2021年高考化學(xué)真題和模擬題分類(lèi)匯編專(zhuān)題13鹽類(lèi)的水解【含答案】

專(zhuān)題13鹽類(lèi)的水解/r/n2021年化學(xué)高考題/r/n一、單選題/r/n1．（全國(guó)高考真題試卷）HA/r/n是一元弱酸，難溶鹽/r/nMA/r/n的飽和溶液中/r/n隨/r/nc(H/r/n+/r/n)/r/n而變化，/r/n不發(fā)生水解。實(shí)驗(yàn)發(fā)現(xiàn)，/r/n時(shí)/r/n為線性關(guān)系，如下圖中實(shí)線所示。/r/n下列敘述錯(cuò)誤的是/r/nA．/r/n溶液/r/n時(shí)，/r/nB．MA/r/n的溶度積度積/r/nC．/r/n溶液/r/n時(shí)，/r/nD．HA/r/n的電離常數(shù)/r/n【KS5U答案】/r/nC/r/n【分析】/r/n本題考查水溶液中離子濃度的關(guān)系，在解題過(guò)程中要注意電荷守恒和物料守恒的應(yīng)用，具體見(jiàn)詳解。/r/n【KS5U解析】/r/nA/r/n．由圖可知/r/npH=4/r/n，即/r/nc(H/r/n+/r/n)=10×10/r/n-5/r/nmol/L/r/n時(shí)，/r/nc/r/n2/r/n(M/r/n+/r/n)=7.5×10/r/n-8/r/nmol/r/n2/r/n/L/r/n2/r/n，/r/nc(M/r/n+/r/n)=/r/nmol/L<3.0×10/r/n-4/r/nmol/L/r/n，/r/nA/r/n正確；/r/nB/r/n．由圖可知，c(H/r/n+/r/n)=0時(shí)，可看作溶液中有較大濃度的OH/r/n-/r/n，此時(shí)A/r/n-/r/n的水解極大地被抑制，溶/r/n液中c(M/r/n+/r/n)=c(A/r/n-/r/n)，則/r/n，/r/nB/r/n正確；/r/nC/r/n．設(shè)調(diào)/r/npH/r/n所用的酸為/r/nH/r/nn/r/nX/r/n，則結(jié)合電荷守恒可知/r/n，題給等式右邊缺陰離子部分/r/nnc(X/r/nn-/r/n)/r/n，/r/nC/r/n錯(cuò)誤；/r/nD/r/n．/r/n當(dāng)/r/n時(shí)，由物料守恒知/r/n，則/r/n，/r/n，則/r/n，對(duì)應(yīng)圖得此時(shí)溶液中/r/n，/r/n，/r/nD/r/n正確；/r/n故選/r/nC/r/n。/r/n2．（浙江）/r/n取兩份/r/n/r/n的/r/n溶液，一份滴加/r/n的鹽酸，另一份滴加/r/n溶液，溶液的/r/npH/r/n隨加入酸/r/n(/r/n或堿/r/n)/r/n體積的變化如圖。/r/n下列說(shuō)法/r/n不正確/r/n的是/r/nA．/r/n由/r/na/r/n點(diǎn)可知：/r/n溶液中/r/n的水解程度大于電離程度/r/nB．/r/n過(guò)程中：/r/n逐漸減小/r/nC．/r/n過(guò)程中：/r/nD．/r/n令/r/nc/r/n點(diǎn)的/r/n，/r/ne/r/n點(diǎn)的/r/n，則/r/n【KS5U答案】/r/nC/r/n【分析】/r/n向/r/n溶液中滴加鹽酸，溶液酸性增強(qiáng)，溶液/r/npH/r/n將逐漸減小，向/r/n溶液中滴加/r/nNaOH/r/n溶液，溶液堿性增強(qiáng)，溶液/r/npH/r/n將逐漸增大，因此/r/nabc/r/n曲線為向/r/n溶液中滴加/r/nNaOH/r/n溶液，/r/nade/r/n曲線為向/r/n溶液中滴加鹽酸。/r/n【KS5U解析】/r/nA/r/n．/r/na/r/n點(diǎn)溶質(zhì)為/r/n，此時(shí)溶液呈堿性，/r/n在溶液中電離使溶液呈酸性，/r/n在溶液中水解使溶液呈堿性，由此可知，/r/n溶液中/r/n的水解程度大于電離程度，故/r/nA/r/n正確；/r/nB/r/n．由電荷守恒可知，/r/n過(guò)程溶液中/r/n，滴加/r/nNaOH/r/n溶液的過(guò)程中/r/n保持不變，/r/n逐漸減小，因此/r/n逐漸減小，故/r/nB/r/n正確；/r/nC/r/n．由物料守恒可知，/r/na/r/n點(diǎn)溶液中/r/n，向/r/n溶液中滴加鹽酸過(guò)程中有/r/nCO/r/n2/r/n逸出，因此/r/n過(guò)程中/r/n，故/r/nC/r/n錯(cuò)誤；/r/nD/r/n．/r/nc/r/n點(diǎn)溶液中/r/n=(0.05+10/r/n-11.3/r/n)mol/L/r/n，/r/ne/r/n點(diǎn)溶液體積增大/r/n1/r/n倍，此時(shí)溶液中/r/n=(0.025+10/r/n-4/r/n)mol/L/r/n，因此/r/nx>y/r/n，故/r/nD/r/n正確；/r/n綜上所述，說(shuō)法不正確的是/r/nC/r/n項(xiàng)，故答案為/r/nC/r/n。/r/n3．（廣東高考真題試卷）/r/n鳥(niǎo)嘌呤/r/n(/r/n)/r/n是一種有機(jī)弱堿，可與鹽酸反應(yīng)生成鹽酸鹽/r/n(/r/n用/r/n表示/r/n)/r/n。已知/r/n水溶液呈酸性，下列敘述正確的是/r/nA．/r/n水溶液的/r/nB．/r/n水溶液加水稀釋?zhuān)?r/n升高/r/nC．/r/n在水中的電離方程式為：/r/nD．/r/n水溶液中：/r/n【KS5U答案】/r/nB/r/n【KS5U解析】/r/nA/r/n．/r/nGHCl/r/n為強(qiáng)酸弱堿鹽，電離出的/r/nGH/r/n+/r/n會(huì)發(fā)生水解，弱離子的水解較為微弱，因此/r/n0.001mol/LGHCl/r/n水溶液的/r/npH>3/r/n，故/r/nA/r/n錯(cuò)誤；/r/nB/r/n．稀釋/r/nGHCl/r/n溶液時(shí)，/r/nGH/r/n+/r/n水解程度將增大，根據(jù)勒夏特列原理可知溶液中/r/nc/r/n(H/r/n+/r/n)/r/n將減小，溶液/r/npH/r/n將升高，故/r/nB/r/n正確；/r/nC/r/n．/r/nGHCl/r/n為強(qiáng)酸弱堿鹽，在水中電離方程式為/r/nGHCl=GH/r/n+/r/n+Cl/r/n-/r/n，故/r/nC/r/n錯(cuò)誤；/r/nD/r/n．根據(jù)電荷守恒可知，/r/nGHCl/r/n溶液中/r/nc/r/n(OH/r/n-/r/n)/r/n+c/r/n(Cl/r/n-/r/n)=/r/nc/r/n(H/r/n+/r/n)+/r/nc/r/n(GH/r/n+/r/n)/r/n，故/r/nD/r/n錯(cuò)誤；/r/n綜上所述，敘述正確的是/r/nB/r/n項(xiàng)，故答案為/r/nB/r/n。/r/n4．（湖南高考真題試卷）/r/n常溫下，用/r/n的鹽酸分別滴定/r/n20.00mL/r/n濃度均為/r/n三種一元弱酸的鈉鹽/r/n溶液，滴定曲線如圖所示。下列判斷錯(cuò)誤的是/r/nA．/r/n該/r/n溶液中：/r/nB．/r/n三種一元弱酸的電離常數(shù)：/r/nC．/r/n當(dāng)/r/n時(shí)，三種溶液中：/r/nD．/r/n分別滴加/r/n20.00mL/r/n鹽酸后，再將三種溶液混合：/r/n【KS5U答案】/r/nC/r/n【分析】/r/n由圖可知，沒(méi)有加入鹽酸時(shí)，/r/nNaX/r/n、/r/nNaY/r/n、/r/nNaZ/r/n溶液的/r/npH/r/n依次增大，則/r/nHX/r/n、/r/nHY/r/n、/r/nHZ/r/n三種一元弱酸的酸性依次減弱。/r/n【KS5U解析】/r/nA/r/n．/r/nNaX/r/n為強(qiáng)堿弱酸鹽，在溶液中水解使溶液呈堿性，則溶液中離子濃度的大小順序?yàn)?r/nc/r/n(Na/r/n+/r/n)/r/n＞/r/nc/r/n(X/r/n-/r/n)/r/n＞/r/nc/r/n(OH/r/n-/r/n)/r/n＞/r/nc/r/n(H/r/n+/r/n)/r/n，故/r/nA/r/n正確；/r/nB/r/n．弱酸的酸性越弱，電離常數(shù)越小，由分析可知，/r/nHX/r/n、/r/nHY/r/n、/r/nHZ/r/n三種一元弱酸的酸性依次減弱，則三種一元弱酸的電離常數(shù)的大小順序?yàn)?r/nK/r/na/r/n(HX)/r/n＞/r/nK/r/na/r/n(HY)/r/n＞/r/nK/r/na/r/n(HZ)/r/n，故/r/nB/r/n正確；/r/nC/r/n．