Verified Solution • 2025–26 Session moderate

QJEEM17Chb2da: JEE Main 2017 Chemistry - Class 11 JEEM17Chb2da Solution: Additin of sodium hydroxide solution to a weak acid (HA)results …

Additin of sodium hydroxide solution to a weak acid (HA)results in a buffer of pH 6. If ionition constant of HA is 10 −- 5, the ratio of salt to acid concentration in the buffer solution will be :

Multiple Choice Options

  • A 4:05
  • B 1:10
  • C 10:01
  • D 5:04

Step-by-Step Verified Solution

HA   ⇌\rightleftharpoons    H+ + A −-

Ka = [H+][A−][HA]{{\left[ {{H^ + }} \right]\left[ {{A^ - }} \right]} \over {\left[ {HA} \right]}} = 10 −- 5

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

⇒\Rightarrow    \,\,\, 6 = −- log [10 −- 5] + log [Salt][Acid]{{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}

⇒\Rightarrow    \,\,\, 6 = 5 + log [Salt][Acid]{{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}

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

⇒\Rightarrow    \,\,\, [Salt][Acid]{{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}} = 10

∴   \therefore\,\,\, Salt : Acid = 10 : 1

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