高中化学鲁科版选修四试题：3.3.1《难溶电解质的溶解平衡》随堂练习（教师版） Word版含解析

1、3.3.1难溶电解质的溶解平衡随堂练习（教师版）1caco3在下列哪种液体中，溶解度最大()ah2o bna2co3溶液www#.zzst%e*ccacl2溶液 d乙醇2下列说法正确的是()zz*step.co#ma溶度积就是溶解平衡时难溶电解质在溶液中的各离子浓度的乘积b溶度积常数是不受任何条件影响的常数，简称溶度积c可用浓度商qc判断沉淀溶解平衡进行的方向d所有物质的溶度积都是随温度的升高而增大的3下列属于微溶物质的是()aagcl bbacl2 ccaso4 dag2s4下列有关agcl沉淀的溶解平衡说法正确的是()aagcl沉淀生成和沉淀溶解达平衡后不再进行bagcl难溶于水，溶液中没

2、有ag和clc升高温度，agcl沉淀的溶解度增大d向agcl沉淀中加入nacl固体，agcl沉淀的溶解度不变5.已知ag2so4的ksp为2.0105 mol3l3，将适量ag2so4固体溶于100 ml水中至刚好饱和，该过程中ag和so浓度随时间变化关系如图所示饱和ag2so4溶液中ag0.034 moll1。若t1时刻在上述体系中加入100 ml 0.020 moll1 na2so4溶液，下列示意图中，能正确表示t1时刻后ag和so浓度随时间变化关系的是() 练方法技巧6agcl和ag2cro4的溶度积分别为1.81010 mol2l2和2.01012 mol3l3，若用难溶盐在溶液中的浓

3、度来表示其溶解度，则下面的叙述中正确的是()来源:zzstep&%.comaagcl和ag2cro4的溶解度相等bagcl的溶解度大于ag2cro4的溶解度c两者类型不同，不能由ksp的大小直接判断溶解能力的大小d都是难溶盐，溶解度无意义www%.zzst*e#7下列说法正确的是()a在一定温度下agcl水溶液中，ag和cl浓度的乘积是一个常数bagcl的ksp1.81010 mol2l2，在任何含agcl固体的溶液中，agcl且ag与cl浓度的乘积等于1.81010 mol2l2c温度一定时，当溶液中ag和cl浓度的乘积等于ksp值时，此溶液为agcl的饱和溶液d向饱和agcl水溶

4、液中加入盐酸，ksp值变大题号1234567答案8.mg(oh)2难溶于水，但它溶解的部分全部电离。室温下时，饱和mg(oh)2溶液的ph11，若不考虑kw的变化，则该温度下mg(oh)2的溶解度是多少？(溶液密度为1.0 gcm3)ww&w.zz*ste%&zzs*ww*w.zzste&#练综合拓展9与hohkw类似，fes饱和溶液中存在：fes(s)fe2(aq)s2(aq)，fe2s2ksp。常温下ksp8.11017 mol2l2。www.z%z&st*(1)理论上fes的溶解度为_，而事实上其溶解度远大于这个数值，其原因可能是_。www.%zzst&ep#.com*(2)又

5、知fes饱和溶液中h与s2之间存在以下限量关系：h2s21.01022 mol3l3，为了使溶液中fe2达到1 moll1，现将适量fes投入其饱和溶液中，应调节溶液中的h为_。来源*:中国教育出&版网10已知：某温度时，ksp(agcl)1.81010 mol2l2，ksp(ag2cro4)1.11012 mol3l3。试求：(1)此温度下agcl饱和溶液和ag2cro4饱和溶液的物质的量浓度，并比较两者的大小。(2)此温度下，在0.010 moll1的agno3溶液中，agcl与ag2cro4分别能达到的最大物质的量浓度，并比较两者的大小。参考答案与解析1a在b、c选项的溶液中，分别含有c

6、o、ca2，会抑制caco3的溶解，而caco3在乙醇中是不溶的。zz#s*tep.co&m2c溶度积不是溶解平衡时难溶电解质在溶液中的各离子浓度的简单乘积，还与平衡式中化学计量数的幂指数有关，溶度积受温度的影响，不受离子浓度的影响。3cbacl2易溶，agcl、ag2s属难溶物。来#源:%中教&网4c难溶物达到溶解平衡时沉淀的生成和溶解都不停止，但溶解速率和生成速率相等；没有绝对不溶的物质；温度越高，一般物质的溶解度越大；向agcl沉淀中加入nacl固体，使溶解平衡左移，agcl的溶解度减小。w*ww.zz&5b已知ag2so4的ksp2.0105 mol3l3ag2so，则

7、饱和溶液中so moll10.017 moll1，当加入100 ml 0.020 moll1 na2so4溶液时，so0.018 5 moll1，ag0.017 moll1(此时q cksp)，因此答案为b。6cagcl和ag2cro4阴、阳离子比类型不同，不能直接利用ksp来比较二者溶解能力的大小，所以只有c对；其余三项叙述均错误。比较溶解度大小，若用溶度积必须是同类型，否则不能比较。7c在agcl的饱和溶液中，ag和cl浓度的乘积是一个常数，故a项叙述错误；只有在饱和agcl溶液中agcl才等于1.81010，故b项叙述错误；当qcksp，则溶液为饱和溶液，故c项叙述正确；在饱和agcl溶

8、液中加入盐酸只会使溶解平衡发生移动，不会影响ksp，故d项叙述错误。中国教育%*出版&网80.002 9 g解析根据mg(oh)2(s)mg2(aq)2oh(aq)，oh103 moll1，则1 l mg(oh)2溶液中，溶解的nmg(oh)21 l103 moll15104 mol，其质量为58 gmol15104 mol0.029 g，再根据可求出s0.002 9 g。凡是此类题都是设溶液的体积为1 l，根据1 l溶液中溶解溶质的质量，计算溶解度。9(1)7.92108gfes在水中存在下列平衡：fes(s)fe2(aq)s2(aq)，而fe2、s2均可以发生水解：fe22h2ofe(oh

9、)22h、s2h2ohsoh，使上述平衡中的fe2、s2降低，平衡向右移动，造成更多的fes溶解(2)1.11103 moll1解析(1)由fes(s)fe2(aq)s2(aq)www.zz&s#tep.c*omfe2 moll19109 moll1来%源:zzst&即1 l水中可溶解9109 mol的fes由100s1 000(910988) g来#&*源:中教网所以s7.92108 g(2)fe21 moll1，则s2 moll18.11017 moll1。又h2s21.01022 mol3l3，所以h moll1 moll11.11103 moll1。10(1)agcl(s)ag(aq)

10、cl(aq)agcl moll11.3105 moll1ag2cro4(s)2ag(aq)cro(aq)www#.zzst%e*2xx(2x)2xkspag2cro4 moll16.5105 moll1所以agclag2cro4来源:*中国教育&出版网qcwa3ptgz7r4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgshls50clmtwn60eo8wgqv7xav2ohum32wgeauwydiawgmer4i30ka1dkaghn3xtkknbycudxqa7fhyi2chhi92tgkqcwa3ptgz7r4i30ka1dkagtg

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