2021年高考化学真题和模拟题分类汇编专题13盐类的水解【含答案】

专题13盐类的水解/r/n2021年化学高考题/r/n一、单选题/r/n1．（全国高考真题试卷）HA/r/n是一元弱酸，难溶盐/r/nMA/r/n的饱和溶液中/r/n随/r/nc(H/r/n+/r/n)/r/n而变化，/r/n不发生水解。实验发现，/r/n时/r/n为线性关系，如下图中实线所示。/r/n下列叙述错误的是/r/nA．/r/n溶液/r/n时，/r/nB．MA/r/n的溶度积度积/r/nC．/r/n溶液/r/n时，/r/nD．HA/r/n的电离常数/r/n【KS5U答案】/r/nC/r/n【分析】/r/n本题考查水溶液中离子浓度的关系，在解题过程中要注意电荷守恒和物料守恒的应用，具体见详解。/r/n【KS5U解析】/r/nA/r/n．由图可知/r/npH=4/r/n，即/r/nc(H/r/n+/r/n)=10×10/r/n-5/r/nmol/L/r/n时，/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正确；/r/nB/r/n．由图可知，c(H/r/n+/r/n)=0时，可看作溶液中有较大浓度的OH/r/n-/r/n，此时A/r/n-/r/n的水解极大地被抑制，溶/r/n液中c(M/r/n+/r/n)=c(A/r/n-/r/n)，则/r/n，/r/nB/r/n正确；/r/nC/r/n．设调/r/npH/r/n所用的酸为/r/nH/r/nn/r/nX/r/n，则结合电荷守恒可知/r/n，题给等式右边缺阴离子部分/r/nnc(X/r/nn-/r/n)/r/n，/r/nC/r/n错误；/r/nD/r/n．/r/n当/r/n时，由物料守恒知/r/n，则/r/n，/r/n，则/r/n，对应图得此时溶液中/r/n，/r/n，/r/nD/r/n正确；/r/n故选/r/nC/r/n。/r/n2．（浙江）/r/n取两份/r/n/r/n的/r/n溶液，一份滴加/r/n的盐酸，另一份滴加/r/n溶液，溶液的/r/npH/r/n随加入酸/r/n(/r/n或碱/r/n)/r/n体积的变化如图。/r/n下列说法/r/n不正确/r/n的是/r/nA．/r/n由/r/na/r/n点可知：/r/n溶液中/r/n的水解程度大于电离程度/r/nB．/r/n过程中：/r/n逐渐减小/r/nC．/r/n过程中：/r/nD．/r/n令/r/nc/r/n点的/r/n，/r/ne/r/n点的/r/n，则/r/n【KS5U答案】/r/nC/r/n【分析】/r/n向/r/n溶液中滴加盐酸，溶液酸性增强，溶液/r/npH/r/n将逐渐减小，向/r/n溶液中滴加/r/nNaOH/r/n溶液，溶液碱性增强，溶液/r/npH/r/n将逐渐增大，因此/r/nabc/r/n曲线为向/r/n溶液中滴加/r/nNaOH/r/n溶液，/r/nade/r/n曲线为向/r/n溶液中滴加盐酸。/r/n【KS5U解析】/r/nA/r/n．/r/na/r/n点溶质为/r/n，此时溶液呈碱性，/r/n在溶液中电离使溶液呈酸性，/r/n在溶液中水解使溶液呈碱性，由此可知，/r/n溶液中/r/n的水解程度大于电离程度，故/r/nA/r/n正确；/r/nB/r/n．由电荷守恒可知，/r/n过程溶液中/r/n，滴加/r/nNaOH/r/n溶液的过程中/r/n保持不变，/r/n逐渐减小，因此/r/n逐渐减小，故/r/nB/r/n正确；/r/nC/r/n．由物料守恒可知，/r/na/r/n点溶液中/r/n，向/r/n溶液中滴加盐酸过程中有/r/nCO/r/n2/r/n逸出，因此/r/n过程中/r/n，故/r/nC/r/n错误；/r/nD/r/n．/r/nc/r/n点溶液中/r/n=(0.05+10/r/n-11.3/r/n)mol/L/r/n，/r/ne/r/n点溶液体积增大/r/n1/r/n倍，此时溶液中/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正确；/r/n综上所述，说法不正确的是/r/nC/r/n项，故答案为/r/nC/r/n。/r/n3．（广东高考真题试卷）/r/n鸟嘌呤/r/n(/r/n)/r/n是一种有机弱碱，可与盐酸反应生成盐酸盐/r/n(/r/n用/r/n表示/r/n)/r/n。已知/r/n水溶液呈酸性，下列叙述正确的是/r/nA．/r/n水溶液的/r/nB．/r/n水溶液加水稀释，/r/n升高/r/nC．/r/n在水中的电离方程式为：/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为强酸弱碱盐，电离出的/r/nGH/r/n+/r/n会发生水解，弱离子的水解较为微弱，因此/r/n0.001mol/LGHCl/r/n水溶液的/r/npH>3/r/n，故/r/nA/r/n错误；/r/nB/r/n．稀释/r/nGHCl/r/n溶液时，/r/nGH/r/n+/r/n水解程度将增大，根据勒夏特列原理可知溶液中/r/nc/r/n(H/r/n+/r/n)/r/n将减小，溶液/r/npH/r/n将升高，故/r/nB/r/n正确；/r/nC/r/n．