n(CO 2 )=
含有0.2mol氢氧化钠和0.1mol氢氧化钙的溶液中:n(OH - )=0.2mol+0.1mol×2=0.4mol,n(Na + )=0.2mol, n(Ca 2+ )=0.1mol, 通入CO 2 ,发生:2OH - +CO 2 =CO 3 2- +H 2 O,OH - +CO 2 =HCO 3 - +H 2 O,Ca 2+ +CO 3 2- =CaCO 3 ↓, 设生成xmolCO 3 2- ,ymolHCO 3 - , 则
x=0.1,y=0.2, 所以反应后溶液中含有:n(Na + )=0.2mol,n(HCO 3 - )=0.2mol, 可依次发生:①2OH - +CO 2 =CO 3 2- +H 2 O, ②Ca 2+ +CO 3 2- =CaCO 3 ↓,离子浓度迅速减小, ③OH - +CO 2 =HCO 3 - , 所以图象C符合, 故选C. |