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AIIMS 201827 May
Exam
AIIMS
Year
2018
Date
27 May
Subject
Chemistry
Chapter
Ionic Equilibrium

AIIMS 2018 · 27 May · Chemistry · Ionic Equilibrium

Calculate ionisation constant for pyridinium hydrogen chloride....

Calculate ionisation constant for pyridinium hydrogen chloride. (Given that H+\mathrm{H}^{+}ion concentration is 3.6×10−4M3.6 \times 10^{-4} \mathrm{M} and its concentration is 0.02M0.02 \mathrm{M}.)

Options

  1. A6.48×10−26.48 \times 10^{-2}
  2. B6×10−66 \times 10^{-6}
  3. C1.5×10−91.5 \times 10^{-9}
  4. D12×10−812 \times 10^{-8}

Correct answer

C. 1.5×10−91.5 \times 10^{-9}

Step-by-step solution

Pyridinium hydrochloride is a salt of weak base and strong acid. pH=−log⁡[H+]=−log⁡(3.6×10−4)=3.44 Now, pH=−12[log⁡Kw−log⁡Ka+log⁡C)⇒3.44=−12[−14−log⁡Ka+log⁡(2×10−2)]⇒6.88=14+log⁡Ka+1.70⇒log⁡Ka=−8.82⇒Ka=antilog⁡(−8.82)=1.5×10−9\begin{aligned} & \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left(3.6 \times 10^{-4}\right)=3.44 \\ & \text{ Now, } \mathrm{pH}=-\frac{1}{2}\left[\log K_w-\log K_a+\log C\right) \\ & \Rightarrow \quad 3.44=-\frac{1}{2}\left[-14-\log K_a+\log \left(2 \times 10^{-2}\right)\right] \\ & \Rightarrow 6.88=14+\log K_a+1.70 \\ & \Rightarrow \log K_a=-8.82 \\ & \Rightarrow K_a=\operatorname{antilog}(-8.82)=1.5 \times 10^{-9}\end{aligned}
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