature is $25^\circ\text{C}$, $K_w$ will always be $10^{-14}$.
Updated On: Jul 1, 2026
  • $1.0 \times 10^{-13} \text{ mol}^2/\text{L}^2$
  • $1.0 \times 10^{-12} \text{ mol}^2/\text{L}^2$
  • $1.0 \times 10^{-14} \text{ mol}^2/\text{L}^2$
  • $1.0 \times 10^{-15} \text{ mol}^2/\text{L}^2$
Show Solution

The Correct Option is C

Solution and Explanation

1. Definition of Ionic Product ($K_w$): The ionic product of water is defined as the product of the molar concentrations of hydrogen ions $[H^+]$ and hydroxyl ions $[OH^-]$ in water or an aqueous solution. $$K_w = [H^+][OH^-]$$

2. Dependence on Temperature: The value of $K_w$ is constant at a fixed temperature, regardless of the acidity or alkalinity of the solution. At the standard temperature of $25^\circ\text{C}$ (298 K), the value of $K_w$ for pure water and all dilute aqueous solutions is: $$K_w = 1.0 \times 10^{-14} \text{ mol}^2/\text{L}^2$$

3. Application to the Problem: The concentration of the $HCl$ solution ($0.01\text{M}$) affects the individual concentrations of $[H^+]$ and $[OH^-]$, but their product remains equal to the constant $K_w$ value for that temperature.

• $[H^+]$ from $HCl = 10^{-2} \text{ M}$

• $[OH^-]$ in solution $= \frac{K_w}{[H^+]} = \frac{10^{-14}}{10^{-2}} = 10^{-12} \text{ M}$

• Product $[H^+][OH^-] = 10^{-2} \times 10^{-12} = 10^{-14}$
Thus, at $25^\circ\text{C}$, the ionic product remains $1.0 \times 10^{-14} \text{ mol}^2/\text{L}^2$.
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