2021年高考化學真題和模擬題分類匯編專題13鹽類的水解【含答案】

專題13鹽類的水解/r/n2021年化學高考題/r/n一、單選題/r/n1．（全國高考真題試卷）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。實驗發(fā)現(xiàn)，/r/n時/r/n為線性關系，如下圖中實線所示。/r/n下列敘述錯誤的是/r/nA．/r/n溶液/r/n時，/r/nB．MA/r/n的溶度積度積/r/nC．/r/n溶液/r/n時，/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系，在解題過程中要注意電荷守恒和物料守恒的應用，具體見詳解。/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，/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，可看作溶液中有較大濃度的OH/r/n-/r/n，此時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．設調/r/npH/r/n所用的酸為/r/nH/r/nn/r/nX/r/n，則結合電荷守恒可知/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錯誤；/r/nD/r/n．/r/n當/r/n時，由物料守恒知/r/n，則/r/n，/r/n，則/r/n，對應圖得此時溶液中/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下列說法/r/n不正確/r/n的是/r/nA．/r/n由/r/na/r/n點可知：/r/n溶液中/r/n的水解程度大于電離程度/r/nB．/r/n過程中：/r/n逐漸減小/r/nC．/r/n過程中：/r/nD．/r/n令/r/nc/r/n點的/r/n，/r/ne/r/n點的/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，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，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點溶質為/r/n，此時溶液呈堿性，/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過程溶液中/r/n，滴加/r/nNaOH/r/n溶液的過程中/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點溶液中/r/n，向/r/n溶液中滴加鹽酸過程中有/r/nCO/r/n2/r/n逸出，因此/r/n過程中/r/n，故/r/nC/r/n錯誤；/r/nD/r/n．/r/nc/r/n點溶液中/r/n=(0.05+10/r/n-11.3/r/n)mol/L/r/n，/r/ne/r/n點溶液體積增大/r/n1/r/n倍，此時溶液中/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，說法不正確的是/r/nC/r/n項，故答案為/r/nC/r/n。/r/n3．（廣東高考真題試卷）/r/n鳥嘌呤/r/n(/r/n)/r/n是一種有機弱堿，可與鹽酸反應生成鹽酸鹽/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，/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為強酸弱堿鹽，電離出的/r/nGH/r/n+/r/n會發(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錯誤；/r/nB/r/n．稀釋/r/nGHCl/r/n溶液時，/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錯誤；/r/n綜上所述，敘述正確的是/r/nB/r/n項，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/r/n時，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，沒有加入鹽酸時，/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為/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)的大小順序為/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．當溶液/r/npH/r/n為/r/n7/r/n時，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為/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，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實驗測得/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/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下列說法/r/n不正確/r/n的是/r/nA．/r/n圖中/r/n實線/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，水解反應為吸熱過程，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，若氯化銨的水解平衡不發(fā)生移動，則其中的/r/nc/r/n(H/r/n+/r/n)/r/n變?yōu)樵瓉淼?r/n，則溶液的/r/npH/r/n將增大/r/nlgx/r/n，但是，加水稀釋時，氯化銨的水解平衡向正反應方向移動，/r/nc/r/n(H/r/n+/r/n)/r/n大于原來的/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說法不正確；/r/nD/r/n．/r/n25℃/r/n時稀釋相同倍數(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/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，溶質的物質的量濃度仍相等，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，下列說法正確的是/r/nA．NaHA/r/n溶液呈酸性，可以推測/r/nH/r/n2/r/nA/r/n為強酸/r/nB．/r/n可溶性正鹽/r/nBA/r/n溶液呈中性，可以推測/r/nBA/r/n為強酸強堿鹽/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為強酸的結論，/r/nA/r/n錯誤；/r/nB/r/n．可溶性正鹽/r/nBA/r/n溶液呈中性，不能推測/r/nBA/r/n為強酸強堿鹽，因為也可能是/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，/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，/r/npH=/r/nD．P/r/n點，/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：/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/r/n，由此可知/r/n，/r/nN/r/n點/r/n，則/r/n，/r/nP/r/n點/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錯誤；/r/nB/r/n．/r/nM/r/n點存在電荷守恒：/r/n，此時/r/n，因此/r/n，故/r/nB/r/n錯誤；/r/nC/r/n．/r/nO/r/n點/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點溶質為/r/nNaCl/r/n、/r/nHR/r/n、/r/nNaR/r/n，此時溶液呈堿性，因此/r/n，溶質濃度大于水/r/n解和電離所產生微粒濃度，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。