为了区别氢氧化钠和氢氧化钙溶液,甲同学设计了如图所示的四组实验方案.(1)其中能达到实验目的是_____

2025年03月15日 22:19
有1个网友回答
网友(1):

(1)区别两种碱,需要借助差异,而不是共性,所以要从钠离子和钙离子上入手.
A、盐酸虽然都可以与氢氧化钠和氢氧化钙反应,但是没有反应现象,没法鉴别,故A错误,
B、酚酞与遇碱变红色,氢氧化钠和氢氧化钙都是碱,没法鉴别,故B错误,
C、碳酸钠与氢氧化钠不反应,和氢氧化钙反应生成白色沉淀,可以鉴别,故C正确,
D、二氧化碳与氢氧化钙反应出现沉淀,与氢氧化钠反应没现象,可以鉴别,故D正确.
(2)根据题意生成的白色沉淀只能是碳酸钙,其化学式为:CaCO3
(3)根据题意可以知道,在溶液中可能含有氢氧化钠,其化学式为:NaOH;
(4)能够和氯化钡溶液反应生成白色沉淀为碳酸钠,而氢氧化钙不能和碳酸钠共存,所以可以判断使溶液变红的是氢氧化钠,即该溶液中含有碳酸钠和氢氧化钠;
(5)氢氧化钠溶液也呈碱性,也可以使酚酞变红,所以不能仅仅根据溶液变红就判断溶液中只是氢氧化钙.
(6)解:设参加反应的Na2CO3的质量为x
Ca(OH)2+Na2CO3═CaCO3↓+2NaOH 
         106      100
          x        5g
106
100
x
5g
           
x═5.3 g            
则该Na2CO3溶液的溶质的质量分数是
5.3g
100g
×100%
=5.3%
答:该Na2CO3溶液的溶质的质量分数是5.3%.
故答案为:(1)C、D(1 分) 
(2)CaCO3(1 分)
(3)NaOH(1 分)
(4)物质是碳酸钠和氢氧化钠的混合物(1 分)
(5)氢氧化钠溶液也显碱性,还可能是氢氧化钠或氢氧化钠和氢氧化钙的混合物
(1分)
(6)解:设参加反应的Na2CO3的质量为x
Ca(OH)2+Na2CO3═CaCO3↓+2NaOH      (2 分)
106          100
x            5g[中
106
100
x
5g
              (1分)
x═5.3 g                 (1 分)
则该Na2CO3溶液的溶质的质量分数是
5.3g
100g
×100%
=5.3%           (1分)
答:该Na2CO3溶液的溶质的质量分数是5.3%.

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