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JEE Main 202628 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2026 · 28 January · Morning Shift · Chemistry · Ionic Equilibrium

Consider a weak base ' B ' of pK_ b =5.699 . ' x ' mL of 0.02 M HCl...

Consider a weak base ' B ' of pKb=5.699\mathrm{pK}_{\mathrm{b}}=5.699. ' xx ' mL of 0.02 M HCl and ' y ' mL of 0.02 M weak base ' B ' are mixed to make 100 mL of a buffer of pH 9 at 25∘C25^{\circ} \mathrm{C}. The values of ' xx ' and ' yy ' respectively are: [Given: log⁡2=0.3010,log⁡3=0.4771,log⁡5=0.699\log 2=0.3010, \log 3=0.4771, \log 5=0.699]Consider a weak base ' B ' of pK_ b =5.699 . ' x ' mL of 0.02 M HCl and ' y ' mL of 0.02 M weak base ' B ' are mixed to

Options

  1. A(\begin{array}{|c|c|}\hline x & y \hline85.7 & 14.3 \hline\end{array}
  2. B(\begin{array}{|c|c|}\hline x & y \hline14.3 & 85.7 \hline\end{array}
  3. C(\begin{array}{|c|c|}\hline x & y \hline11.1 & 88.9 \hline\end{array}
  4. D(\begin{array}{|c|c|}\hline x & y \hline42.7 & 57.3 \hline\end{array}

Correct answer

B. (\begin{array}{|c|c|}\hline x & y \hline14.3 & 85.7 \hline\end{array}

Step-by-step solution

Consider a weak base ' B ' of pK_ b =5.699 . ' x ' mL of 0.02 M HCl and ' y ' mL of 0.02 M weak base ' B ' are mixed to
pOH=pKb+log⁡[SaltBase]pOH = pK_b + \log \left[ \dfrac{Salt}{Base} \right]
5=5.699+log⁡[SaltBase]5 = 5.699 + \log \left[ \dfrac{Salt}{Base} \right]
xy−x=15\dfrac{x}{y - x} = \dfrac{1}{5}
6x=y6x = y
7x=1007x = 100
x=1007 mlx = \dfrac{100}{7} \text{ ml}
y=6007 mly = \dfrac{600}{7} \text{ ml}
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