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Question

Calculate amount of NH4Cl required to be dissolved in 500 mL water to have pH=5. [Kb NH4OH=1.8×105].
(Molar mass of NH4Cl=53.5 g mol1)

A
1.855 g
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B
2.543 g
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C
4.815 g
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D
6.066 g
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Solution

The correct option is C 4.815 g
[H+]=10pH=105M
NH+4+H2OC(1h)NH4OHCh+H+Ch
Kh=[NH4OH][H+][NH+4]=(Ch)2C(1h)
Since weak base, h<0.11h1

NH4OHNH+4+OHKb=[NH+4][OH][NH4OH]

H2OH++OHKw=[H+][OH]
Kh=Ch2=KwKb=10141.8×105=5.5×1010

h=KhCh=Kh[H+]=5.5×1010105=5.5×105
C=[H+]h=1055.5×105=0.18 mol/L
500 ml of NH4Cl contains 0.182=0.9 mol
Mass in g =0.9×53.5=48.15 g

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