At a certain temperature, the ph of a neutral solution is 7.78. what is the value of kw at that temperature? express your answer numerically using two significant figures.

Given the pH of the solution, we can determine [H⁺]. 

pH = -log[H⁺]
[H⁺] = 10⁻⁷·⁷⁸
[H⁺] = 1.659586907 x 10⁻⁸ M

We also know that pOH = 14 - pH. Thus, the pOH is  6.22. Given the pOH of the solution, we can determine [OH⁻]. 

pOH = -log[OH⁻]
[OH⁻] = 10⁻6.22
[OH⁻] =  6.02559586 x 10⁻⁷ M

The ionic product of water Kw, can then be calculated using the following equation.

Kw = [H⁺][OH⁻]
Kw = (1.659586907 x 10⁻⁸)(6.02559586 x 10⁻⁷)
Kw = 1.0 x 10⁻¹⁴

Ur e n d is not real

Must be negative.substance increases