/r/nGHCl/r/n为强酸弱碱盐，在水中电离方程式为/r/nGHCl=GH/r/n+/r/n+Cl/r/n-/r/n，故/r/nC/r/n错误；/r/nD/r/n．根据电荷守恒可知，/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错误；/r/n综上所述，叙述正确的是/r/nB/r/n项，故答案为/r/nB/r/n。/r/n4．（湖南高考真题试卷）/r/n常温下，用/r/n的盐酸分别滴定/r/n20.00mL/r/n浓度均为/r/n三种一元弱酸的钠盐/r/n溶液，滴定曲线如图所示。下列判断错误的是/r/nA．/r/n该/r/n溶液中：/r/nB．/r/n三种一元弱酸的电离常数：/r/nC．/r/n当/r/n时，三种溶液中：/r/nD．/r/n分别滴加/r/n20.00mL/r/n盐酸后，再将三种溶液混合：/r/n【KS5U答案】/r/nC/r/n【分析】/r/n由图可知，没有加入盐酸时，/r/nNaX/r/n、/r/nNaY/r/n、/r/nNaZ/r/n溶液的/r/npH/r/n依次增大，则/r/nHX/r/n、/r/nHY/r/n、/r/nHZ/r/n三种一元弱酸的酸性依次减弱。/r/n【KS5U解析】/r/nA/r/n．/r/nNaX/r/n为强碱弱酸盐，在溶液中水解使溶液呈碱性，则溶液中离子浓度的大小顺序为/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正确；/r/nB/r/n．弱酸的酸性越弱，电离常数越小，由分析可知，/r/nHX/r/n、/r/nHY/r/n、/r/nHZ/r/n三种一元弱酸的酸性依次减弱，则三种一元弱酸的电离常数的大小顺序为/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正确；/r/nC/r/n．当溶液/r/npH/r/n为/r/n7/r/n时，酸越弱，向盐溶液中加入盐酸的体积越大，酸根离子的浓度越小，则三种盐溶液中酸根的浓度大小顺序为/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错误；/r/nD/r/n．向三种盐溶液中分别滴加/r/n20.00mL/r/n盐酸，三种盐都完全反应，溶液中钠离子浓度等于氯离子浓度，将三种溶液混合后溶液中存在电荷守恒关系/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正确；/r/n故选/r/nC/r/n。/r/n5．（浙江高考真题试卷）/r/n实验测得/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分别随温度与稀释加水量的变化如图所示。已知/r/n25/r/n℃时/r/nCH/r/n3/r/nCOOH/r/n和/r/nNH/r/n3/r/n·H/r/n2/r/nO/r/n的电离常数均为/r/n1.8×10/r/n-5./r/n下列说法/r/n不正确/r/n的是/r/nA．/r/n图中/r/n实线/r/n表示/r/npH/r/n随加水量的变化，/r/n虚线/r/n表示/r/npH/r/n随温度的变化/r/n'/r/nB．/r/n将/r/nNH/r/n4/r/nCl/r/n溶液加水稀释至浓度/r/nmol·L/r/n-1/r/n，溶液/r/npH/r/n变化值小于/r/nlgx/r/nC．/r/n随温度升高，/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减小，/r/nc/r/n(H/r/n+/r/n)/r/n增大，/r/npH/r/n减小/r/nD．25/r/n℃时稀释相同倍数的/r/nNH/r/n4/r/nCl/r/n溶液与/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由题中信息可知，图中两条曲线为/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分别随温度与稀释加水量的变化曲线，由于两种盐均能水解，水解反应为吸热过程，且温度越高、浓度越小其水解程度越大。氯化铵水解能使溶液呈酸性，浓度越小，虽然水程度越大，但其溶液的酸性越弱，故其/r/npH/r/n越大；醋酸钠水解能使溶液呈碱性，浓度越小，其水溶液的碱性越弱，故其/r/npH/r/n越小。温度越高，水的电离度越大。因此，图中的实线为/r/npH/r/n随加水量的变化，虚线表示/r/npH/r/n随温度的变化。/r/n【KS5U解析】/r/nA/r/n．由分析可知，图中实线表示/r/npH/r/n随加水量的变化，虚线表示/r/npH/r/n随温度的变化，/r/nA/r/n说法正确；/r/nB/r/n．将/r/nNH/r/n4/r/nCl/r/n溶液加水稀释至浓度/r/nmol·L/r/n-1/r/n时，若氯化铵的水解平衡不发生移动，则其中的/r/nc/r/n(H/r/n+/r/n)/r/n变为原来的/r/n，则溶液的/r/npH/r/n将增大/r/nlgx/r/n，但是，加水稀释时，氯化铵的水解平衡向正反应方向移动，/r/nc/r/n(H/r/n+/r/n)/r/n大于原来的/r/n，因此，溶液/r/npH/r/n的变化值小于/r/nlgx/r/n，/r/nB/r/n说法正确；/r/nC/r/n．