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Harman [31]
3 years ago
10

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

Chemistry
1 answer:
sashaice [31]3 years ago
4 0

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

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Answer: The [OH^-] of a solution is 10^{-12} M

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Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

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Now put all the given values in the formula of molality, we get

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pH or pOH is the measure of acidity or alkalinity of a solution.

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Putting in the values:

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