當(dāng)溶液/r/npH/r/n為/r/n7/r/n時(shí)，酸越弱，向鹽溶液中加入鹽酸的體積越大，酸根離子的濃度越小，則三種鹽溶液中酸根的濃度大小順序?yàn)?r/nc/r/n(X/r/n-/r/n)/r/n＞/r/nc/r/n(Y/r/n-/r/n)/r/n＞/r/nc/r/n(Z/r/n-/r/n)/r/n，故/r/nC/r/n錯(cuò)誤；/r/nD/r/n．向三種鹽溶液中分別滴加/r/n20.00mL/r/n鹽酸，三種鹽都完全反應(yīng)，溶液中鈉離子濃度等于氯離子濃度，將三種溶液混合后溶液中存在電荷守恒關(guān)系/r/nc/r/n(Na/r/n+/r/n)+/r/nc/r/n(H/r/n+/r/n)=/r/nc/r/n(X/r/n-/r/n)+/r/nc/r/n(Y/r/n-/r/n)+/r/nc/r/n(Z/r/n-/r/n)+/r/nc/r/n(Cl/r/n-/r/n)+/r/nc/r/n(OH/r/n-/r/n)/r/n，由/r/nc/r/n(Na/r/n+/r/n)=/r/nc/r/n(Cl/r/n-/r/n)/r/n可得：/r/nc/r/n(X/r/n-/r/n)+/r/nc/r/n(Y/r/n-/r/n)+/r/nc/r/n(Z/r/n-/r/n)=/r/nc/r/n(H/r/n+/r/n)—/r/nc/r/n(OH/r/n-/r/n)/r/n，故/r/nD/r/n正確；/r/n故選/r/nC/r/n。/r/n5．（浙江高考真題試卷）/r/n實(shí)驗(yàn)測(cè)得/r/n10mL0.50mol·L/r/n-1/r/nNH/r/n4/r/nCl/r/n溶液、/r/n10mL0.50mol·L/r/n-1/r/nCH/r/n3/r/nCOONa/r/n溶液的/r/npH/r/n分別隨溫度與稀釋加水量的變化如圖所示。已知/r/n25/r/n℃時(shí)/r/nCH/r/n3/r/nCOOH/r/n和/r/nNH/r/n3/r/n·H/r/n2/r/nO/r/n的電離常數(shù)均為/r/n1.8×10/r/n-5./r/n下列說(shuō)法/r/n不正確/r/n的是/r/nA．/r/n圖中/r/n實(shí)線/r/n表示/r/npH/r/n隨加水量的變化，/r/n虛線/r/n表示/r/npH/r/n隨溫度的變化/r/n'/r/nB．/r/n將/r/nNH/r/n4/r/nCl/r/n溶液加水稀釋至濃度/r/nmol·L/r/n-1/r/n，溶液/r/npH/r/n變化值小于/r/nlgx/r/nC．/r/n隨溫度升高，/r/nK/r/nw/r/n增大，/r/nCH/r/n3/r/nCOONa/r/n溶液中/r/nc(OH/r/n-/r/n)/r/n減小，/r/nc/r/n(H/r/n+/r/n)/r/n增大，/r/npH/r/n減小/r/nD．25/r/n℃時(shí)稀釋相同倍數(shù)的/r/nNH/r/n4/r/nCl/r/n溶液與/r/nCH/r/n3/r/nCOONa/r/n溶液中：/r/nc/r/n(Na/r/n+/r/n)-/r/nc/r/n(CH/r/n3/r/nCOO/r/n-/r/n)=/r/nc/r/n(Cl/r/n-/r/n)-/r/nc/r/n(NH/r/n)/r/n【KS5U答案】/r/nC/r/n【分析】/r/n由題中信息可知，圖中兩條曲線為/r/n10mL0.50mol·L/r/n-1/r/nNH/r/n4/r/nCl/r/n溶液、/r/n10mL0.50mol·L/r/n-1/r/nCH/r/n3/r/nCOONa/r/n溶液的/r/npH/r/n分別隨溫度與稀釋加水量的變化曲線，由于兩種鹽均能水解，水解反應(yīng)為吸熱過(guò)程，且溫度越高、濃度越小其水解程度越大。氯化銨水解能使溶液呈酸性，濃度越小，雖然水程度越大，但其溶液的酸性越弱，故其/r/npH/r/n越大；醋酸鈉水解能使溶液呈堿性，濃度越小，其水溶液的堿性越弱，故其/r/npH/r/n越小。溫度越高，水的電離度越大。因此，圖中的實(shí)線為/r/npH/r/n隨加水量的變化，虛線表示/r/npH/r/n隨溫度的變化。/r/n【KS5U解析】/r/nA/r/n．由分析可知，圖中實(shí)線表示/r/npH/r/n隨加水量的變化，虛線表示/r/npH/r/n隨溫度的變化，/r/nA/r/n說(shuō)法正確；/r/nB/r/n．將/r/nNH/r/n4/r/nCl/r/n溶液加水稀釋至濃度/r/nmol·L/r/n-1/r/n時(shí)，若氯化銨的水解平衡不發(fā)生移動(dòng)，則其中的/r/nc/r/n(H/r/n+/r/n)/r/n變?yōu)樵瓉?lái)的/r/n，則溶液的/r/npH/r/n將增大/r/nlgx/r/n，但是，加水稀釋時(shí)，氯化銨的水解平衡向正反應(yīng)方向移動(dòng)，/r/nc/r/n(H/r/n+/r/n)/r/n大于原來(lái)的/r/n，因此，溶液/r/npH/r/n的變化值小于/r/nlgx/r/n，/r/nB/r/n說(shuō)法正確；/r/nC/r/n．隨溫度升高，水的電離程度變大，因此水的離子積變大，即/r/nK/r/nw/r/n增大；隨溫度升高，/r/nCH/r/n3/r/nCOONa/r/n的水解程度變大，溶液中/r/nc/r/n(OH/r/n-/r/n)/r/n增大，因此，/r/nC/r/n說(shuō)法不正確；/r/nD/r/n．/r/n25℃/r/n時(shí)稀釋相同倍數(shù)的/r/nNH/r/n4/r/nC1/r/n溶液與/r/nCH/r/n3/r/nCOONa/r/n溶液中均分別存在電荷守恒，/r/nc/r/n(Na/r/n+/r/n)+/r/nc/r/n(H/r/n+/r/n)=/r/nc/r/n(OH/r/n-/r/n)+/r/nc/r/n(CH/r/n3/r/nCOO/r/n-/r/n)/r/n，/r/nc/r/n(NH/r/n4/r/n+/r/n)+/r/nc/r/n(H/r/n+/r/n)=/r/nc/r/n(Cl/r/n-/r/n)+/r/nc/r/n(OH/r/n-/r/n)/r/n。因此，氯化銨溶液中，/r/nc/r/n(Cl/r/n-/r/n)-/r/nc/r/n(NH/r/n4/r/n+/r/n)=/r/nc/r/n(H/r/n+/r/n)-/r/nc/r/n(OH/r/n-/r/n)/r/n，醋酸鈉溶液中，/r/nc/r/n(Na/r/n+/r/n)-/r/nc/r/n(CH/r/n3/r/nCOO/r/n-/r/n)=/r/nc/r/n(OH/r/n-/r/n)-/r/nc/r/n(H/r/n+/r/n)/r/n。由于/r/n25℃/r/n時(shí)/r/nCH/r/n3/r/nCOOH/r/n和/r/nNH/r/n3/r/n·/r/nH/r/n2/r/nO/r/n的電離常數(shù)均為/r/n1.8/r/n×/r/n10/r/n-5/r/n，因此，由于原溶液的物質(zhì)的量濃度相同，稀釋相同倍數(shù)后的/r/nNH/r/n4/r/nC1/r/n溶液與/r/nCH/r/n3/r/nCOONa/r/n溶液，溶質(zhì)的物質(zhì)的量濃度仍相等，由于電離常數(shù)相同，其中鹽的水解程度是相同的，因此，兩溶液中/r/n/r/nc/r/n(OH/r/n-/r/n)-/r/nc/r/n(H/r/n+/r/n)/r/n/r/n（兩者差的絕對(duì)值）相等，故/r/nc/r/n(Na/r/n+/r/n)-/r/nc/r/n(CH/r/n3/r/nCOO/r/n-/r/n)=/r/nc/r/n(Cl/r/n-/r/n)-/r/nc/r/n(NH/r/n4/r/n+/r/n)/r/n，/r/nD/r/n說(shuō)法正確。/r/n綜上所述，本題選/r/nC/r/n。/r/n6．（浙江高考真題試卷）25/r/n℃時(shí)，下列說(shuō)法正確的是/r/nA．NaHA/r/n溶液呈酸性，可以推測(cè)/r/nH/r/n2/r/nA/r/n為強(qiáng)酸/r/nB．/r/n可溶性正鹽/r/nBA/r/n溶液呈中性，可以推測(cè)/r/nBA/r/n為強(qiáng)酸強(qiáng)堿鹽/r/nC．0/r/n./r/n010/r/n/r/nmol/r/n·/r/nL/r/n-/r/n1/r/n、/r/n0/r/n./r/n10mol/r/n·/r/nL/r/n-/r/n1/r/n的醋酸溶液的電離度分別為/r/nα/r/n1/r/n、/r/nα/r/n2/r/n，則/r/nα/r/n1/r/n＜/r/nα/r/n2/r/nD．