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Question

To 1 litre solution containing 0.1 mole each of NH3 and NH4Cl,0.05 mol of NaOH is pH will be (pKb for NH3=4.74).

A
1.48
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B
0.48
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C
0.20
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D
0.80
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Solution

The correct option is B 0.48
Initial pH of the solution can be calculated as
[NH3]=No.ofmolesVolume=0.11=0.1 M
Similarly, [NH4Cl]=0.1 M
According to Henderson's equation
pOH=Kb+log[salt][base]
pOH=4.74+log0.10.1
pOH=4.74
Now, pH=14pOH=144.74=9.26
When o0.05 mole of NAOH is added
NaOH+NH4ClNaCl+NH4OH
Initial moles 0.05 0.1 0 0.1
Final moles 0 0.05 0.05 0.15
Now, pOH2=4.74+log0.050.15
pOH2=4.74+log0.3333
pOH2=4.74+0.4771
=4.26
pH2=144.74
Change in pH=pHpH2=0.48
Thus we can say that the pH of the solution increases by 0.48

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