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

The concentration of KOH and Calcium hydroxide are 7.76×10⁻³ M  and 2.39×10⁻³ M respectively.

Given Here,

pH of KOH - 11.89

Since,

pH + pOH = 14

So,

11.89 + pOH = 14

pOH = 14 - 11.89

pOH = 2.11

Since, the ratio of K and Hydroxide ion in KOH is 1:1.

2.11 = - log [OH⁻]  

 [OH⁻]  = 7.76×10⁻³ M  

pOH = 14 - 11.68 = 2.32  

= [OH⁻] = 4.78×10⁻³ M

Since, the ratio of calcium and Hydroxide ion in  is 2:1.

Thus,  

4.78×10⁻³  /2 = 2.39×10⁻³ M

Therefore, the concentration of KOH and Calcium hydroxide are 7.76×10⁻³ M  and 2.39×10⁻³ M respectively.

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[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

2b2t

explanation:

2b2t