100/r/n/r/nmL/r/n/r/npH/r/n=/r/n10/r/n./r/n00/r/n的/r/nNa/r/n2/r/nCO/r/n3/r/n溶液中水電離出/r/nH/r/n+/r/n的物質(zhì)的量為/r/n1/r/n./r/n0/r/n×/r/n10/r/n-/r/n5/r/nmol/r/n【KS5U答案】/r/nD/r/n【KS5U解析】/r/nA/r/n．/r/nNaHA/r/n溶液呈酸性，可能是/r/nHA/r/n－/r/n的電離程度大于其水解程度，不能據(jù)此得出/r/nH/r/n2/r/nA/r/n為強(qiáng)酸的結(jié)論，/r/nA/r/n錯(cuò)誤；/r/nB/r/n．可溶性正鹽/r/nBA/r/n溶液呈中性，不能推測(cè)/r/nBA/r/n為強(qiáng)酸強(qiáng)堿鹽，因?yàn)橐部赡苁?r/nB/r/n＋/r/n和/r/nA/r/n－/r/n的水解程度相同，即也可能是弱酸弱堿鹽，/r/nB/r/n錯(cuò)誤；/r/nC/r/n．弱酸的濃度越小，其電離程度越大，因此/r/n0/r/n./r/n010/r/n/r/nmol/r/n·/r/nL/r/n-/r/n1/r/n、/r/n0/r/n./r/n10/r/n/r/nmol/r/n·/r/nL/r/n-/r/n1/r/n的醋酸溶液的電離度分別為/r/nα/r/n1/r/n、/r/nα/r/n2/r/n，則/r/nα/r/n1/r/n＞/r/nα/r/n2/r/n，/r/nC/r/n錯(cuò)誤；/r/nD/r/n．/r/n100/r/n/r/nmL/r/n/r/npH/r/n=/r/n10/r/n./r/n00/r/n的/r/nNa/r/n2/r/nCO/r/n3/r/n溶液中氫氧根離子的濃度是/r/n1/r/n×/r/n10/r/n－/r/n4/r/nmol/r/n//r/nL/r/n，碳酸根水解促進(jìn)水的電離，則水電離出/r/nH/r/n＋/r/n的濃度是/r/n1/r/n×/r/n10/r/n－/r/n4/r/nmol/r/n//r/nL/r/n，其物質(zhì)的量為/r/n0/r/n./r/n1L/r/n×/r/n1/r/n×/r/n10/r/n－/r/n4/r/nmol/r/n//r/nL/r/n＝/r/n1/r/n×/r/n10/r/n－/r/n5/r/nmol/r/n，/r/nD/r/n正確；/r/n答案選/r/nD/r/n。/r/n二、多選題/r/n7．（山東高考真題試卷）/r/n賴(lài)氨酸/r/n[H/r/n3/r/nN/r/n+/r/n(CH/r/n2/r/n)/r/n4/r/nCH(NH/r/n2/r/n)COO/r/n-/r/n，用/r/nHR/r/n表示/r/n]/r/n是人體必需氨基酸，其鹽酸鹽/r/n(H/r/n3/r/nRCl/r/n2/r/n)/r/n在水溶液中存在如下平衡：/r/nH/r/n3/r/nR/r/n2+/r/nH/r/n2/r/nR/r/n+/r/nHR/r/nR/r/n-/r/n。向一定濃度的/r/nH/r/n3/r/nRCl/r/n2/r/n溶液中滴加/r/nNaOH/r/n溶液，溶液中/r/nH/r/n3/r/nR/r/n2+/r/n、/r/nH/r/n2/r/nR/r/n+/r/n、/r/nHR/r/n和/r/nR/r/n-/r/n的分布系數(shù)/r/nδ(x)/r/n隨/r/npH/r/n變化如圖所示。已知/r/nδ(x)=/r/n，下列表述正確的是/r/n

/r/nA．/r/n>/r/nB．M/r/n點(diǎn)，/r/nc/r/n(Cl/r/n-/r/n)+/r/nc/r/n(OH/r/n-/r/n)+/r/nc/r/n(R/r/n-/r/n)=2/r/nc/r/n(H/r/n2/r/nR/r/n+/r/n)+/r/nc/r/n(Na/r/n+/r/n)+/r/nc/r/n(H/r/n+/r/n)/r/nC．O/r/n點(diǎn)，/r/npH=/r/nD．P/r/n點(diǎn)，/r/nc/r/n(Na/r/n+/r/n)>/r/nc/r/n(Cl/r/n-/r/n)>/r/nc/r/n(OH/r/n-/r/n)>/r/nc/r/n(H/r/n+/r/n)/r/n【KS5U答案】/r/nCD/r/n【分析】/r/n向/r/nH/r/n3/r/nRCl/r/n2/r/n溶液中滴加/r/nNaOH/r/n溶液，依次發(fā)生離子反應(yīng)：/r/n、/r/n、/r/n，溶液中/r/n逐漸減小，/r/n和/r/n先增大后減小，/r/n逐漸增大。/r/n，/r/n，/r/n，/r/nM/r/n點(diǎn)/r/n，由此可知/r/n，/r/nN/r/n點(diǎn)/r/n，則/r/n，/r/nP/r/n點(diǎn)/r/n，則/r/n。/r/n【KS5U解析】/r/nA/r/n．/r/n，/r/n，因此/r/n，故/r/nA/r/n錯(cuò)誤；/r/nB/r/n．/r/nM/r/n點(diǎn)存在電荷守恒：/r/n，此時(shí)/r/n，因此/r/n，故/r/nB/r/n錯(cuò)誤；/r/nC/r/n．/r/nO/r/n點(diǎn)/r/n，因此/r/n，即/r/n，因此/r/n，溶液/r/n，故/r/nC/r/n正確；/r/nD/r/n．/r/nP/r/n點(diǎn)溶質(zhì)為/r/nNaCl/r/n、/r/nHR/r/n、/r/nNaR/r/n，此時(shí)溶液呈堿性，因此/r/n，溶質(zhì)濃度大于水/r/n解和電離所產(chǎn)生微粒濃度，因此/r/n，故/r/nD/r/n正確；/r/n綜上所述，正確的是/r/nCD/r/n，故答案為/r/nCD/r/n。/r/n三、工業(yè)流程題/r/n8．（湖南高考真題試卷）/r/n可用于催化劑載體及功能材料的制備。天然獨(dú)居石中，鈰/r/n(Ce)/r/n主要以/r/n形式存在，還含有/r/n、/r/n、/r/n、/r/n等物質(zhì)。以獨(dú)居石為原料制備/r/n的工藝流程如下：/r/n回答下列問(wèn)題：/r/n(1)/r/n鈰的某種核素含有/r/n58/r/n個(gè)質(zhì)子和/r/n80/r/n個(gè)中子，該核素的符號(hào)為/r/n_______/r/n；/r/n(2)/r/n為提高/r/n“/r/n水浸/r/n”/r/n效率，可采取的措施有/r/n_______(/r/n至少寫(xiě)兩條/r/n)/r/n；/r/n(3)/r/n濾渣Ⅲ的主要成分是/r/n_______(/r/n填化學(xué)式/r/n)/r/n；/r/n(4)/r/n加入絮凝劑的目的是/r/n_______/r/n；/r/n(5)“/r/n沉鈰/r/n”/r/n過(guò)程中，生成/r/n的離子方程式為/r/n_______/r/n，常溫下加入的/r/n溶液呈/r/n_______(/r/n填/r/n“/r/n酸性/r/n”“/r/n堿性/r/n”/r/n或/r/n“/r/n中性/r/n”)(/r/n已知：/r/n的/r/n，/r/n的/r/n，/r/n)/r/n；/r/n(6)/r/n濾渣Ⅱ的主要成分為/r/n，在高溫條件下，/r/n、葡萄糖/r/n(/r/n)/r/n和/r/n可制備電極材料/r/n，同時(shí)生成/r/n和/r/n，該反應(yīng)的化學(xué)方程式為/r/n_______/r/n【KS5U答案】/r/n/r/n適當(dāng)升高溫度，將獨(dú)居石粉碎等/r/nAl(OH)/r/n3/r/n/r/n促使鋁離子沉淀/r/n/r/n↑/r/n堿性/r/n6/r/n+/r/n+12/r/n=12/r/n+6CO↑+6H/r/n2/r/nO+6CO/r/n2/r/n↑/r/n【分析】/r/n焙燒濃硫酸和獨(dú)居石的混合物、水浸，/r/n轉(zhuǎn)化為/r/nCe/r/n2/r/n(SO/r/n4/r/n)/r/n3/r/n和/r/nH/r/n3/r/nPO/r/n4/r/n，/r/n與硫酸不反應(yīng)，/r/n轉(zhuǎn)化為/r/nAl/r/n2/r/n(SO/r/n4/r/n)/r/n3/r/n，/r/n轉(zhuǎn)化為/r/nFe/r/n2/r/n(SO/r/n4/r/n)/r/n3/r/n，/r/n轉(zhuǎn)化為/r/nCaSO/r/n4/r/n和/r/nHF/r/n，酸性廢氣含/r/nHF/r/n；后過(guò)濾，濾渣Ⅰ為/r/n和磷酸鈣、/r/nFePO/r/n4/r/n，濾液主要含/r/nH/r/n3/r/nPO/r/n4/r/n，/r/nCe/r/n2/r/n(SO/r/n4/r/n)/r/n3/r/n，/r/nAl/r/n2/r/n(SO/r/n4/r/n)/r/n3/r/n，/r/nFe/r/n2/r/n(SO/r/n4/r/n)/r/n3/r/n，加氯化鐵溶液除磷，濾渣Ⅱ?yàn)?r/nFePO/r/n4/r/n；聚沉將鐵離子、鋁離子轉(zhuǎn)化為沉淀，過(guò)濾除去，濾渣Ⅲ主要為氫氧化鋁，還含氫氧化鐵；加碳酸氫銨沉鈰得/r/nCe/r/n2/r/n(CO/r/n3/r/n)/r/n3/r/n·nH/r/n2/r/nO/r/n。