Determine the h + of 0.005 KOH

[H⁺] = 2.0 * 10⁻¹² M

Explanation:

Given dа ta:

Concentration of KOH = 0.005M

H⁺ = ?

Solution:

pOH = - log[ OH⁻]

pOH = - log[0.005]

pOH = 2.3

pH + pOH = 14

14 - pOH = pH

pH = 14 - 2.3

pH = 11.7

pH = - log[H⁺]

11.7 = - log[H⁺]

[H⁺] = Antilog (-11.7)

[H⁺] = 2.0 * 10⁻¹² M