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Calculate the hydrolysis constant of ammonium acetate (NH4CH3COO), degree of hydrolysis and ph.Take M=0.01 and 0.04(pka=4.76 and pkb=4.75) |
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Answer» ANSWERBoth NH 4+ and CH 3 COO − ion of CH 3 COONH 4 CH 3 COONH 4 +H 2 O⇋CH 3 COO − +NH 4+ 1 0 0 1−h h hpK a (CH−3COOH)=4.76⟹K a =1.74×10 −5 PK b (NH 4 OH)=4.75⟹K b =1.78×10 −5 K H for CH 3 COONH 4 = K a K b K w = (1.74×10 −5 )(1.78×10 −5 )10 −14 =3.23×10 −5 also, h for CH 3 COONH 4 = K H = 3.23×10 −5 =5.68×10 −3 CH 3 COOH⇋CH C OO − +H + K a = [CH 3 COOH][CH 3 COO − ][H + ] = ChC(1−h)[H + ] [H + ]=K a 1−h1 =K a K H =K a K a ×K b K w − K b K w K a = 1.78×10 5 10 −14 ×1.74×10 −5 [H + ]=9.88×10 −8 pH=7.005[pH and h are independent of initial concentration of Salt] |
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