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 = [H₃O⁺] · [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.
