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QJEEM22Chec51: JEE Main 2022 Chemistry - Class 11 JEEM22Chec51 Solution: of is mixed with of . The of the mixture is nearest to : (Given :

20 mL20 \mathrm{~mL} of 0.1 M NH4OH0.1\, \mathrm{M} \,\mathrm{NH}_{4} \mathrm{OH} is mixed with 40 mL40 \mathrm{~mL} of 0.05MHCl0.05 \mathrm{M} \mathrm{HCl} . The pH\mathrm{pH} of the mixture is nearest to :

(Given : Kb(NH4OH)=1×10−5,log⁡2=0.30,log⁡3=0.48,log⁡5=0.69,log⁡7=0.84,log⁡11=1.04)\mathrm{K}_{\mathrm{b}}\left(\mathrm{NH}_{4} \mathrm{OH}\right)=1 \times 10^{-5}, \log 2=0.30, \log 3=0.48, \log 5=0.69, \log 7=0.84, \log 11= 1.04)

Multiple Choice Options

  • A 3.2
  • B 4.2
  • C 5.2
  • D 6.2

Step-by-Step Verified Solution

JEE Main 2022 (Online) 25th July Morning Shift Chemistry - Ionic Equilibrium Question 42 English Explanation

[NH4+]=2mmole60ml=130M\left[\mathrm{NH}_4^{+}\right]=\frac{2 \mathrm{mmole}}{60 \mathrm{ml}}=\frac{1}{30} \mathrm{M}

pH=pKw−pKb−log⁡C2=14−5+1.482=5.24\mathrm{pH}=\frac{\mathrm{pK}_{\mathrm{w}}-\mathrm{pK}_{\mathrm{b}}-\log \mathrm{C}}{2}=\frac{14-5+1.48}{2}=5.24

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