高中化學(xué)魯科版選修四試題：第三章《物質(zhì)在水溶液中的行為》測(cè)試卷（能力卷） Word版含解析

1、第三章物質(zhì)在水溶液中的行為測(cè)試卷（能力卷）一、選擇題(每小題3分，共24分，每小題只有一個(gè)選項(xiàng)符合題意)1用0.01 moll1的h2so4滴定0.01 moll1的naoh溶液，中和后加水至100 ml。若滴定時(shí)的終點(diǎn)判斷有誤差：多加1滴h2so4；少加1滴h2so4(設(shè)1滴是0.05 ml)。則中，ch的比值是()a5103 b104 c50 d102在一定條件下，na2co3溶液存在水解平衡：coh2ohcooh。下列說(shuō)法正確的是()a稀釋溶液，水解平衡常數(shù)增大 b通入co2，平衡朝正反應(yīng)方向移動(dòng)c升高溫度，減小 d加入naoh固體，溶液ph減小325 時(shí)，向純水中加入naoh，使溶液的

2、ph是11，則由naoh電離出的oh濃度與水電離出的oh濃度之比是()a10101 b51091 c1081 d114室溫下，在ph12的某溶液中，分別有甲、乙、丙、丁四位同學(xué)計(jì)算出由水電離出的oh的數(shù)據(jù)分別為：甲：1.0107 moll1；乙：1.0106 moll1；丙：1.0102 moll1；丁：1.01012 moll1。其中你認(rèn)為可能正確的數(shù)據(jù)是()a甲、乙 b乙、丙 c丙、丁 d乙、丁來(lái)源:#中國(guó)%教育出版網(wǎng)5以mno2為原料制得的mncl2溶液中常含有cu2、pb2、cd2等金屬離子，通過(guò)添加難溶電解質(zhì)mns，可使這些金屬離子形成硫化物沉淀，過(guò)濾除去包括mns在內(nèi)的沉淀，再經(jīng)蒸

3、發(fā)、結(jié)晶，可得純凈物mncl2。根據(jù)上述實(shí)驗(yàn)事實(shí)，可推知mns具有相關(guān)性質(zhì)是()www.zzs&tep.co%ma具有吸附性b溶解度與cus、pbs、cds等相同c溶解度大于cus、pbs、cds中國(guó)教育出&版*#網(wǎng)d溶解度小于cus、pbs、cds來(lái)源:%*中教網(wǎng)#625時(shí)，將稀氨水逐滴加入到稀硫酸中，當(dāng)溶液的ph7時(shí)，下列關(guān)系正確的是()anh2so bnhso來(lái)源:z#zstep%&.comcnhso dohsohnh7下列溶液中有關(guān)物質(zhì)的量濃度關(guān)系正確的是()aph2的ha溶液與ph12的moh溶液任意比混合：hmoha中&*國(guó)教育出#版網(wǎng)bph相等的ch3coona、naoh和na2

4、co3三種溶液：naohch3coona hh2aa28在25時(shí)，在濃度為1 moll1的(nh4)2so4、(nh4)2co3、(nh4)2fe(so4)2的溶液中，測(cè)得其cnh分別為a、b、c(單位為 moll1)。下列判斷正確的是()aabc babc cacb dcab二、非選擇題(本題包括2個(gè)大題，共26分)9(9分)某學(xué)生用0.100 moll1的koh標(biāo)準(zhǔn)溶液滴定未知濃度的鹽酸，其操作可分解為如下幾步：a移取20 ml待測(cè)鹽酸溶液注入潔凈的錐形瓶，并加入23滴酚酞；b用標(biāo)準(zhǔn)溶液潤(rùn)洗滴定管23次；c把盛有標(biāo)準(zhǔn)溶液的堿式滴定管固定好，調(diào)節(jié)滴定管尖嘴使之充滿溶液；d取標(biāo)準(zhǔn)koh溶液注入

5、堿式滴定管至刻度0以上23 cm；來(lái)源:zzs*te%#e調(diào)節(jié)液面至0或0以下刻度，記下讀數(shù)；來(lái)源:*&中教%網(wǎng)f把錐形瓶放在滴定管的下面，用標(biāo)準(zhǔn)koh溶液滴定至終點(diǎn)并記下滴定管液面的刻度。中國(guó)教#育*出版網(wǎng)就此實(shí)驗(yàn)完成填空：(1)正確操作步驟的順序是(用序號(hào)字母填寫(xiě))_。(2)上述b步驟操作的目的是_。(3)上述a步驟操作之前，若先用待測(cè)溶液潤(rùn)洗錐形瓶，則滴定結(jié)果_(填“偏高”、“偏低”或“不變”)。(4)判斷到達(dá)滴定終點(diǎn)的實(shí)驗(yàn)現(xiàn)象是_。10(8分)已知難溶電解質(zhì)在水溶液中存在溶解平衡：mman(s) mmn(aq)nam(aq)kspmnmamn，稱為溶度積。某學(xué)習(xí)小組欲探究caso4沉淀

