按要求作答:(1)FeCl3水溶液呈酸性,原因是(用离子方程式表示):______,实验室配制FeCl3溶液的方法

2024-12-20 00:09:42
推荐回答(1个)
回答1:

(1)Fe3+易水解,水解生成H+,水解的离子方程式为Fe3++3H2O?Fe(OH)3+3H+,配制溶液时,可加入盐酸抑制FeCl3水解,防止生成沉淀而导致溶液变浑浊;
故答案为:Fe3++3H2O?Fe(OH)3+3H+;配制溶液时,可加入盐酸抑制FeCl3水解,防止生成沉淀而导致溶液变浑浊;
(1)Al2(SO43溶液水解成酸性,NaHCO3溶液水解成碱性,二者发生互促水解生成Al(OH)3和CO2,反应的离子方程式为:Al3++3HCO3-=Al(OH)3↓+3CO2↑;
故答案为:Al3++3HCO3-=Al(OH)3↓+3CO2↑;
(3)①N2(g)+2O2(g)═2NO2(g),△H=+67.7KJ?mol-1;②N2H4(g)+O2(g)═N2(g)+2H2O (g),△H=-534KJ?mol-1
依据盖斯定律,②×2-①得到:2N2H4(g)+2NO2(g)=3N2(g)+4H2O(g)△H=-1135,7KJ/mol;
故答案为:2N2H4(g)+2NO2(g)=3N2(g)+4H2O(g)△H=-1135,7KJ/mol;
(4)常温下,水溶液中存在离子积常数,[H+]?[OH-]=10-14;pH 均为6的H2SO4和溶液中由水电离出的氢离子浓度C1=10-8mol/L;A12(SO43溶液中,铝离子水解生成氢氧化铝和盐酸,由水电离出的C2=10-6mol/L;则

C1
C2
=
10?8
10?6
=
1
100

故答案为:1:100;
(5)常温下,pH=13的Ba(OH)2溶液aL,溶液中氢氧根离子物质的量=10-1mol/L×aL=0.1amol;PH=1的H2SO4溶液bL中含氢离子物质的量=0.1mol/L×bL=0.1bmol;混合后溶液体积变化忽略不计,溶液体积为(a+b)L;混合溶液中氢氧根离子浓度=10-2mol/L;所以得到
0.1a?0.1b
a+b
mol/L=10-2mol/L;计算得到a:b=11:9,
故答案为:11:9.

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