高中化学鲁科版选修四试题：第三章《物质在水溶液中的行为》测试卷（能力卷） Word版含解析

1、第三章物质在水溶液中的行为测试卷（能力卷）一、选择题(每小题3分，共24分，每小题只有一个选项符合题意)1用0.01 moll1的h2so4滴定0.01 moll1的naoh溶液，中和后加水至100 ml。若滴定时的终点判断有误差：多加1滴h2so4；少加1滴h2so4(设1滴是0.05 ml)。则中，ch的比值是()a5103 b104 c50 d102在一定条件下，na2co3溶液存在水解平衡：coh2ohcooh。下列说法正确的是()a稀释溶液，水解平衡常数增大 b通入co2，平衡朝正反应方向移动c升高温度，减小 d加入naoh固体，溶液ph减小325 时，向纯水中加入naoh，使溶液的

2、ph是11，则由naoh电离出的oh浓度与水电离出的oh浓度之比是()a10101 b51091 c1081 d114室温下，在ph12的某溶液中，分别有甲、乙、丙、丁四位同学计算出由水电离出的oh的数据分别为：甲：1.0107 moll1；乙：1.0106 moll1；丙：1.0102 moll1；丁：1.01012 moll1。其中你认为可能正确的数据是()a甲、乙 b乙、丙 c丙、丁 d乙、丁来源:#中国%教育出版网5以mno2为原料制得的mncl2溶液中常含有cu2、pb2、cd2等金属离子，通过添加难溶电解质mns，可使这些金属离子形成硫化物沉淀，过滤除去包括mns在内的沉淀，再经蒸

3、发、结晶，可得纯净物mncl2。根据上述实验事实，可推知mns具有相关性质是()www.zzs&tep.co%ma具有吸附性b溶解度与cus、pbs、cds等相同c溶解度大于cus、pbs、cds中国教育出&版*#网d溶解度小于cus、pbs、cds来源:%*中教网#625时，将稀氨水逐滴加入到稀硫酸中，当溶液的ph7时，下列关系正确的是()anh2so bnhso来源:z#zstep%&.comcnhso dohsohnh7下列溶液中有关物质的量浓度关系正确的是()aph2的ha溶液与ph12的moh溶液任意比混合：hmoha中&*国教育出#版网bph相等的ch3coona、naoh和na2

4、co3三种溶液：naohch3coona hh2aa28在25时，在浓度为1 moll1的(nh4)2so4、(nh4)2co3、(nh4)2fe(so4)2的溶液中，测得其cnh分别为a、b、c(单位为 moll1)。下列判断正确的是()aabc babc cacb dcab二、非选择题(本题包括2个大题，共26分)9(9分)某学生用0.100 moll1的koh标准溶液滴定未知浓度的盐酸，其操作可分解为如下几步：a移取20 ml待测盐酸溶液注入洁净的锥形瓶，并加入23滴酚酞；b用标准溶液润洗滴定管23次；c把盛有标准溶液的碱式滴定管固定好，调节滴定管尖嘴使之充满溶液；d取标准koh溶液注入

5、碱式滴定管至刻度0以上23 cm；来源:zzs*te%#e调节液面至0或0以下刻度，记下读数；来源:*&中教%网f把锥形瓶放在滴定管的下面，用标准koh溶液滴定至终点并记下滴定管液面的刻度。中国教#育*出版网就此实验完成填空：(1)正确操作步骤的顺序是(用序号字母填写)_。(2)上述b步骤操作的目的是_。(3)上述a步骤操作之前，若先用待测溶液润洗锥形瓶，则滴定结果_(填“偏高”、“偏低”或“不变”)。(4)判断到达滴定终点的实验现象是_。10(8分)已知难溶电解质在水溶液中存在溶解平衡：mman(s) mmn(aq)nam(aq)kspmnmamn，称为溶度积。某学习小组欲探究caso4沉淀

6、转化为caco3沉淀的可能性，查得如下资料(25)：难溶电解质caco3caso4mgco3mg(oh)2ksp2.81097.11056.81065.61012实验步骤如下：往100 ml 0.1 moll1的cacl2溶液中加入100 ml 0.1 moll1的na2so4溶液，立即有白色沉淀生成。向上述悬浊液中加入固体na2co3 3 g，搅拌，静置，沉淀后弃去上层清液。再加入蒸馏水搅拌，静置、沉淀后再弃去上层清液。_。(1)由题中信息ksp越大，表示电解质的溶解度越_(填“大”或“小”)。(2)写出第步发生反应的化学方程式：_。(3)设计第步的目的是_。(4)请补充第步操作及发生的现象

