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Ionic Equilibrium question

2007 · Shift 0 · Q4
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  5. /2007 · Shift 0 · Q4

Ionic Equilibrium question

2007 · Shift 0 · Q4

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA in which 50% of the acid is ionized is :
  1. A
    7.0
  2. B
    4.5
  3. C
    2.5
  4. D
    9.5
View written solutionFree

Correct answer: D

  1. Use Henderson–Hasselbalch equation for the weak acid buffer:
pH=pKa+log⁡[A−][HA]\text{pH} = \text{p}K_a + \log\frac{[A^-]}{[HA]}pH=pKa​+log[HA][A−]​
  1. Interpret “50% of the acid is ionized”:

If 50% of HAHAHA is ionized, then half remains as HAHAHA and half is converted to A−A^-A−. Hence,

[A−]=[HA][A^-] = [HA][A−]=[HA]

So,

[A−][HA]=1\frac{[A^-]}{[HA]} = 1[HA][A−]​=1
  1. Substitute into the Henderson–Hasselbalch equation:
pH=pKa+log⁡1=4.5+0=4.5\text{pH} = \text{p}K_a + \log 1 = 4.5 + 0 = 4.5pH=pKa​+log1=4.5+0=4.5
  1. Now calculate pOH:

At 25∘C25^\circ C25∘C,

pH+pOH=14\text{pH} + \text{pOH} = 14pH+pOH=14

Therefore,

pOH=14−4.5=9.5\text{pOH} = 14 - 4.5 = 9.5pOH=14−4.5=9.5
  1. Match with the options:
9.5\boxed{9.5}9.5​

So the correct option is D.

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