What is the pH of a neutral solution at a temperature where Kw=9.9×10−14?

At such temperature, Kw = [H₃O⁺][OH⁻] = 9.9 × 10⁻¹⁴

As the solution is neutral, [H₃O⁺] = [OH⁻]
Then, [H₃O⁺]²= 9.9 × 10⁻¹⁴
[H₃O⁺] = √(9.9 × 10⁻¹⁴) = 3.1 × 10⁻⁷

pH = -log[H₃O⁺]²= -log(3.1 × 10⁻⁷) = 6.5

Since Kw = [H+][OH-] = 9.9X10^-14
And since [H+] = [OH-],
[H+] = square root (9.9X10^-14) = 3.15X10^-7
pH = -log 3.15X10^-7 = 6.50