JEE Main 26-Aug-2021 Shift 1 Solved Paper

Number of moles of NH4‌Cl= concentration of NH4‌Cl× volume of NH4‌Cl
=0.0504M×5‌mL=0.2520 millimoles
Total volume of solution in mixture
= Volume of NH4‌Cl solution + volume of NH3solution
=5+2=7‌mL
Concentration of NH4‌Cl in mixture
=

Number‌of‌moles‌of‌NH4‌Cl

Total‌volume

=

0.2520‌millimole

7‌mL

=0.036M
Number‌of‌moles‌of‌NH3 =concentration‌of‌NH3×volume‌of‌NH3
=0.0210×2=0.042 millimole
Concentration of NH3 in mixture =

Number‌of‌moles‌of‌NH3

Total‌volume

=

0.042

7

=0.006M
According to Handerson’s equation
pOH=pKb+log‌

[Salt]

[Base]

[Salt]=[NH4‌Cl]
[Base]=[NH3]
pOH=−log‌Kb+log‌

0.036

0.006

pOH=−log(1.8×10−5)+0.7782
pOH=5.523
∵ pOH=−log[OH−]
So, [OH−]=2.9×10−6≈3×10−6
∴ x=3