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Zorluk: KolaypH and pOH Scale and Calculations

An aqueous solution at 25∘C25^\circ\text{C} has a hydrogen ion concentration, [H+][\text{H}^+], of 1.0×10−5 mol dm−31.0 \times 10^{-5}\text{ mol dm}^{-3}. What is the pOH of this solution?

  1. A
    5.05.0
  2. 9.09.0Cevap
  3. C
    14.014.0
  4. D
    1.0×10−9 mol dm−31.0 \times 10^{-9}\text{ mol dm}^{-3}

Cevap

The pOH of the solution is 9.09.0.
First, find the pH using pH=−log⁡10([H+])=−log⁡10(1.0×10−5)=5.0\text{pH} = -\log_{10}([\text{H}^+]) = -\log_{10}(1.0 \times 10^{-5}) = 5.0. Then, apply the relationship pH+pOH=14.0\text{pH} + \text{pOH} = 14.0 at 25∘C25^\circ\text{C} to calculate pOH=14.0−5.0=9.0\text{pOH} = 14.0 - 5.0 = 9.0. Thus, the option specifying 9.09.0 is correct.

Adım Adım Çözüm

1
Calculate the pH of the solution from the given hydrogen ion concentration
pH=−log⁡10([H+])=−log⁡10(1.0×10−5)=5.0\text{pH} = -\log_{10}([\text{H}^+]) = -\log_{10}(1.0 \times 10^{-5}) = 5.0
pH is defined as the negative logarithm to base 10 of the hydrogen ion concentration.
2
Use the relationship between pH and pOH at 25∘C25^\circ\text{C} to find pOH
pOH=14.0−pH=14.0−5.0=9.0\text{pOH} = 14.0 - \text{pH} = 14.0 - 5.0 = 9.0
For any aqueous solution at 25°C, pH+pOH=14.0\text{pH} + \text{pOH} = 14.0.

Anahtar Kavram

Relationship between pH, pOH, and ion concentrations in aqueous solutions at 25°C
Tahmini Süre:45s
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Konular

pH and pOH Scale and Calculations

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