7、_。11(9分)水体中重金属铅的污染问题备受关注。水溶液中铅的存在形态主要有pb2、pb(oh)、pb(oh)2、pb(oh) 、pb(oh)，各形态的浓度分数随溶液ph变化的关系如图所示：注：1表示pb2，2表示pb(oh)、3表示pb(oh)2、4表示pb(oh)、5表示pb(oh)。(1)pb(no3)2溶液中，_2(填“”、“”或“h，溶液中的oh由水电离所得，所以丙也是正确的。来源:*&中教网5、c【解析】加入mns后可以除去cu2、pb2、cd2说明mns的溶解度大于cus、pbs、cds，才能实现沉淀的转化。中国#教*%育出版网6、a【解析】由电荷守恒知：nhhoh2so，又ph

8、7，hoh，故nh2so。7、ac【解析】由电荷守恒知a正确；b项中ph相等，则oh相同，弱离子越弱水解程度越大，浓度弱酸对应盐越小，所以正确顺序应为naohna2co3h2a，d错误。8、d【解析】三种溶液中均存在水解平衡：nhh2onh3h2oh，对于(nh4)2co3，因coh2ohcooh，使上述平衡向右移动；对于(nh4)2fe(so4)2，fe22h2oe(oh)22h，h增大，抑制nh的水解。9、(1)bdceaf(2)洗去附在滴定管壁上的水，防止因将标准溶液稀释而带来误差(3)偏高(4)溶液由无色变为浅红色，且半分钟内不退色【解析】(2)无论是盛放标准液还是待测液的滴定管均应润

9、洗，因为滴定管的内壁上附着的蒸馏水会将放入的溶液稀释而引起测定误差。(3)a步骤操作之前，先用待测液洗涤锥形瓶，会使待测液的体积大于20.00 ml，消耗标准液多，则测定结果偏高。(4)由于用酚酞作指示剂，故当溶液由无色变为浅红色，且半分钟内不退色，即达到了滴定终点。zz#*10、(1)大(2)na2co3(aq)caso4(s)=na2so4(aq)caco3(s)(3)洗去沉淀中附着的so(4)向沉淀中加入足量的盐酸，沉淀完全溶解，并生成无色无味气体【解析】由ksp表达式不难看出其与溶解度的关系，在硫酸钙的悬浊液中存在着：caso4(s)so(aq)ca2(aq)，而加入n

10、a2co3后，溶液中co浓度较大，而caco3的ksp较小，故co与ca2结合生成沉淀，即coca2=caco3。既然是探究性实验，必须验证所推测结果的正确性，故设计了步操作，即验证所得固体是否为碳酸钙。11、(1)pb2与cl反应，pb2减小(2)pb2、pb(oh)、hpb(oh)oh=pb(oh)2中教网&%(3)fe3(4)b【解析】本题以水体铅污染为背景考查有关水解离子反应等问题。(1)根据题中铅的多种存在方式，可知pb2易水解：pb2h2opb(oh)h，pb(oh)h2ob(oh)2h，故在pb(no3)2溶液中2，当加入nh4cl后，pb2与cl反应使pb2浓度降低，故增加。中

11、国教育出版网#%&(2)由题所给图示可知，当ph8时溶液中存在的阳离子为pb2、pb(oh)和h，而在ph9时图中pb(oh)迅速降低，而pb(oh)2浓度增大，故发生的反应为pb(oh)oh=pb(oh)2。(3)分析比较表中所给数据可知cu2、mn2、cl的浓度都降低不到原浓度的1/2，而fe3却降低为原浓度的1/3，故对fe3的去除效果最好。(4)由所给平衡可知，要使脱铅效果好，铅应以pb2的形式存在于ph7的溶液中，而h增大时，2eh(s)pb2e2pb(s)2h又会向逆反应方向移动，h不宜过大，根据本题所示，不难知道脱铅最适宜的ph为67。qcwa3ptgz7r4i30ka1dkag

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