以獨居石為原料制備/r/n的工藝流程如下：/r/n回答下列問題：/r/n(1)/r/n鈰的某種核素含有/r/n58/r/n個質子和/r/n80/r/n個中子，該核素的符號為/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，同時生成/r/n和/r/n，該反應的化學方程式為/r/n_______/r/n【KS5U答案】/r/n/r/n適當升高溫度，將獨居石粉碎等/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焙燒濃硫酸和獨居石的混合物、水浸，/r/n轉化為/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，/r/n轉化為/r/nAl/r/n2/r/n(SO/r/n4/r/n)/r/n3/r/n，/r/n轉化為/r/nFe/r/n2/r/n(SO/r/n4/r/n)/r/n3/r/n，/r/n轉化為/r/nCaSO/r/n4/r/n和/r/nHF/r/n，酸性廢氣含/r/nHF/r/n；后過濾，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為/r/nFePO/r/n4/r/n；聚沉將鐵離子、鋁離子轉化為沉淀，過濾除去，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個質子和/r/n80/r/n個中子，則質量數(shù)為/r/n58+80=138/r/n，該核素的符號為/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升高溫度，將獨居石粉碎等；/r/n(3)/r/n結合流程可知，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，同時生成/r/n和/r/n可知，該反應中/r/nFe/r/n價態(tài)降低，/r/nC/r/n價態(tài)升高，結合得失電子守恒、原子守恒可知該反應的化學方程式為/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忽略溶液混合時的體積變化/r/n)/r/n。下列說法正確的是/r/nA．/r/n本實驗應該選擇酚酞作指示劑/r/nB．/r/n室溫下，/r/nC．a/r/n點溶液中水的電離程度最大/r/nD．b/r/n點溶液中存在：/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是強酸弱堿鹽，其水溶液呈酸性，結合電荷守恒關系分析解答。/r/n【KS5U解析】/r/nA/r/n．用二甲胺溶液滴定鹽酸溶液達到滴定終點時生成強酸弱堿鹽/r/n(CH/r/n3/r/n)/r/n2/r/nNH/r/n2/r/nCl/r/n，溶液呈酸性，應該選擇甲基橙作指示劑，故/r/nA/r/n錯誤；/r/nB/r/n．/r/n(CH/r/n3/r/n)/r/n2/r/nNH/r/n2/r/nCl/r/n是強酸弱堿鹽，其水溶液呈酸性，電荷關系為/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點時溶液呈中性，/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．強酸弱堿鹽能夠促進水的電離，并且其濃度越大、促進作用越強，二者恰好完全反應時生成/r/n(CH/r/n3/r/n)/r/n2/r/nNH/r/n2/r/nCl/r/n，此時溶液呈酸性，圖中/r/na/r/n點呈中性、二甲胺過量，所以/r/na/r/n點以前的某點溶液中水的電離程度最大，故/r/nC/r/n錯誤；/r/nD/r/n．圖中/r/nb/r/n點溶液的/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，各種微粒物質的量分數(shù)與溶液的/r/npH/r/n關系如圖所示。下列說法不正確的是/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/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，當溶液/r/npH/r/n在/r/n7.35-7.45/r/n之間時，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，溶液/r/npH/r/n增大，氫氧根離子濃度增大，平衡向逆反應方向移動，/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，當溶液中砷元素的主要存在形式為/r/nH/r/n2/r/nAsO/r/n時，溶液/r/npH/r/n約為/r/n11/r/n，溶液呈堿性，說明/r/nH/r/n2/r/nAsO/r/n的水解程度大于電離程度，故/r/nC/r/n錯誤；/r/nD/r/n．由圖可知，交點/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，充分反應后，四種離子的濃度不變的是/r/n(/r/n忽略反應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氧化性的過氧化鈉固體，過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增大，故/r/nA/r/n錯誤；/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，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，不能大量共存，故/r/nB/r/n錯誤；/r/nC/r/n．氫氧化鐵溶于氫碘酸溶液的反應為氫氧化鐵與氫碘酸溶液反應生成碘化亞鐵、碘和水，反應的離子方程式為/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發(fā)生雙水解反應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錯誤；/r/n故選/r/nC/r/n。/r/n4．（重慶市第十一中學校高三二模）/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，測得溶液的電阻率/r/n(/r/n溶液的電阻率越大，導電能力越弱/r/n)/r/n與加入氨水的體積/r/n(V)/r/n的關系如圖。/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→c/r/n過程，水的電離程度逐漸減小/r/nC．c/r/n點時，/r/nD．d/r/n點時，/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過程中溶液的導電性減弱，向混合溶液中加入等物質的量濃度的/r/nNH/r/n3/r/n?H/r/n2/r/nO/r/n溶液時，發(fā)生反應先后順序為：/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、導電性減弱，/r/nb/r/n點最小，原因是溶液體積增大導致/r/nb/r/n點離子濃度減小，/r/nb/r/n點溶液中溶質為/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，生成的/r/nCH/r/n3/r/nCOONH/r/n4/r/n是強電解質，導致溶液中離子濃度增大，溶液的電導性增大，/r/nc/r/n點時醋酸和一水合氨恰好完全反應生成醋酸銨，/r/nc/r/n點溶液中溶質為/r/nNH/r/n4/r/nX/r/n、/r/nCH/r/n3/r/nCOONH/r/n4/r/n，且二者的物質的量相等；/r/nd/r/n點溶液中溶質為等物質的量濃度的/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，酸堿反應生成鹽，溶液導電性將增強、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、導電性反而減弱，說明原混合溶液中離子濃度更大，即/r/nHX/r/n為強電解質，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溶液的過程：/r/na→c/r/n為/r/nHX/r/n和/r/nCH/r/n3/r/nCOOH/r/n轉化為/r/nNH/r/n4/r/nX/r/n、/r/nCH/r/n3/r/nCOONH/r/n4/r/n的過程，溶液的酸性減弱，水的電離程度增大，故/r/nB/r/n錯誤；/r/n

