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Zorluk: OrtapH and pOH Scale and Calculations

An aqueous solution of barium hydroxide, Ba(OH)2\text{Ba(OH)}_2, has a molar concentration of 0.0005 mol dm−30.0005\text{ mol dm}^{-3}. Assuming complete dissociation of the base at 25∘C25^\circ\text{C}, what is the pH of this solution?

  1. 11.011.0Cevap
  2. B
    3.03.0
  3. C
    10.710.7
  4. D
    3.33.3

Cevap

The pH of the solution is 11.011.0.
Barium hydroxide (Ba(OH)2\text{Ba(OH)}_2) produces two hydroxide ions (OH−\text{OH}^-) per formula unit upon complete dissociation. Thus, a 0.0005 mol dm−30.0005\text{ mol dm}^{-3} solution produces [OH−]=2×0.0005=0.001 mol dm−3=1.0×10−3 mol dm−3[\text{OH}^-] = 2 \times 0.0005 = 0.001\text{ mol dm}^{-3} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}. The pOH is −log⁡10(1.0×10−3)=3.0-\log_{10}(1.0 \times 10^{-3}) = 3.0. Subtracting pOH from 14.0 yields a pH of 11.011.0.

Adım Adım Çözüm

1
Determine the concentration of hydroxide ions [OH−][\text{OH}^-] from the mole ratio of Ba(OH)2\text{Ba(OH)}_2.
Since each mole of Ba(OH)2\text{Ba(OH)}_2 dissociates into two moles of OH−\text{OH}^-, [OH−]=2×0.0005 mol dm−3=0.001 mol dm−3=1.0×10−3 mol dm−3[\text{OH}^-] = 2 \times 0.0005\text{ mol dm}^{-3} = 0.001\text{ mol dm}^{-3} = 1.0 \times 10^{-3}\text{ mol dm}^{-3}.
Barium hydroxide is a strong dibasic base that fully dissociates in aqueous solution.
2
Calculate the pOH of the solution.
pOH=−log⁡10[OH−]=−log⁡10(1.0×10−3)=3.0\text{pOH} = -\log_{10}[\text{OH}^-] = -\log_{10}(1.0 \times 10^{-3}) = 3.0.
The pOH scale is defined as the negative logarithm of the hydroxide ion concentration.
3
Calculate the pH using the relationship pH+pOH=14.0\text{pH} + \text{pOH} = 14.0.
pH=14.0−3.0=11.0\text{pH} = 14.0 - 3.0 = 11.0.
At 25∘C25^\circ\text{C}, the sum of pH and pOH for any dilute aqueous solution equals 14.0.

Anahtar Kavram

pH and pOH calculations for strong basic solutions considering stoichiometry
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pH and pOH Scale and Calculations

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