/r/n【KS5U解析】/r/n(1)/r/n鈰的某種核素含有/r/n58/r/n個(gè)質(zhì)子和/r/n80/r/n個(gè)中子，則質(zhì)量數(shù)為/r/n58+80=138/r/n，該核素的符號(hào)為/r/n；/r/n(2)/r/n為提高/r/n“/r/n水浸/r/n”/r/n效率，可采取的措施有適當(dāng)升高溫度，將獨(dú)居石粉碎等；/r/n(3)/r/n結(jié)合流程可知，濾渣Ⅲ的主要成分是/r/nAl(OH)/r/n3/r/n；/r/n(4)/r/n加入絮凝劑的目的是促使鋁離子沉淀；/r/n(5)/r/n用碳酸氫銨/r/n“/r/n沉鈰/r/n”/r/n，則結(jié)合原子守恒、電荷守恒可知生成/r/n的離子方程式為/r/n↑/r/n；銨根離子的水解常數(shù)/r/nK/r/nh/r/n(/r/n)=/r/n≈5.7×10/r/n-10/r/n，碳酸氫根的水解常數(shù)/r/nK/r/nh/r/n(/r/n)==/r/n≈2.3×10/r/n-8/r/n，則/r/nK/r/nh/r/n(/r/n)<K/r/nh/r/n(/r/n)/r/n，因此常溫下加入的/r/n溶液呈堿性；/r/n(6)/r/n由在高溫條件下，/r/n、葡萄糖/r/n(/r/n)/r/n和/r/n可制備電極材料/r/n，同時(shí)生成/r/n和/r/n可知，該反應(yīng)中/r/nFe/r/n價(jià)態(tài)降低，/r/nC/r/n價(jià)態(tài)升高，結(jié)合得失電子守恒、原子守恒可知該反應(yīng)的化學(xué)方程式為/r/n6/r/n+/r/n+12/r/n=12/r/n+6CO↑+6H/r/n2/r/nO+6CO/r/n2/r/n↑/r/n。/r/n2021年化學(xué)高考模擬題/r/n一、單選題/r/n1．（九龍坡區(qū)·重慶市育才中學(xué)高三三模）/r/n室溫下，用/r/nmmol/L/r/n的二甲胺/r/n[(CH/r/n3/r/n)/r/n2/r/nNH]/r/n溶/r/n液/r/n(/r/n二甲胺在水中的電離與一水合氨相似/r/n)/r/n滴定/r/n10.00mL0.1mol/L/r/n的鹽酸溶液。溶液/r/npH/r/n隨加入二甲胺溶液體積變化曲線如圖所示/r/n(/r/n忽略溶液混合時(shí)的體積變化/r/n)/r/n。下列說(shuō)法正確的是/r/nA．/r/n本實(shí)驗(yàn)應(yīng)該選擇酚酞作指示劑/r/nB．/r/n室溫下，/r/nC．a(chǎn)/r/n點(diǎn)溶液中水的電離程度最大/r/nD．b/r/n點(diǎn)溶液中存在：/r/nc/r/n[(CH/r/n3/r/n)/r/n2/r/nNH/r/n]/r/n＞/r/nc/r/n(OH/r/n-/r/n)/r/n＞/r/nc/r/n(Cl/r/n-/r/n)/r/n＞/r/nc/r/n(H/r/n＋/r/n)/r/n【KS5U答案】/r/nB/r/n【分析】/r/n二甲胺/r/n[(CH/r/n3/r/n)/r/n2/r/nNH]/r/n為一元弱堿，二甲胺在水中的電離與一水合氨相似，二甲胺/r/n[(CH/r/n3/r/n)/r/n2/r/nNH]/r/n電離方程式為/r/n(CH/r/n3/r/n)/r/n2/r/nNH?H/r/n2/r/nO/r/n?/r/n(CH/r/n3/r/n)/r/n2/r/nNH/r/n+OH/r/n-/r/n，電離平衡常數(shù)/r/nK/r/nb/r/n[(CH/r/n3/r/n)/r/n2/r/nNH?H/r/n2/r/nO]=/r/n，/r/n(CH/r/n3/r/n)/r/n2/r/nNH/r/n2/r/nCl/r/n是強(qiáng)酸弱堿鹽，其水溶液呈酸性，結(jié)合電荷守恒關(guān)系分析解答。/r/n【KS5U解析】/r/nA/r/n．用二甲胺溶液滴定鹽酸溶液達(dá)到滴定終點(diǎn)時(shí)生成強(qiáng)酸弱堿鹽/r/n(CH/r/n3/r/n)/r/n2/r/nNH/r/n2/r/nCl/r/n，溶液呈酸性，應(yīng)該選擇甲基橙作指示劑，故/r/nA/r/n錯(cuò)誤；/r/nB/r/n．/r/n(CH/r/n3/r/n)/r/n2/r/nNH/r/n2/r/nCl/r/n是強(qiáng)酸弱堿鹽，其水溶液呈酸性，電荷關(guān)系為/r/nc/r/n[(CH/r/n3/r/n)/r/n2/r/nNH/r/n]+/r/nc/r/n(H/r/n+/r/n)=/r/nc/r/n(OH/r/n-/r/n)+/r/nc/r/n(Cl/r/n-/r/n)/r/n，圖中/r/na/r/n點(diǎn)時(shí)溶液呈中性，/r/nc/r/n(OH/r/n-/r/n)=/r/nc/r/n(H/r/n+/r/n)=10/r/n-7/r/nmol/L/r/n、并且二甲胺溶液過(guò)量、/r/nc/r/n[(CH/r/n3/r/n)/r/n2/r/nNH/r/n]=/r/nc/r/n(Cl/r/n-/r/n)=/r/n=0.05mol/L/r/n，剩余/r/n(CH/r/n3/r/n)/r/n2/r/nNH/r/n的濃度/r/nc/r/n[(CH/r/n3/r/n)/r/n2/r/nNH?H/r/n2/r/nO]=/r/n=0.5(m-0.1)mol/L/r/n，電離方程式為/r/n(CH/r/n3/r/n)/r/n2/r/nNH?H/r/n2/r/nO/r/n?/r/n(CH/r/n3/r/n)/r/n2/r/nNH/r/n+OH/r/n-/r/n，則電離平衡常數(shù)/r/nK/r/nb/r/n[(CH/r/n3/r/n)/r/n2/r/nNH?H/r/n2/r/nO]=/r/n=/r/n=/r/n×10/r/n-7/r/n，故/r/nB/r/n正確；/r/nC/r/n．強(qiáng)酸弱堿鹽能夠促進(jìn)水的電離，并且其濃度越大、促進(jìn)作用越強(qiáng)，二者恰好完全反應(yīng)時(shí)生成/r/n(CH/r/n3/r/n)/r/n2/r/nNH/r/n2/r/nCl/r/n，此時(shí)溶液呈酸性，圖中/r/na/r/n點(diǎn)呈中性、二甲胺過(guò)量，所以/r/na/r/n點(diǎn)以前的某點(diǎn)溶液中水的電離程度最大，故/r/nC/r/n錯(cuò)誤；/r/nD/r/n．圖中/r/nb/r/n點(diǎn)溶液的/r/npH=8/r/n、呈堿性，/r/nc/r/n(OH/r/n-/r/n)/r/n＞/r/nc/r/n(H/r/n+/r/n)/r/n，溶液中主要離子為/r/n(CH/r/n3/r/n)/r/n2/r/nNH/r/n和/r/nCl/r/n-/r/n，電荷關(guān)系為/r/nc/r/n[(CH/r/n3/r/n)/r/n2/r/nNH/r/n]+/r/nc/r/n(H/r/n+/r/n)=/r/nc/r/n(OH/r/n-/r/n)+/r/nc/r/n(Cl/r/n-/r/n)/r/n，所以有/r/nc/r/n[(CH/r/n3/r/n)/r/n2/r/nNH/r/n]/r/n＞/r/nc/r/n(Cl/r/n-/r/n)/r/n＞/r/nc/r/n(OH/r/n-/r/n)/r/n＞/r/nc/r/n(H/r/n+/r/n)/r/n，故/r/nD/r/n錯(cuò)誤；/r/n故選/r/nB/r/n。/r/n2．（福建省南安高三二模）/r/n亞砷酸/r/n(H/r/n3/r/nAsO/r/n3/r/n)/r/n可以用于治療白血病，其在溶液中存在多種微粒形態(tài)，將/r/nKOH/r/n溶液滴入亞砷酸溶液，各種微粒物質(zhì)的量分?jǐn)?shù)與溶液的/r/npH/r/n關(guān)系如圖所示。下列說(shuō)法不正確的是/r/nA．/r/n人體血液的/r/npH/r/n在/r/n7.35-7.45/r/n之間，患者用藥后人體中含/r/nAs/r/n元素的主要微粒是/r/nH/r/n3/r/nAsO/r/n3/r/nB．pH/r/n在/r/n10～13/r/n之間，隨/r/npH/r/n增大/r/nHAsO/r/n水解程度減小/r/nC．