【正确答案】解：( 1 ) aM(NH3) = 1 + b1[NH3] +…… + b4[NH3]4 = 1 + 102.0 ′ 10-2.0 + 105.0 ′ 10-4.0 + 107.0 ′ 10-6.0 + 1010.0 ′ 10-8.0 = 122 aM(OH) = 1 + b1[OH] + LL + b4[OH]4 = 1 + 104.0 ′ 10-5.0 + 108.0 ′10-10.0 + 1014.0 ′ 10-15.0 + 1015.0 ′ 10-20.0 ? 1 aM(NH3) >> aM(OH) , 由aM(NH3)可知，主要存在形式为M(NH3)42+。 [ M(NH3)42+ ] = b4[NH3]4×cM / aM(NH3) = (102 / 122) ′ 0.10 = 8 ′ 10-2 mol / L ( 2 ) aM(OH) = 1 + 104.0 ′ 10-1.0 + 108.0 ′ 10-2.0 + 1014.0 ′ 10-3.0 + 1015.0 ′ 10-4.0 = 1011.3 aM(NH3) = 122 (同(1) ) << aM(OH) 由aM(OH)可知，主要存在形式为M(OH)3-和M(OH)42-。