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Ionic Equilibrum question

2013 · Q114
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Ionic Equilibrum question

2013 · Q114

NEETChemistryIonic EquilibrumMCQ+4 / −1
The dissociation constant of weak acid is 1 ×\times× 10−-−4. In order to prepare a buffer solution with a pH = 5, the [Salt]/[Acid] ratio should be
  1. A
    4 : 5
  2. B
    10 : 1
  3. C
    5 : 4
  4. D
    1 : 10
View written solutionFree

Correct answer: B

Given, Ka = 1 × 10–4

pKa = – log (1× 10–4) = 4

pH = pKa + log [Salt][Acid]{{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}[Acid][Salt]​

⇒\Rightarrow⇒ 5 = 4 + log [Salt][Acid]{{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}[Acid][Salt]​

⇒\Rightarrow⇒ log [Salt][Acid]{{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}[Acid][Salt]​ = 1

⇒\Rightarrow⇒ [Salt][Acid]{{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}[Acid][Salt]​ = 10 = 10 : 1

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