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
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⁻]
[tex]\bold {10^{2.11} =}[/tex] [OH⁻] = 7.76×10⁻³ M
pOH = 14 - 11.68 = 2.32
[tex]\bold {10^{-2.32}}[/tex] = [OH⁻] = 4.78×10⁻³ M
Since, the ratio of calcium and Hydroxide ion in [tex]\bold {Ca(OH)_2 }[/tex] 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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