Calculate amount of ammonium chloride required to dissolve in 500 ml water to make pH = 4.5 (KbforNH3=1.8×10−5)
A
48.15 gm
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B
47.15 gm
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C
48.12 gm
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D
48.18 gm
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Solution
The correct option is A 48.15 gm [H+]=10−pH=10−4.5=3.162×10−5M NH+⇌NH3+H+ C(1−h)chch Kh=Ch2[assuming h≪1] Kh=KwKb=10−41.8×10−5=5.5×10−10 ∴h=Khch=Kn[H+]=5.5×10−103.162×10−5=1.74×10−5 ∴C=[H+]h=3.162×10−51.74×10−5=1.8mol/L 500 ml of H2O contain 1.82=0.9 mole Mass in gm =0.9×53.5=48.15 gm