随温度升高，水的电离程度变大，因此水的离子积变大，即/r/nK/r/nw/r/n增大；随温度升高，/r/nCH/r/n3/r/nCOONa/r/n的水解程度变大，溶液中/r/nc/r/n(OH/r/n-/r/n)/r/n增大，因此，/r/nC/r/n说法不正确；/r/nD/r/n．/r/n25℃/r/n时稀释相同倍数的/r/nNH/r/n4/r/nC1/r/n溶液与/r/nCH/r/n3/r/nCOONa/r/n溶液中均分别存在电荷守恒，/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，氯化铵溶液中，/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，醋酸钠溶液中，/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时/r/nCH/r/n3/r/nCOOH/r/n和/r/nNH/r/n3/r/n·/r/nH/r/n2/r/nO/r/n的电离常数均为/r/n1.8/r/n×/r/n10/r/n-5/r/n，因此，由于原溶液的物质的量浓度相同，稀释相同倍数后的/r/nNH/r/n4/r/nC1/r/n溶液与/r/nCH/r/n3/r/nCOONa/r/n溶液，溶质的物质的量浓度仍相等，由于电离常数相同，其中盐的水解程度是相同的，因此，两溶液中/r/n/r/nc/r/n(OH/r/n-/r/n)-/r/nc/r/n(H/r/n+/r/n)/r/n/r/n（两者差的绝对值）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说法正确。/r/n综上所述，本题选/r/nC/r/n。/r/n6．（浙江高考真题试卷）25/r/n℃时，下列说法正确的是/r/nA．NaHA/r/n溶液呈酸性，可以推测/r/nH/r/n2/r/nA/r/n为强酸/r/nB．/r/n可溶性正盐/r/nBA/r/n溶液呈中性，可以推测/r/nBA/r/n为强酸强碱盐/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的醋酸溶液的电离度分别为/r/nα/r/n1/r/n、/r/nα/r/n2/r/n，则/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溶液中水电离出/r/nH/r/n+/r/n的物质的量为/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的电离程度大于其水解程度，不能据此得出/r/nH/r/n2/r/nA/r/n为强酸的结论，/r/nA/r/n错误；/r/nB/r/n．可溶性正盐/r/nBA/r/n溶液呈中性，不能推测/r/nBA/r/n为强酸强碱盐，因为也可能是/r/nB/r/n＋/r/n和/r/nA/r/n－/r/n的水解程度相同，即也可能是弱酸弱碱盐，/r/nB/r/n错误；/r/nC/r/n．弱酸的浓度越小，其电离程度越大，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的醋酸溶液的电离度分别为/r/nα/r/n1/r/n、/r/nα/r/n2/r/n，则/r/nα/r/n1/r/n＞/r/nα/r/n2/r/n，/r/nC/r/n错误；/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溶液中氢氧根离子的浓度是/r/n1/r/n×/r/n10/r/n－/r/n4/r/nmol/r/n//r/nL/r/n，碳酸根水解促进水的电离，则水电离出/r/nH/r/n＋/r/n的浓度是/r/n1/r/n×/r/n10/r/n－/r/n4/r/nmol/r/n//r/nL/r/n，其物质的量为/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正确；/r/n答案选/r/nD/r/n。/r/n二、多选题/r/n7．（山东高考真题试卷）/r/n赖氨酸/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是人体必需氨基酸，其盐酸盐/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。向一定浓度的/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的分布系数/r/nδ(x)/r/n随/r/npH/r/n变化如图所示。已知/r/nδ(x)=/r/n，下列表述正确的是/r/n

