What is the concentration of a solution of (a) KOH for which the pH is 11.89; (b) Ca (OH) 2 for which the pH is 11.68

[KOH] = 7.76*10⁻³ M

[Ca (OH) ₂] = 2.39*10⁻³ M

Explanation:

KOH → K⁺ + OH⁻

pH = - log [H⁺]

14 = pH + pOH

pOH = - log [OH⁻]

10^-pOH = [OH⁻]

14 - 11.89 = 2.11 → pOH

2.11 = - log [OH⁻]

10⁻²°¹¹ = [OH⁻] → 7.76*10⁻³ M

As ratio is 1:1, [KOH] = 7.76*10⁻³ M

14 - 11.68 = 2.32 → pOH

10⁻²°³² = [OH⁻] → 4.78*10⁻³ M

Ca (OH) ₂ → Ca²⁺ + 2OH⁻

Ratio is 2:1, so I will have the half of base.

4.78*10⁻³ / 2 = 2.39*10⁻³ M