6、轉(zhuǎn)化為caco3沉淀的可能性，查得如下資料(25)：難溶電解質(zhì)caco3caso4mgco3mg(oh)2ksp2.81097.11056.81065.61012實(shí)驗(yàn)步驟如下：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。向上述懸濁液中加入固體na2co3 3 g，攪拌，靜置，沉淀后棄去上層清液。再加入蒸餾水?dāng)嚢?，靜置、沉淀后再棄去上層清液。_。(1)由題中信息ksp越大，表示電解質(zhì)的溶解度越_(填“大”或“小”)。(2)寫(xiě)出第步發(fā)生反應(yīng)的化學(xué)方程式：_。(3)設(shè)計(jì)第步的目的是_。(4)請(qǐng)補(bǔ)充第步操作及發(fā)生的現(xiàn)象

7、_。11(9分)水體中重金屬鉛的污染問(wèn)題備受關(guān)注。水溶液中鉛的存在形態(tài)主要有pb2、pb(oh)、pb(oh)2、pb(oh) 、pb(oh)，各形態(tài)的濃度分?jǐn)?shù)隨溶液ph變化的關(guān)系如圖所示：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由水電離所得，所以丙也是正確的。來(lái)源:*&中教網(wǎng)5、c【解析】加入mns后可以除去cu2、pb2、cd2說(shuō)明mns的溶解度大于cus、pbs、cds，才能實(shí)現(xiàn)沉淀的轉(zhuǎn)化。中國(guó)#教*%育出版網(wǎng)6、a【解析】由電荷守恒知：nhhoh2so，又ph

8、7，hoh，故nh2so。7、ac【解析】由電荷守恒知a正確；b項(xiàng)中ph相等，則oh相同，弱離子越弱水解程度越大，濃度弱酸對(duì)應(yīng)鹽越小，所以正確順序應(yīng)為naohna2co3h2a，d錯(cuò)誤。8、d【解析】三種溶液中均存在水解平衡：nhh2onh3h2oh，對(duì)于(nh4)2co3，因coh2ohcooh，使上述平衡向右移動(dòng)；對(duì)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?qū)?biāo)準(zhǔn)溶液稀釋而帶來(lái)誤差(3)偏高(4)溶液由無(wú)色變?yōu)闇\紅色，且半分鐘內(nèi)不退色【解析】(2)無(wú)論是盛放標(biāo)準(zhǔn)液還是待測(cè)液的滴定管均應(yīng)潤(rùn)

9、洗，因?yàn)榈味ü艿膬?nèi)壁上附著的蒸餾水會(huì)將放入的溶液稀釋而引起測(cè)定誤差。(3)a步驟操作之前，先用待測(cè)液洗滌錐形瓶，會(huì)使待測(cè)液的體積大于20.00 ml，消耗標(biāo)準(zhǔn)液多，則測(cè)定結(jié)果偏高。(4)由于用酚酞作指示劑，故當(dāng)溶液由無(wú)色變?yōu)闇\紅色，且半分鐘內(nèi)不退色，即達(dá)到了滴定終點(diǎn)。zz#*10、(1)大(2)na2co3(aq)caso4(s)=na2so4(aq)caco3(s)(3)洗去沉淀中附著的so(4)向沉淀中加入足量的鹽酸，沉淀完全溶解，并生成無(wú)色無(wú)味氣體【解析】由ksp表達(dá)式不難看出其與溶解度的關(guān)系，在硫酸鈣的懸濁液中存在著：caso4(s)so(aq)ca2(aq)，而加入n

10、a2co3后，溶液中co濃度較大，而caco3的ksp較小，故co與ca2結(jié)合生成沉淀，即coca2=caco3。既然是探究性實(shí)驗(yàn)，必須驗(yàn)證所推測(cè)結(jié)果的正確性，故設(shè)計(jì)了步操作，即驗(yàn)證所得固體是否為碳酸鈣。11、(1)pb2與cl反應(yīng)，pb2減小(2)pb2、pb(oh)、hpb(oh)oh=pb(oh)2中教網(wǎng)&%(3)fe3(4)b【解析】本題以水體鉛污染為背景考查有關(guān)水解離子反應(yīng)等問(wèn)題。(1)根據(jù)題中鉛的多種存在方式，可知pb2易水解：pb2h2opb(oh)h，pb(oh)h2ob(oh)2h，故在pb(no3)2溶液中2，當(dāng)加入nh4cl后，pb2與cl反應(yīng)使pb2濃度降低，故增加。中

11、國(guó)教育出版網(wǎng)#%&(2)由題所給圖示可知，當(dāng)ph8時(shí)溶液中存在的陽(yáng)離子為pb2、pb(oh)和h，而在ph9時(shí)圖中pb(oh)迅速降低，而pb(oh)2濃度增大，故發(fā)生的反應(yīng)為pb(oh)oh=pb(oh)2。(3)分析比較表中所給數(shù)據(jù)可知cu2、mn2、cl的濃度都降低不到原濃度的1/2，而fe3卻降低為原濃度的1/3，故對(duì)fe3的去除效果最好。(4)由所給平衡可知，要使脫鉛效果好，鉛應(yīng)以pb2的形式存在于ph7的溶液中，而h增大時(shí)，2eh(s)pb2e2pb(s)2h又會(huì)向逆反應(yīng)方向移動(dòng)，h不宜過(guò)大，根據(jù)本題所示，不難知道脫鉛最適宜的ph為67。qcwa3ptgz7r4i30ka1dkag

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