Question

What is the concentration of a solution of (a) KOH for which the pH is 11.89

Answer 1

Answer:

[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