/r/n通常情況下，/r/nH/r/n2/r/nAsO/r/n電離程度大于水解程度/r/nD．/r/n交點(diǎn)/r/nb/r/n的溶液中：/r/n2/r/nc/r/n(HAsO/r/n)+4/r/nc/r/n(AsO/r/n)</r/nc/r/n(K/r/n+/r/n)/r/n【KS5U答案】/r/nC/r/n【KS5U解析】/r/nA/r/n．由圖可知，當(dāng)溶液/r/npH/r/n在/r/n7.35-7.45/r/n之間時(shí)，砷元素的主要存在形式為/r/nH/r/n3/r/nAsO/r/n3/r/n，故/r/nA/r/n正確；/r/nB/r/n．/r/nHAsO/r/n在溶液中存在如下水解平衡：/r/nHAsO/r/n+H/r/n2/r/nO/r/nH/r/n2/r/nAsO/r/n+OH/r/n—/r/n，/r/npH/r/n在/r/n10～13/r/n之間時(shí)，溶液/r/npH/r/n增大，氫氧根離子濃度增大，平衡向逆反應(yīng)方向移動(dòng)，/r/nHAsO/r/n水解程度減小，故/r/nB/r/n正確；/r/nC/r/n．由圖可知，當(dāng)溶液中砷元素的主要存在形式為/r/nH/r/n2/r/nAsO/r/n時(shí)，溶液/r/npH/r/n約為/r/n11/r/n，溶液呈堿性，說(shuō)明/r/nH/r/n2/r/nAsO/r/n的水解程度大于電離程度，故/r/nC/r/n錯(cuò)誤；/r/nD/r/n．由圖可知，交點(diǎn)/r/nb/r/n的溶液為堿性溶液，溶液中/r/nc/r/n(H/r/n+/r/n)/r/n/r/n</r/nc/r/n(OH/r/n—/r/n)/r/n、/r/nc/r/n(AsO/r/n)=/r/nc/r/n(H/r/n2/r/nAsO/r/n)/r/n，由電荷守恒關(guān)系/r/nc/r/n(K/r/n+/r/n)+/r/nc/r/n(H/r/n+/r/n)=3/r/nc/r/n(AsO/r/n)+2/r/nc/r/n(HAsO/r/n)+/r/nc/r/n(H/r/n2/r/nAsO/r/n)+/r/nc/r/n(OH/r/n—/r/n)/r/n可得/r/n2/r/nc/r/n(HAsO/r/n)+4/r/nc/r/n(AsO/r/n)</r/nc/r/n(K/r/n+/r/n)/r/n，故/r/nD/r/n正確；/r/n故選/r/nC/r/n。/r/n3．（福建省南安高三二模）/r/n常溫下，下列說(shuō)法正確的是/r/nA．/r/n某溶液中含有/r/n、/r/n、/r/n和/r/nNa/r/n+/r/n，若向其中加入/r/nNa/r/n2/r/nO/r/n2/r/n，充分反應(yīng)后，四種離子的濃度不變的是/r/n(/r/n忽略反應(yīng)前后溶液體積的變化/r/n)/r/nB．/r/n水電離的/r/nc/r/n水/r/n(H/r/n+/r/n)=10/r/n-12/r/nmol·L/r/n—/r/n1/r/n的溶液中，下列離子能大量共存：/r/n、/r/nNa/r/n+/r/n、/r/n、/r/nC．/r/n氫氧化鐵溶于/r/nHI/r/n溶液中的離子方程式為：/r/n2Fe(OH)/r/n3/r/n+6H/r/n+/r/n+2I/r/n—/r/n=2Fe/r/n2+/r/n+I/r/n2/r/n+6H/r/n2/r/nO/r/nD．NaHS/r/n溶液中，下列離子能大量共存：/r/nK/r/n+/r/n、/r/nAl/r/n3+/r/n、/r/nCl/r/n—/r/n、/r/n【KS5U答案】/r/nC/r/n【KS5U解析】/r/nA/r/n．向溶液中加入具有強(qiáng)氧化性的過(guò)氧化鈉固體，過(guò)氧化鈉能將溶液中的亞硫酸根離子氧化為硫酸根離子，溶液中硫酸根離子濃度會(huì)增大，故/r/nA/r/n錯(cuò)誤；/r/nB/r/n．水電離的/r/nc/r/n水/r/n(H/r/n+/r/n)=10/r/n-12/r/nmol·L/r/n—/r/n1/r/n的溶液可能為酸溶液，也可能為堿溶液，酸溶液中，氫離子與碳酸氫根離子反應(yīng)，不能大量共存，堿溶液中，氫氧根離子與銨根離子、碳酸氫根離子反應(yīng)，不能大量共存，故/r/nB/r/n錯(cuò)誤；/r/nC/r/n．氫氧化鐵溶于氫碘酸溶液的反應(yīng)為氫氧化鐵與氫碘酸溶液反應(yīng)生成碘化亞鐵、碘和水，反應(yīng)的離子方程式為/r/n2Fe(OH)/r/n3/r/n+6H/r/n+/r/n+2I/r/n—/r/n=2Fe/r/n2+/r/n+I/r/n2/r/n+6H/r/n2/r/nO/r/n，故/r/nC/r/n正確；/r/nD/r/n．在硫氫化鈉溶液中，硫氫根離子與鋁離子會(huì)發(fā)生雙水解反應(yīng)生成氫氧化鋁沉淀和硫化氫氣體，不能大量共存，故/r/nD/r/n錯(cuò)誤；/r/n故選/r/nC/r/n。/r/n4．（重慶市第十一中學(xué)校高三二模）/r/n常溫下，向/r/n20mL/r/n濃度均為/r/n0.1mol/LHX/r/n和/r/nCH/r/n3/r/nCOOH/r/n的混合溶液中滴加/r/n0.1mol/L/r/n的氨水，測(cè)得溶液的電阻率/r/n(/r/n溶液的電阻率越大，導(dǎo)電能力越弱/r/n)/r/n與加入氨水的體積/r/n(V)/r/n的關(guān)系如圖。/r/n(CH/r/n3/r/nCOOH/r/n的/r/nK/r/na/r/n＝/r/n1.8×10/r/n-5/r/n，/r/nNH/r/n3/r/n·H/r/n2/r/nO/r/n的/r/nK/r/nb/r/n＝/r/n1.8×10/r/n-5/r/n)/r/n下列說(shuō)法正確的是/r/nA．/r/n同濃度的/r/nHX/r/n比/r/nCH/r/n3/r/nCOOH/r/n的/r/npH/r/n大/r/nB．a(chǎn)→c/r/n過(guò)程，水的電離程度逐漸減小/r/nC．c/r/n點(diǎn)時(shí)，/r/nD．d/r/n點(diǎn)時(shí)，/r/n【KS5U答案】/r/nC/r/n【分析】/r/n電阻率與離子濃度成反比，即/r/na→b/r/n過(guò)程中溶液的導(dǎo)電性減弱，向混合溶液中加入等物質(zhì)的量濃度的/r/nNH/r/n3/r/n?H/r/n2/r/nO/r/n溶液時(shí)，發(fā)生反應(yīng)先后順序?yàn)椋?r/nHX+NH/r/n3/r/n?H/r/n2/r/nO=NH/r/n4/r/nX+H/r/n2/r/nO/r/n、/r/nNH/r/n3/r/n?H/r/n2/r/nO+CH/r/n3/r/nCOOH=CH/r/n3/r/nCOONH/r/n4/r/n+H/r/n2/r/nO/r/n，/r/n0/r/n～/r/n20mL/r/n溶液中電阻率增大、導(dǎo)電性減弱，/r/nb/r/n點(diǎn)最小，原因是溶液體積增大導(dǎo)致/r/nb/r/n點(diǎn)離子濃度減小，/r/nb/r/n點(diǎn)溶液中溶質(zhì)為/r/nNH/r/n4/r/nX/r/n、/r/nCH/r/n3/r/nCOOH/r/n；繼續(xù)加入/r/nNH/r/n3/r/n?H/r/n2/r/nO/r/n溶液，/r/nNH/r/n3/r/n?H/r/n2/r/nO/r/n是弱電解質(zhì)，生成的/r/nCH/r/n3/r/nCOONH/r/n4/r/n是強(qiáng)電解質(zhì)，導(dǎo)致溶液中離子濃度增大，溶液的電導(dǎo)性增大，/r/nc/r/n點(diǎn)時(shí)醋酸和一水合氨恰好完全反應(yīng)生成醋酸銨，/r/nc/r/n點(diǎn)溶液中溶質(zhì)為/r/nNH/r/n4/r/nX/r/n、/r/nCH/r/n3/r/nCOONH/r/n4/r/n，且二者的物質(zhì)的量相等；/r/nd/r/n點(diǎn)溶液中溶質(zhì)為等物質(zhì)的量濃度的/r/nNH/r/n4/r/nX/r/n、/r/nCH/r/n3/r/nCOONH/r/n4/r/n、/r/nNH/r/n3/r/n?H/r/n2/r/nO/r/n，據(jù)此分析解答。/r/n【KS5U解析】/r/nA/r/n．若/r/nHX/r/n和/r/nCH/r/n3/r/nCOOH/r/n都是弱酸，則隨著/r/nNH/r/n3/r/n?H/r/n2/r/nO/r/n的加入，酸堿反應(yīng)生成鹽，溶液導(dǎo)電性將增強(qiáng)、電阻率將減小，但圖象上隨著/r/nNH/r/n3/r/n?H/r/n2/r/nO/r/n的加入溶液電阻率增大、導(dǎo)電性反而減弱，說(shuō)明原混合溶液中離子濃度更大，即/r/nHX/r/n為強(qiáng)電解質(zhì)，同濃度的/r/nHX/r/n的/r/npH/r/n比/r/nCH/r/n3/r/nCOOH/r/n的/r/npH/r/n小，故/r/nA/r/n錯(cuò)誤；/r/n

