Question

Find out the values of ionic product Kw of water at
various temperatures.
273 K, 283 K, 293 K, 303 K, 313 K, 323 K​

Answer 1

Answer:

273k- 0.11×10-¹⁴

283k- 0.292×10-¹⁴

293k - 0.681×10-¹⁴

303k - 1.47×10-¹⁴

313k - 2.92×10-¹⁴

323- 5.47× 10-¹⁴

Answer 2

Given:

Temperatures: 273K, 283K, 293K, 303K, 313K, 323K​

To find:

Ionic product of water at various temperatures.

Solution:

According to Le Chatelier's principle, if the temperature of water increases, water absorbs heat to move the equillibrium in order to lower the temperature.

At 273K, $K_{w}=0.11$×$10^{-14}$ $mol^2 dm^-6$

At 283K, $K_{w}=0.293$×$10^{-14}$ $mol^2 dm^-6$

At 293K, $K_{w}=0.681$×$10^{-14}$ $mol^2 dm^-6$

At 303K, $K_{w}=1.471$×$10^{-14}$ $mol^2 dm^-6$

At 313K, $K_{w}=2.916$×$10^{-14}$ $mol^2 dm^-6$

At 323K​, $K_w=51.3$×$10^{-14}$ $mol^2 dm^-6$.