At 80°C, pure distilled water has [H₃O⁺] concentration $1\times 10{}^{-6}$ mole/L. What is the value of K_w at this temperature?

CONCEPT:

Ionization of Water and Kw

K_w = [H₃O⁺] × [OH^-]

• Water ionizes slightly into hydronium ions (H₃O⁺) and hydroxide ions (OH^-).
• The ionization constant of water, K_w, is given by the product of the concentrations of H₃O⁺ and OH^- ions:
• At 25°C (standard temperature), K_w is 1 × 10^-14, but it changes with temperature.

EXPLANATION:

Thus, [H₃O⁺] = [OH^-] = 1 × 10^-6 M.

K_w = [H₃O⁺] × [OH^-]

K_w = (1 × 10^-6) × (1 × 10^-6)

K_w = 1 × 10^-12

• At 80°C, the problem states that the [H₃O⁺] concentration is 1 × 10^-6 M.
• For pure water, the concentrations of H₃O⁺ and OH^- are equal because water ionizes symmetrically.
• K_w is calculated as the product of these concentrations:

CONCLUSION:

The value of K_w at 80°C is 1 × 10^-12, and the correct answer is option 3.