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Mr.NH₄Cl = 53,5
⇔ n NH₄Cl= 1,07/53,5 = 0,02 mol
Volume (V) = 100 ml = 0,1 L
⇔ M = [NH₄Cl] = n/V = 0,02/0,1 = 0,2 M
pH = 5 - log 1,05 ----> [H⁺] = 1,05 x 10⁻⁵ M
Kb = .....?