At 100°C, the Kw of water is 5.6x10-13. What is the pOH of a solution that has a pH of 6.25 at 100°C? A) 6.00 B) 6.25 C) 7.00 D) 7.75 E) 8.00

D. 7.75.

Explanation:

Equation for the dissociation of water:

H2O (aq) - -> H + (aq) + OH - (aq)

[H+][OH-] = Kw

The constant, Kw, is the ionic product of water. The product of concentrations of H + and OH - ions in water at a particular temperature is known as ionic product of water.

Kw = 5.6x10-13

Since,

pH + pOH = 14

pH = 6.25

6.25 + pOH = 14

Therefore,

pOH = 14 - 6.25

= 7.75.

The pOH of a solution that has a pH of 6.25 at 100°C is 7.75 (option D)

Explanation:

pOH + pH = 14

pOH = 14 - 6.25

pOH = 7.75

Therefore, at 100°C, the Kw of water is 5.6x10⁻¹³ and the pOH of a solution that has a pH of 6.25 at 100°C is 7.75.

Thus, option D, is the correct answer.