B/r/n．酸或堿都抑制水的電離，滴加/r/nNH/r/n3/r/n?H/r/n2/r/nO/r/n溶液的過(guò)程：/r/na→c/r/n為/r/nHX/r/n和/r/nCH/r/n3/r/nCOOH/r/n轉(zhuǎn)化為/r/nNH/r/n4/r/nX/r/n、/r/nCH/r/n3/r/nCOONH/r/n4/r/n的過(guò)程，溶液的酸性減弱，水的電離程度增大，故/r/nB/r/n錯(cuò)誤；/r/n

/r/nC/r/n．根據(jù)分析可知，/r/nc/r/n點(diǎn)溶液中溶質(zhì)為/r/nNH/r/n4/r/nX/r/n、/r/nCH/r/n3/r/nCOONH/r/n4/r/n，且二者的物質(zhì)的量相等，溶液為酸性，則/r/nc(H/r/n+/r/n)/r/n＞/r/nc(OH/r/n-/r/n)/r/n，根據(jù)電荷守恒/r/nc(NH/r/n)+c(H/r/n+/r/n)=c(OH/r/n-/r/n)+c(X/r/n-/r/n)+c(CH/r/n3/r/nCOO/r/n-/r/n)/r/n可知：/r/n，故/r/nC/r/n正確；/r/n

D/r/n．/r/nd/r/n點(diǎn)溶液中溶質(zhì)為等物質(zhì)的量濃度的/r/nNH/r/n4/r/nX/r/n、/r/nCH/r/n3/r/nCOONH/r/n4/r/n、/r/nNH/r/n3/r/n?H/r/n2/r/nO/r/n，/r/n0/r/n～/r/n40mL/r/n時(shí)，/r/nNH/r/n3/r/n?H/r/n2/r/nO/r/n轉(zhuǎn)化為/r/nNH/r/n，/r/n40/r/n～/r/n60mL/r/n時(shí)，/r/nNH/r/n3/r/n?H/r/n2/r/nO/r/n過(guò)量，/r/nd/r/n點(diǎn)時(shí)，溶液體積共為/r/n80mL/r/n，/r/n2c(NH/r/n3/r/n?H/r/n2/r/nO)+2c(NH/r/n)=2×0.1mol/L×0.02L/0.08L+2×0.01mol/L×0.04L/0.08L=0.15mol/L/r/n，故/r/nD/r/n錯(cuò)誤；/r/n

/r/n故選：/r/nC/r/n。/r/n5．（青海高三三模）/r/n用下列實(shí)驗(yàn)裝置/r/n(/r/n部分夾持裝置略去/r/n)/r/n，能達(dá)到實(shí)驗(yàn)?zāi)康牡氖?r/nA．/r/n加熱裝置/r/nI/r/n中的燒杯，分離/r/nI/r/n2/r/n和高錳酸鉀固體/r/nB．/r/n利用裝置/r/nII/r/n除去/r/nCO/r/n中的/r/nCO/r/n2/r/nC．/r/n利用裝置/r/nIII/r/n制備/r/nFe(OH)/r/n3/r/n膠體/r/nD．/r/n利用裝置/r/nIV/r/n蒸干/r/nAlCl/r/n3/r/n溶液制無(wú)水/r/nAlCl/r/n3/r/n固體/r/n【KS5U答案】/r/nB/r/n【KS5U解析】/r/nA/r/n．/r/n/r/n加熱裝置/r/nI/r/n中的燒杯，高錳酸鉀固體會(huì)分解，故/r/nA/r/n錯(cuò)誤；/r/nB/r/n．/r/n/r/n利用裝置/r/nII/r/n除去/r/nCO/r/n中的/r/nCO/r/n2/r/n，只有二氧化碳能與/r/nNaOH/r/n溶液反應(yīng)，故/r/nB/r/n正確；/r/nC/r/n．/r/n/r/n利用裝置/r/nIII/r/n制備/r/nFe(OH)/r/n3/r/n膠體，應(yīng)使用酒精燈的外焰加熱，加熱至液體呈紅褐色即可，故/r/nC/r/n錯(cuò)誤；/r/nD/r/n．/r/n/r/n利用裝置/r/nIV/r/n蒸干/r/nAlCl/r/n3/r/n溶液制無(wú)水/r/nAlCl/r/n3/r/n固體，氯化鋁會(huì)水解，應(yīng)在氯化氫氣流中加熱，故/r/nD/r/n錯(cuò)誤；/r/n故選/r/nB/r/n。/r/n6．（四川成都市·成都七中高二零模）/r/n常溫下，向/r/n20mL0.1mol·L/r/n-1/r/nNa/r/n2/r/nCO/r/n3/r/n溶液中滴加/r/n0.1mol·L/r/n-1/r/nCaCl/r/n2/r/n溶液，碳酸根離子濃度與氯化鈣溶液體積的關(guān)系如圖所示。已知：/r/npC=-lg/r/nc(CO/r/n)/r/n，/r/nK/r/nsp/r/n(CdCO/r/n3/r/n)=1.0×10/r/n-12/r/n，/r/nK/r/nsp/r/n(CaCO/r/n3/r/n)=3.6×10/r/n-9./r/n下列說(shuō)法正確的是/r/n

/r/nA．/r/n圖像中/r/nV/r/n0/r/n=20/r/n，/r/nm=5/r/nB．a(chǎn)/r/n點(diǎn)溶液：/r/nc(OH/r/n-/r/n)>2c(HCO/r/n)+2c(H/r/n2/r/nCO/r/n3/r/n)/r/nC．/r/n若/r/nNa/r/n2/r/nCO/r/n3/r/n溶液的濃度變?yōu)?r/n0.05mol·L/r/n-1/r/n，則/r/nn/r/n點(diǎn)向/r/nc/r/n點(diǎn)方向遷移/r/nD．/r/n若用/r/nCdCl/r/n2/r/n溶液替代/r/nCaCl/r/n2/r/n溶液，則/r/nn/r/n點(diǎn)向/r/nb/r/n點(diǎn)方向遷移/r/n【KS5U答案】/r/nC/r/n【KS5U解析】/r/nA/r/n．/r/n/r/n圖像中/r/nV/r/n0/r/n=20/r/n，/r/nNa/r/n2/r/nCO/r/n3/r/n溶液與/r/n/r/nCaCl/r/n2/r/n溶液恰好完全反應(yīng)/r/nc(Ca/r/n2+/r/n)=c(CO/r/n)=/r/nmol/L/r/n，/r/npC=-lgc(CO/r/n)=-lg6×10/r/n-5/r/n，/r/nm/r/n不等于/r/n5/r/n，故/r/nA/r/n錯(cuò)誤；/r/nB/r/n．/r/na/r/n點(diǎn)溶液：溶質(zhì)為/r/nNa/r/n2/r/nCO/r/n3/r/n，存在物料守恒/r/nc(Na/r/n+/r/n)=2c(CO/r/n)+2c(HCO/r/n)+2c(H/r/n2/r/nCO/r/n3/r/n)/r/n，溶液中存在電荷守恒/r/nc(Na/r/n+/r/n)+c(H/r/n+/r/n)=2c(CO/r/n)+c(HCO/r/n)+c(OH/r/n-/r/n)/r/n，/r/n2c(HCO/r/n)+2c(H/r/n2/r/nCO/r/n3/r/n)=c(HCO/r/n)+c(OH/r/n-/r/n)-c(H/r/n+/r/n)/r/n，/r/nc(HCO/r/n)-c(H/r/n+/r/n)>0/r/n，/r/nc(OH/r/n-/r/n)<2c(HCO/r/n)+2c(H/r/n2/r/nCO/r/n3/r/n)/r/n，故/r/nB/r/n錯(cuò)誤；/r/nC/r/n．/r/n/r/n若/r/nNa/r/n2/r/nCO/r/n3/r/n溶液的濃度變?yōu)?r/n0.05mol·L/r/n-1/r/n，用的/r/n0.1mol·L/r/n-1/r/nCaCl/r/n2/r/n溶液體積減小，則/r/nn/r/n點(diǎn)向/r/nc/r/n點(diǎn)方向遷移，故/r/nC/r/n正確；/r/nD/r/n．/r/n/r/n若用/r/nCdCl/r/n2/r/n溶液替代/r/nCaCl/r/n2/r/n溶液，反應(yīng)后，碳酸根離子濃度減小，/r/npC=-lgc(CO/r/n)/r/n增大，則/r/nn/r/n點(diǎn)向/r/nd/r/n點(diǎn)方向遷移，故/r/nD/r/n錯(cuò)誤；/r/n故選/r/nC/r/n。/r/n7．（四川成都市·成都七中高二零模）/r/n下列有關(guān)電解質(zhì)溶液的說(shuō)法正確的是/r/nA．/r/n加水稀釋?zhuān)?r/nNa/r/n2/r/nS/r/n溶液中離子濃度均減小/r/nB．0.1mol/LNaOH/r/n溶液中滴加等體積等濃度醋酸溶液，溶液的導(dǎo)電性增強(qiáng)/r/nC．