/r/nC/r/n．根據(jù)分析可知，/r/nc/r/n點溶液中溶質為/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點溶液中溶質為等物質的量濃度的/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，/r/nNH/r/n3/r/n?H/r/n2/r/nO/r/n轉化為/r/nNH/r/n，/r/n40/r/n～/r/n60mL/r/n時，/r/nNH/r/n3/r/n?H/r/n2/r/nO/r/n過量，/r/nd/r/n點時，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用下列實驗裝置/r/n(/r/n部分夾持裝置略去/r/n)/r/n，能達到實驗目的的是/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，高錳酸鉀固體會分解，故/r/nA/r/n錯誤；/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，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溶液制無水/r/nAlCl/r/n3/r/n固體，氯化鋁會水解，應在氯化氫氣流中加熱，故/r/nD/r/n錯誤；/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/r/n點溶液：/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點向/r/nc/r/n點方向遷移/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點向/r/nb/r/n點方向遷移/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/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錯誤；/r/nB/r/n．/r/na/r/n點溶液：溶質為/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點向/r/nc/r/n點方向遷移，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點向/r/nd/r/n點方向遷移，故/r/nD/r/n錯誤；/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下列有關電解質溶液的說法正確的是/r/nA．/r/n加水稀釋，/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，溶液的導電性增強/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，/r/nNa/r/n2/r/nS/r/n溶液中氫離子濃度增大，故/r/nA/r/n錯誤；/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，溶液的導電性減小，故/r/nB/r/n錯誤；/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，/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，/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變化關系如圖所示。下列說法正確的是/r/nA．a/r/n點溶液中：/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當溶液呈中性時：/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/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/r/nA/r/n錯誤；/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/r/nB/r/n錯誤；/r/nC/r/n．/r/nb/r/n點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/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/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，當溶液呈中性時/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/r/np/r/n=0/r/n，即為/r/nb/r/n點，但/r/nb/r/n點溶液呈酸性，不符合，選項/r/nD/r/n錯誤；/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為測定某二元弱酸/r/nH/r/n2/r/nA/r/n與/r/nNaOH/r/n溶液反應過程中溶液/r/npH/r/n與粒子關系，在/r/n25℃/r/n時進行實驗，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的變化關系如圖所示。下列說法正確的是/r/nA．/r/n直線/r/nII/r/n中/r/nX/r/n表示的是/r/nB．/r/n當/r/npH=3.81/r/n時，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/r/npH=6.91/r/n時，對應的溶液中，/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/r/npH=0/r/n時，/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錯誤；/r/nB/r/n．當/r/nlgX=0/r/n時溶液的/r/npH=1.81/r/n，帶入/r/nK/r/na1/r/n計算式中可求出/r/nK/r/na1/r/n=1×10/r/n-1.81/r/n，當/r/npH=3.81/r/n時，/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錯誤；/r/nC/r/n．與/r/nB/r/n項同理，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錯誤；/r/nD/r/n．當/r/npH=6.91/r/n時，對應的溶液中/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與微粒濃度的對數(shù)值/r/n(lgc)/r/n和反應物物質的量之比/r/nX[X=/r/n]/r/n的關系如圖所示/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為恰好完全反應點，/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，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錯誤；/r/nB/r/n．由圖可知，/r/nP/r/n2/r/n點對應的溶液/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錯誤；/r/nD/r/n．/r/nP/r/n3/r/n點為恰好完全反應，溶質為/r/nNH/r/n4/r/nCl/r/n，故之后，加入的/r/nHCl/r/n越來越多，由于/r/nH+/r/n對水解的抑制作用，水的電離程度減小，當/r/nHCl/r/n達到飽和溶液時，水的電離程度將不再改變，故不是一直減小，/r/nD/r/n錯誤；/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，下列說法正確的是/r/nA．H/r/n、/r/nF/r/n點溶液顯中性/r/nB．E/r/n點溶液中由水電離的/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點溶液中離子濃度關系為/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點溶液中各離子濃度關系為/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，對水的電離的抑制作用逐漸減弱，當溶液為醋酸鈉溶液時，水的電離程度最大，則/r/nG/r/n點為醋酸鈉溶液；從/r/nE/r/n點到/r/nG/r/n點的反應過程中，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點為醋酸和醋酸鈉混合溶液，溶液呈中性，/r/nF/r/n點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錯誤；/r/nC/r/n．由分析可知，/r/nH/r/n點為醋酸和醋酸鈉混合溶液，溶液呈中性，由電荷守恒關系/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錯誤；/r/nD/r/n．由分析可知，/r/nG/r/n點為醋酸鈉溶液，由電荷守恒關系/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