Find the h3o + and the oh - from the following values

Missing question on internet:

a) pH = 5.

pH = - log[H₃O⁺ ].

[H₃O⁺] = 10∧ (-pH).

[H₃O⁺] = 10⁻⁵ M.

The Kw (the ionic product for water) at 25°C is 1·10⁻¹⁴ mol²/dm⁶ or 10⁻¹⁴ M².

Kw = [H ₃O⁺] · [OH⁻].

[OH⁻] = Kw : [H₃O⁺].

[OH⁻] = 10⁻¹⁴ M² : 10⁻⁵ M = 10⁻⁹ M.

b) pH = 1.82.

[H₃O⁺] = 10∧ (-1.82) = 1.5·10⁻² M.

[OH⁻] = 10⁻¹⁴ M² : 1.5·10⁻² M = 6.6·10⁻¹³ M.

c) pH = 10.65.

[H₃O⁺] = 10∧ (-10.65) = 2.23·10⁻¹¹ M.

[OH⁻] = 10⁻¹⁴ M² : 2.23·10⁻¹¹ M = 4.46·10⁻⁴ M.