高中化學(xué)魯科版選修四試題：3.2.4《離子濃度大小的比較》隨堂練習(xí)（教師版） Word版含解析

1、3.2.4離子濃度大小的比較隨堂練習(xí)（教師版）1下列各組離子在溶液中能大量共存的是()aca2、hco、cl、kbal3、al(oh)4、hco、nacfe2、h、so、s2dfe3、scn、na、co來&源:#中教*網(wǎng)20.02 moll1的hcn溶液與0.02 moll1 nacn溶液等體積混合，已知混合溶液中cncnhohbhcncn0.04 moll1cnahcnohdcnhcn3下列各溶液中，粒子的物質(zhì)的量濃度關(guān)系正確的是()a0.1 moll1 na2co3溶液：ohhhco2h2co3中國教%*育&出版網(wǎng)b0.1 moll1 nh4cl溶液：nhclc向醋酸鈉溶液中加入適量醋酸，

2、得到酸性的混合溶液：nach3coohohd向硝酸鈉溶液中滴加稀鹽酸得到ph5的混合溶液：nanocl4在0.1 moll1的nahco3溶液中，下列關(guān)系正確的是()anahcohohbnahcoohhcnahhcooh2codnahhcoohco5在25時，將ph11的naoh溶液與ph3的ch3cooh溶液等體積混合后，下列關(guān)系式中正確的是()anach3cooch3coohbhc(ch3coo)ohcnach3cooohhdch3coonahoh6把0.02 moll1 ch3cooh溶液和0.01 moll1 naoh溶液以等體積混合，溶液呈酸性，混合溶液中粒子濃度關(guān)系正確的是()來源

3、:中國&*教育出版網(wǎng)#ach3coonabch3coohch3cooc2hch3cooch3coohdch3coohch3coo0.02 moll17下列各種情況下一定能大量共存的離子組為()aph7的溶液中：fe3、cl、na、nob水電離出的h1103 moll1的水溶液中：na、co、cl、kcph1的水溶液中：nh、cl、mg2、sodal3、hco、i、ca28.物質(zhì)的量濃度相同的氨水、氯化銨、碳酸氫銨、硫酸氫銨、硫酸銨5種溶液中nh的大小順序是_，溶液ph的大小順序是_。9已知在0.1 moll1的nahso3溶液中有關(guān)微粒濃度由大到小的順序為nahsosoh2so3。(1)該溶液

4、中h_oh(填“”、“7，則na_a(填“”、“cn，所以hcnohh。中國%*教育出版網(wǎng)根據(jù)cn物料守恒有hcncn0.02 moll1。因為hoh，故溶液顯堿性，nacn的水解程度大于hcn的電離程度，故cnhcn。3a在0.1 moll1 na2co3溶液中，存在如下平衡：coh2ohcooh，hcoh2oh2co3oh，h2ohoh，根據(jù)電荷守恒有：hnaohhco2co；根據(jù)物料守恒有：na2hco2co2h2co3，綜合上述兩式可得：ohhhco2h2co3。在0.1 moll1 nh4cl溶液中，nh發(fā)生水解，故nhoh，根據(jù)電荷守恒有nahch3coooh，則ch3coona，

5、因此溶液中的離子濃度關(guān)系為ch3coonahoh。硝酸鈉與稀鹽酸混合得到ph5的溶液，硝酸鈉不發(fā)生水解，忽略水的電離，根據(jù)電荷守恒則有nano。4c在nahco3溶液中存在平衡hcohco，hcoh2oh2co3oh，h2ohoh，由于hco的水解程度大于電離程度，故溶液中hoh，nahco；由電荷守恒知c項正確，d項不正確。5dph11的naoh溶液中naohoh103 moll1，而ph3的ch3cooh溶液中h103 moll1，ch3cooh遠遠大于103 moll1。兩者等體積混合后，形成ch3cooh與ch3coona混合液，且ch3cooh遠大于ch3coona，溶液呈酸性，電荷

6、恒等關(guān)系為nahch3coooh，離子濃度大小順序為ch3coonahoh。6a溶液混合后，二者反應(yīng)，但ch3cooh過量，故為ch3coona和ch3cooh的混合體系。ch3cooh和ch3coo相互抑制，但以ch3cooh電離為主，溶液顯酸性，即hoh，由電荷守恒nahch3coooh，則有ch3coona，ch3cooh電離程度大于ch3coo水解程度，b不正確；對于c項：由于存在hnaohch3coo的電荷守恒和2nach3cooch3cooh的物料守恒，聯(lián)立兩式可得：2hch3cooch3cooh2oh，所以c錯；d項看似是物料守恒，但溶液的體積是原來的2倍，則ch3coohch3

7、coo0.01 moll1，d錯誤。7ca中由于fe33h2ofe(oh)33h，所以有fe3大量存在的溶液一定為酸性；b中因為由水電離出的h1103 moll1，此時溶液可能顯酸性，co在酸性溶液中不能大量共存；c中因nhh2onh3h2oh，mg22h2omg(oh)22h均使溶液呈酸性，c符合題意；d中發(fā)生al33hco=al(oh)33co2的雙水解反應(yīng)。本題考查離子共存，看清楚題干中的條件。8解析本題可分3種情況討論：(1)氨水電離產(chǎn)生nh、oh；(2)nh水解產(chǎn)生h及剩余nh；(3)nh水解受到影響，促進或抑制。在分析時要注意兩點：氨水電離程度不是很大，nh水解程度也不是很大。另外

8、，nh4hso4溶液，h抑制nh水解，而nh4hco3中，hco比氨水更弱，hco水解呈堿性，會促進nh水解，但影響不是很大，(nh4)2so4溶液中，nh的濃度最大。所以，nh的濃度由大到小順序為(nh4)2so4、nh4hso4、nh4cl、nh4hco3、nh3h2o，其中ph可根據(jù)nh水解程度及電離的情況得出ph由大到小的順序是nh3h2o、nh4hco3、nh4cl、(nh4)2so4、nh4hso4。9(1)由已知條件hsohso，hsoh2oh2so3oh，知hso電離程度大于水解程度，故hoh(2)紅色逐漸變淺直至變?yōu)闊o色hsooh=soh2o解析由溶液中離子濃度大小可判斷na

9、hso3溶液顯酸性(電離大于水解)，則hoh。當加入少量含有酚酞的naoh溶液后，ohhso=soh2o，則紅色逐漸變淺，甚至褪去。中國&教育出#*版網(wǎng)10(1)小于(2)(3)a、d解析(1)據(jù)題意知，ha與naoh恰好完全反應(yīng)，若ha為強酸，則溶液m的ph7；若ha為弱酸，則溶液m的ph7，故無ph7，即ha。(3)a項，若m呈中性，即hoh，因t25，則hoh1107 moll11107 moll12107 moll1，a正確；b項，若v1v2時，如果ha為強酸，則二者恰好反應(yīng)，ph7，如果ha為弱酸，則ha過量，溶液顯酸性，ph7，b錯誤；結(jié)合b項分析知c錯誤，同樣結(jié)合b項分析知若m呈

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