Method 1 J<Kp° c(OH) K(Mg(OH)2) 5.1X1012 4.5x10-6mol.11 c(Mg") 0.25 NH3(aq)+H,O(1)-NH(aq)+OH(aq) Equi./moL)0.050-4.5X106 c。+4.5x1064.5×10f ≈0.050 ≈Co 4.5×106.c。=1.8×105 C(NH)=0.20mol.L 0.050 M(NH,C1)=53.5 NH,C1)=(0.20×0.40×53.5)g=4.3g 0.050  co 6 6 o 6 /(mol L ) 0.050 4.5 10 4.5 10 4.5 10 1 - × c + × × - - - -  Equi. m(NH4 Cl) = (0.2 00.4 05 3.5)g = 4.3g 0.25 5.1 10 (Mg ) (Mg(OH) ) (OH )＜ 12 2 2 - + - × = c c Ksp J ＜ Ksp  3 2 4 NH (aq) + H O(l) NH (aq) + OH (aq) + - 1 o 4 o 5 6 1.8 1 0 (N H ) 0.2 0mol L 0.050 4.5 1 0 - + - - =  =    c c Mr (NH4 Cl) = 53.5 =4.5x10-6mol.l-1 Method 1