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Dear student , NH3 (aq) + H2O (l) ---------> NH4+ (aq) + OH- (aq) t=0 c mol L-1 – – t=eqbm c( 1 – α ) – cα cα ( c ---> is the concentration in mol/ L; α----> degree of dissociation ) Now , Kb = [ NH4+] [ OH-] / [ NH3] = (cα * cα ) / c( 1 – α ) = cα2 / ( 1 – α ) Assuming the value of ‘ α ‘ is very less than 1 we can take (1 – α ) ≈ 1 So, Kb = c α2 1.8 *10-5 = 1.75 α2 α = 0.32 * 10-2 (our assumption is correct ) now , [ OH- ] = cα = 1.75 * 0.32 *10-2 = 0.56 * 10-2 So, pOH = -log [OH-] = – log ( 0.56 * 10-2 ) = 2 – log( 0.56) pH + pOH = 14 Therefore , pH = 14 – {2 – log(0.56)} = 12+ (-0.251) = 11.749 Chemistry ConsultantAskiitians
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