pH/r/n相同的①/r/nCH/r/n3/r/nCOONa/r/n②/r/nNaHCO/r/n3/r/n③/r/nNaClO/r/n三種溶液的/r/nc(Na/r/n+/r/n)/r/n：①/r/n>/r/n②/r/n>/r/n③/r/nD．/r/n向/r/n0.1mol·L/r/n-1/r/n的氨水中加入少量硫酸銨固體，則溶液中/r/nc(OH/r/n﹣/r/n)/c(NH/r/n3/r/n·H/r/n2/r/nO)/r/n增大/r/n【KS5U答案】/r/nC/r/n【KS5U解析】/r/nA/r/n．/r/nNa/r/n2/r/nS/r/n溶液呈堿性，加水稀釋?zhuān)?r/nNa/r/n2/r/nS/r/n溶液中氫離子濃度增大，故/r/nA/r/n錯(cuò)誤；/r/nB/r/n．/r/n0.1mol/LNaOH/r/n溶液中滴加等體積等濃度醋酸溶液，產(chǎn)物為/r/n0.05mol/L/r/n的/r/nCH/r/n3/r/nCOONa/r/n溶液，離子濃度減小，溶液的導(dǎo)電性減小，故/r/nB/r/n錯(cuò)誤；/r/nC/r/n．/r/nCH/r/n3/r/nCOO/r/n-/r/n、/r/n、/r/nClO/r/n-/r/n水解程度依次增強(qiáng)，/r/npH/r/n相同的①/r/nCH/r/n3/r/nCOONa/r/n②/r/nNaHCO/r/n3/r/n③/r/nNaClO/r/n三種溶液的濃度/r/nc(CH/r/n3/r/nCOONa)>c(NaHCO/r/n3/r/n)>c(NaClO)/r/n，所以/r/nc(Na/r/n+/r/n)/r/n：①/r/n>/r/n②/r/n>/r/n③，故/r/nC/r/n正確；/r/nD/r/n．向/r/n0.1mol·L/r/n-1/r/n的氨水中加入少量硫酸銨固體，銨根離子濃度增大，氨水電離平衡逆向移動(dòng)，/r/nc(OH/r/n﹣/r/n)/r/n減小、/r/nc(NH/r/n3/r/n·H/r/n2/r/nO)/r/n增大，所以溶液中/r/nc(OH/r/n﹣/r/n)/c(NH/r/n3/r/n·H/r/n2/r/nO)/r/n減小，故/r/nD/r/n錯(cuò)誤；/r/n選/r/nC/r/n。/r/n8．（浙江高三其他模擬）/r/n已知：/r/np/r/n=-lg/r/n。室溫下向/r/nHX/r/n溶液中滴加等物質(zhì)的量濃度的/r/nNaOH/r/n溶液，溶液/r/npH/r/n隨/r/np/r/n變化關(guān)系如圖所示。下列說(shuō)法正確的是/r/nA．a(chǎn)/r/n點(diǎn)溶液中：/r/n10c(Na/r/n+/r/n)=c(HX)/r/nB．/r/n溶液中由水電離出的/r/nc(H/r/n+/r/n)/r/n：/r/na>b>c/r/nC．b/r/n點(diǎn)溶液中：/r/nc(Na/r/n+/r/n)+c(H/r/n+/r/n)=c(HX)+c(OH/r/n-/r/n)/r/nD．/r/n當(dāng)溶液呈中性時(shí)：/r/nc(Na/r/n+/r/n)=c(HX)/r/n【KS5U答案】/r/nC/r/n【KS5U解析】/r/nA/r/n．/r/na/r/n點(diǎn)溶液中存在電荷守恒：/r/nc(Na/r/n+/r/n)+c(H/r/n+/r/n)=c(X/r/n-/r/n)+c(OH/r/n-/r/n)/r/n，此時(shí)/r/np/r/n=-1/r/n，則/r/nc(X/r/n-/r/n)=10c(HX)/r/n，代入電荷守恒得/r/nc(Na/r/n+/r/n)+c(H/r/n+/r/n)=10c(HX)+c(OH/r/n-/r/n)/r/n，由于溶液呈酸性/r/nc(H/r/n+/r/n)>c(OH/r/n-/r/n)/r/n，/r/nc(Na/r/n+/r/n)<10c(HX)/r/n，選項(xiàng)/r/nA/r/n錯(cuò)誤；/r/nB/r/n．根據(jù)圖示可知，/r/na/r/n、/r/nb/r/n、/r/nc/r/n均為酸性溶液，則溶質(zhì)為/r/nHX/r/n和/r/nNaX/r/n，/r/npH<7/r/n的溶液中，/r/nHX/r/n的電離程度大于/r/nX/r/n-/r/n的水解程度，可只考慮/r/nH/r/n+/r/n對(duì)水的電離的抑制，溶液/r/npH/r/n越大氫離子濃度越小，水的電離程度越大，則溶液中水的電離程度：/r/na<b<c/r/n，選項(xiàng)/r/nB/r/n錯(cuò)誤；/r/nC/r/n．/r/nb/r/n點(diǎn)溶液中存在電荷守恒：/r/nc(Na/r/n+/r/n)+c(H/r/n+/r/n)=c(X/r/n-/r/n)+c(OH/r/n-/r/n)/r/n，此時(shí)/r/np/r/n=0/r/n，則/r/nc(X/r/n-/r/n)=c(HX)/r/n，代入電荷守恒有/r/nc(Na/r/n+/r/n)+c(H/r/n+/r/n)=c(HX)+c(OH/r/n-/r/n)/r/n，選項(xiàng)/r/nC/r/n正確；/r/nD/r/n．溶液中存在電荷守恒：/r/nc(Na/r/n+/r/n)+c(H/r/n+/r/n)=c(X/r/n-/r/n)+c(OH/r/n-/r/n)/r/n，當(dāng)溶液呈中性時(shí)/r/nc(Na/r/n+/r/n)=c(X/r/n-/r/n)/r/n，若/r/nc(Na/r/n+/r/n)=c(HX)/r/n，則/r/nc(X/r/n-/r/n)=c(HX)/r/n，此時(shí)/r/np/r/n=0/r/n，即為/r/nb/r/n點(diǎn)，但/r/nb/r/n點(diǎn)溶液呈酸性，不符合，選項(xiàng)/r/nD/r/n錯(cuò)誤；/r/n答案選/r/nC/r/n。/r/n9．（河南新鄉(xiāng)市·新鄉(xiāng)縣一中高三其他模擬）/r/n為測(cè)定某二元弱酸/r/nH/r/n2/r/nA/r/n與/r/nNaOH/r/n溶液反應(yīng)過(guò)程中溶液/r/npH/r/n與粒子關(guān)系，在/r/n25℃/r/n時(shí)進(jìn)行實(shí)驗(yàn)，向/r/nH/r/n2/r/nA/r/n溶液中滴加/r/nNaOH/r/n溶液，混合溶液中/r/nlgX[X/r/n表示/r/n或/r/n]/r/n隨溶液/r/npH/r/n的變化關(guān)系如圖所示。下列說(shuō)法正確的是/r/nA．/r/n直線/r/nII/r/n中/r/nX/r/n表示的是/r/nB．/r/n當(dāng)/r/npH=3.81/r/n時(shí)，溶液中/r/nc(HA/r/n-/r/n)/r/n：/r/nc(H/r/n2/r/nA)=10/r/n：/r/n1/r/nC．0.1mol·L/r/n-1/r/nNaHA/r/n溶液中：/r/nc(Na/r/n＋/r/n)/r/n＞/r/nc(HA/r/n-/r/n)/r/n＞/r/nc(H/r/n2/r/nA)/r/n＞/r/nc(A/r/n2-/r/n)/r/nD．/r/n當(dāng)/r/npH=6.91/r/n時(shí)，對(duì)應(yīng)的溶液中，/r/n3c(A/r/n2-/r/n)=c(Na/r/n＋/r/n)/r/n＋/r/nc(H/r/n＋/r/n)-c(OH/r/n-/r/n)/r/n【KS5U答案】/r/nD/r/n【KS5U解析】/r/nA/r/n．當(dāng)/r/npH=0/r/n時(shí)，/r/nc/r/n(H/r/n+/r/n)=1mol/L/r/n，/r/nK/r/na1/r/n=/r/n=/r/n，/r/nK/r/na2/r/n=/r/n=/r/n，由于/r/nK/r/na2/r/n＜/r/nK/r/na1/r/n，故直線/r/nⅡ/r/n中/r/nX/r/n表示/r/n，/r/nA/r/n錯(cuò)誤；/r/nB/r/n．當(dāng)/r/nlgX=0/r/n時(shí)溶液的/r/npH=1.81/r/n，帶入/r/nK/r/na1/r/n計(jì)算式中可求出/r/nK/r/na1/r/n=1×10/r/n-1.81/r/n，當(dāng)/r/npH=3.81/r/n時(shí)，/r/nc(H/r/n+/r/n)=1×10/r/n-3.81/r/nmol/L/r/n，所以有/r/nK/r/na1/r/n=1×10/r/n-1.81/r/n=/r/n=1×10/r/n-3.81/r/n×/r/n，解得/r/nc/r/n(HA/r/n-/r/n)/r/n：/r/nc/r/n(H/r/n2/r/nA)=100/r/n：/r/n1/r/n，/r/nB/r/n錯(cuò)誤；/r/nC/r/n．