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JEE Main 202622 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2026 · 22 January · Evening Shift · Chemistry · Ionic Equilibrium

Which of the following mixture gives a buffer solution with pH=9.25...

Which of the following mixture gives a buffer solution with pH=9.25\mathrm{pH}=9.25 ?
Given : $\mathrm{pK}_{\mathrm{b}}\left(\mathrm{NH}_{4} \mathrm{OH}\right)=4.75

Options

  1. A0.2MNH4OH(0.5 L)+0.1MHCl(0.5 L)0.2 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}(0.5 \mathrm{~L})+0.1 \mathrm{M} \mathrm{HCl}(0.5 \mathrm{~L})
  2. B0.2MNH4OH(0.4 L)+0.1MHCl(1 L)0.2 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}(0.4 \mathrm{~L})+0.1 \mathrm{M} \mathrm{HCl}(1 \mathrm{~L})
  3. C 0.4 MNH4OH(1L)+0.1MHCl(1L)\text{ 0.4 } \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}(1 \mathrm{ L})+0.1 \mathrm{M} \mathrm{HCl}(1 \mathrm{ L})
  4. D0.5MNH4OH(0.2 L)+0.2MHCl(0.5 L)0.5 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}(0.2 \mathrm{~L})+0.2 \mathrm{M} \mathrm{HCl}(0.5 \mathrm{~L})

Correct answer

A. 0.2MNH4OH(0.5 L)+0.1MHCl(0.5 L)0.2 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}(0.5 \mathrm{~L})+0.1 \mathrm{M} \mathrm{HCl}(0.5 \mathrm{~L})

Step-by-step solution

pOH=pKb+log⁡SaltBasepOH = pK_b + \log \dfrac{Salt}{Base}
4.75=4.75+log⁡SaltBase4.75 = 4.75 + \log \dfrac{Salt}{Base}
Milimoles of [Salt] = milimoles of [Base]
NH4OH+HCl→NH4Cl+H2ONH_4OH + HCl \rightarrow NH_4Cl + H_2O
0.2M,0.5L0.1M,0.5L0.2M, 0.5L \quad 0.1M, 0.5L
100 mmole50 mmole100 \text{ mmole} \quad 50 \text{ mmole}
50 mmole−50 mmole50 \text{ mmole} \quad - \quad 50 \text{ mmole}
Milimoles of NH4OHNH_4OH = milimoles of NH4ClNH_4Cl
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