Kerjakan soal berikut 1. Diketahui larutan 0,1 M NH4Cl dan Kb NH4OH = 10-5 a. Tuliskan hidrolisis garam NH4Cl b. Hitung harga Kh, [H+], [OH-] dan pH larutan 2.

a [tex] NH_{4}OH  + HCl [/tex] →[tex] NH_{4}Cl + H_{2}O [/tex]
b Kh = Kw/Kb = 10^-14/10^-5 = 10^-9
 H+ dari [tex] NH_{4}OH [/tex] = [tex] \sqrt{10^{-5} . 10^{-1} } [/tex] = [tex] 10^{-4} [/tex]
OH- = √10^-9 x 10^-1 x 1 = 10^-5

POH = 5
PH larutan = 14-5 = 9

2 OH = √(10^-9/10^-5)x10^-2x1 = √10^-6 =  10^-3
POH = 3
PH = 11


KASI YANG THE BEST YA :'D