與/r/nB/r/n項(xiàng)同理，可求出/r/nK/r/na2/r/n=1×10/r/n-6.91/r/n＞/r/n10/r/n-7/r/n，由此可知/r/nHA/r/n-/r/n的電離能力強(qiáng)于其水解能力，電離生成的/r/nc/r/n(A/r/n2-/r/n)/r/n比水解生成的/r/nc/r/n(H/r/n2/r/nA)/r/n大，/r/nC/r/n錯(cuò)誤；/r/nD/r/n．當(dāng)/r/npH=6.91/r/n時(shí)，對(duì)應(yīng)的溶液中/r/nc/r/n(HA/r/n-/r/n)=/r/nc/r/n(A/r/n2-/r/n)/r/n，又因電荷守恒/r/nc/r/n(Na/r/n＋/r/n)/r/n＋/r/nc/r/n(H/r/n＋/r/n)=c(OH/r/n-/r/n)+/r/nc/r/n(HA/r/n-/r/n)+2/r/nc/r/n(A/r/n2-/r/n)/r/n，所以/r/n3/r/nc/r/n(A/r/n2-/r/n)=/r/nc/r/n(Na/r/n＋/r/n)/r/n＋/r/nc/r/n(H/r/n＋/r/n)-/r/nc/r/n(OH/r/n-/r/n)/r/n，/r/nD/r/n正確；/r/n綜上所述答案為/r/nD/r/n。/r/n10．（安徽高三其他模擬）/r/n常溫下，將/r/nHCl/r/n氣體通入/r/n0.1mol/L/r/n氨水中，混合溶液中/r/npH/r/n與微粒濃度的對(duì)數(shù)值/r/n(lgc)/r/n和反應(yīng)物物質(zhì)的量之比/r/nX[X=/r/n]/r/n的關(guān)系如圖所示/r/n(/r/n忽略溶液體積的變化/r/n)/r/n，下列說(shuō)法正確的是/r/nA．NH/r/n3/r/n·H/r/n2/r/nO/r/n的電離平衡常數(shù)為/r/n10/r/n-9.25/r/nB．P/r/n2/r/n點(diǎn)由水電離出的/r/nc(H/r/n+/r/n)=1.0×10/r/n-7/r/nmol/L/r/nC．P/r/n3/r/n為恰好完全反應(yīng)點(diǎn)，/r/nc(Cl/r/n-/r/n)+c(NH/r/n)=0.2mol/L/r/nD．P/r/n3/r/n之后，水的電離程度一直減小/r/n【KS5U答案】/r/nB/r/n【KS5U解析】/r/nA/r/n．一水合氨電離平衡狀態(tài)下，溶液中銨根離子和氫氧根離子濃度相同時(shí)，氨水濃度為/r/n0.1mol/L/r/n，圖象分析可知，/r/nc(/r/n)=c(OH/r/n-/r/n)≈10/r/n-3/r/nmol/L/r/n，/r/nNH/r/n3/r/n?H/r/n2/r/nO/r/n的電離平衡常數(shù)/r/nK/r/nb/r/n=/r/n=/r/n=10/r/n-5/r/n，/r/nA/r/n錯(cuò)誤；/r/nB/r/n．由圖可知，/r/nP/r/n2/r/n點(diǎn)對(duì)應(yīng)的溶液/r/npH=7/r/n，故由水電離出的/r/nc(H/r/n+/r/n)=1.0×10/r/n-7/r/nmol/L/r/n，/r/nB/r/n正確；/r/nC/r/n．/r/nP/r/n3/r/n所示溶液，/r/nt=/r/n=1/r/n，/r/nn(HCl)=n(NH/r/n3/r/n?H/r/n2/r/nO)/r/n，溶液中存在物料守恒得到：/r/nc(/r/n)+c(NH/r/n3/r/n?H/r/n2/r/nO)=c(Cl/r/n-/r/n)/r/n＝/r/n0.1mol/L/r/n，故/r/nc(Cl/r/n-/r/n)+c(/r/n)=/r/n２/r/nc(/r/n)+c(NH/r/n3/r/n?H/r/n2/r/nO)/r/n＜/r/n0.2mol/L/r/n，/r/nC/r/n錯(cuò)誤；/r/nD/r/n．/r/nP/r/n3/r/n點(diǎn)為恰好完全反應(yīng)，溶質(zhì)為/r/nNH/r/n4/r/nCl/r/n，故之后，加入的/r/nHCl/r/n越來(lái)越多，由于/r/nH+/r/n對(duì)水解的抑制作用，水的電離程度減小，當(dāng)/r/nHCl/r/n達(dá)到飽和溶液時(shí)，水的電離程度將不再改變，故不是一直減小，/r/nD/r/n錯(cuò)誤；/r/n故/r/nB/r/n。/r/n11．（陜西寶雞市·高三其他模擬）/r/n常溫下，向/r/n20mL0.01mol·L/r/n-1/r/n的/r/nNaOH/r/n溶液中逐滴加入/r/n0.01mol·L/r/n-1/r/n的/r/nCH/r/n3/r/nCOOH/r/n溶液，溶液中由水電離出的/r/nc/r/n水/r/n(OH/r/n-/r/n)/r/n的對(duì)數(shù)隨加入/r/nCH/r/n3/r/nCOOH/r/n溶液體積的變化如圖所示，下列說(shuō)法正確的是/r/nA．H/r/n、/r/nF/r/n點(diǎn)溶液顯中性/r/nB．E/r/n點(diǎn)溶液中由水電離的/r/nc/r/n水/r/n(OH/r/n—/r/n)=1×10/r/n-3/r/nmol·L/r/n-1/r/nC．H/r/n點(diǎn)溶液中離子濃度關(guān)系為/r/nc/r/n(CH/r/n3/r/nCOO/r/n—/r/n)>/r/nc/r/n(Na/r/n+/r/n)>/r/nc/r/n(H/r/n+/r/n)>/r/nc/r/n(OH/r/n—/r/n)/r/nD．G/r/n點(diǎn)溶液中各離子濃度關(guān)系為/r/nc/r/n(CH/r/n3/r/nCOO/r/n—/r/n)=/r/nc/r/n(Na/r/n+/r/n)+/r/nc/r/n(H/r/n+/r/n)/r/n—/r/nc/r/n(OH/r/n—/r/n)/r/n【KS5U答案】/r/nD/r/n【分析】/r/n氫氧化鈉在溶液中抑制水的電離，向氫氧化鈉溶液中加入醋酸，對(duì)水的電離的抑制作用逐漸減弱，當(dāng)溶液為醋酸鈉溶液時(shí)，水的電離程度最大，則/r/nG/r/n點(diǎn)為醋酸鈉溶液；從/r/nE/r/n點(diǎn)到/r/nG/r/n點(diǎn)的反應(yīng)過(guò)程中，所得溶液為氫氧化鈉和醋酸鈉的混合溶液，溶液為堿性；/r/nH/r/n點(diǎn)為醋酸和醋酸鈉混合溶液，溶液呈中性。/r/n【KS5U解析】/r/nA/r/n．由分析可知，/r/nH/r/n點(diǎn)為醋酸和醋酸鈉混合溶液，溶液呈中性，/r/nF/r/n點(diǎn)為氫氧化鈉和醋酸鈉的混合溶液，溶液為堿性，故/r/nA/r/n錯(cuò)誤；/r/nB/r/n．氫氧化鈉在溶液中抑制水的電離，/r/n0.01mol/r/n·/r/nL/r/n-1/r/n的氫氧化鈉溶液中氫氧根離子的濃度為/r/n0.01mol/r/n·/r/nL/r/n-1/r/n，則溶液中水電離的氫離子濃度為/r/n10/r/n-12/r/nmol/r/n·/r/nL/r/n-1/r/n，故/r/nB/r/n錯(cuò)誤；/r/nC/r/n．由分析可知，/r/nH/r/n點(diǎn)為醋酸和醋酸鈉混合溶液，溶液呈中性，由電荷守恒關(guān)系/r/nc/r/n(CH/r/n3/r/nCOO/r/n—/r/n)+/r/nc/r/n(OH/r/n—/r/n)=/r/nc/r/n(Na/r/n+/r/n)+/r/nc/r/n(H/r/n+/r/n)/r/n可知，溶液中離子濃度關(guān)系為/r/nc/r/n(CH/r/n3/r/nCOO/r/n—/r/n)=/r/nc/r/n(Na/r/n+/r/n)>/r/nc/r/n(H/r/n+/r/n)=/r/nc/r/n(OH/r/n—/r/n)/r/n，故/r/nC/r/n錯(cuò)誤；/r/nD/r/n．由分析可知，/r/nG/r/n點(diǎn)為醋酸鈉溶液，由電荷守恒關(guān)系/r/nc/r/n(CH/r/n3/r/nCOO/r/n—/r/n)+/r/nc/r/n(OH/r/n—/r/n)=/r/nc/r/n(Na/r/n+/r/n)+/r/nc/r/n(H/r/n+/r/n)/r/n可知，溶液中各離子濃度關(guān)系為/r/nc/r/n(CH/r/n3/r/nCOO/r/n—/r/n)=/r/nc/r/n(Na/r/n+/r/n)+/r/nc/r/n(H/r/n+/r/n)/r/n—/r/nc/r/n(OH/r/n—/r/n)/r/n，故/r/nD/r/n正