/r/nA．/r/n>/r/nB．M/r/n点，/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点，/r/npH=/r/nD．P/r/n点，/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，依次发生离子反应：/r/n、/r/n、/r/n，溶液中/r/n逐渐减小，/r/n和/r/n先增大后减小，/r/n逐渐增大。/r/n，/r/n，/r/n，/r/nM/r/n点/r/n，由此可知/r/n，/r/nN/r/n点/r/n，则/r/n，/r/nP/r/n点/r/n，则/r/n。/r/n【KS5U解析】/r/nA/r/n．/r/n，/r/n，因此/r/n，故/r/nA/r/n错误；/r/nB/r/n．/r/nM/r/n点存在电荷守恒：/r/n，此时/r/n，因此/r/n，故/r/nB/r/n错误；/r/nC/r/n．/r/nO/r/n点/r/n，因此/r/n，即/r/n，因此/r/n，溶液/r/n，故/r/nC/r/n正确；/r/nD/r/n．/r/nP/r/n点溶质为/r/nNaCl/r/n、/r/nHR/r/n、/r/nNaR/r/n，此时溶液呈碱性，因此/r/n，溶质浓度大于水/r/n解和电离所产生微粒浓度，因此/r/n，故/r/nD/r/n正确；/r/n综上所述，正确的是/r/nCD/r/n，故答案为/r/nCD/r/n。/r/n三、工业流程题/r/n8．（湖南高考真题试卷）/r/n可用于催化剂载体及功能材料的制备。天然独居石中，铈/r/n(Ce)/r/n主要以/r/n形式存在，还含有/r/n、/r/n、/r/n、/r/n等物质。以独居石为原料制备/r/n的工艺流程如下：/r/n回答下列问题：/r/n(1)/r/n铈的某种核素含有/r/n58/r/n个质子和/r/n80/r/n个中子，该核素的符号为/r/n_______/r/n；/r/n(2)/r/n为提高/r/n“/r/n水浸/r/n”/r/n效率，可采取的措施有/r/n_______(/r/n至少写两条/r/n)/r/n；/r/n(3)/r/n滤渣Ⅲ的主要成分是/r/n_______(/r/n填化学式/r/n)/r/n；/r/n(4)/r/n加入絮凝剂的目的是/r/n_______/r/n；/r/n(5)“/r/n沉铈/r/n”/r/n过程中，生成/r/n的离子方程式为/r/n_______/r/n，常温下加入的/r/n溶液呈/r/n_______(/r/n填/r/n“/r/n酸性/r/n”“/r/n碱性/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滤渣Ⅱ的主要成分为/r/n，在高温条件下，/r/n、葡萄糖/r/n(/r/n)/r/n和/r/n可制备电极材料/r/n，同时生成/r/n和/r/n，该反应的化学方程式为/r/n_______/r/n【KS5U答案】/r/n/r/n适当升高温度，将独居石粉碎等/r/nAl(OH)/r/n3/r/n/r/n促使铝离子沉淀/r/n/r/n↑/r/n碱性/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焙烧浓硫酸和独居石的混合物、水浸，/r/n转化为/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与硫酸不反应，/r/n转化为/r/nAl/r/n2/r/n(SO/r/n4/r/n)/r/n3/r/n，/r/n转化为/r/nFe/r/n2/r/n(SO/r/n4/r/n)/r/n3/r/n，/r/n转化为/r/nCaSO/r/n4/r/n和/r/nHF/r/n，酸性废气含/r/nHF/r/n；后过滤，滤渣Ⅰ为/r/n和磷酸钙、/r/nFePO/r/n4/r/n，滤液主要含/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，加氯化铁溶液除磷，滤渣Ⅱ为/r/nFePO/r/n4/r/n；聚沉将铁离子、铝离子转化为沉淀，过滤除去，滤渣Ⅲ主要为氢氧化铝，还含氢氧化铁；加碳酸氢铵沉铈得/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铈的某种核素含有/r/n58/r/n个质子和/r/n80/r/n个中子，则质量数为/r/n58+80=138/r/n，该核素的符号为/r/n；/r/n(2)/r/n为提高/r/n“/r/n水浸/r/n”/r/n效率，可采取的措施有适当升高温度，将独居石粉碎等；/r/n(3)/r/n结合流程可知，滤渣Ⅲ的主要成分是/r/nAl(OH)/r/n3/r/n；/r/n(4)/r/n加入絮凝剂的目的是促使铝离子沉淀；/r/n(5)/r/n用碳酸氢铵/r/n“/r/n沉铈/r/n”/r/n，则结合原子守恒、电荷守恒可知生成/r/n的离子方程式为/r/n↑/r/n；铵根离子的水解常数/r/nK/r/nh/r/n(/r/n)=/r/n≈5.7×10/r/n-10/r/n，碳酸氢根的水解常数/r/nK/r/nh/r/n(/r/n)==/r/n≈2.3×10/r/n-8/r/n，则/r/nK/r/nh/r/n(/r/n)<K/r/nh/r/n(/r/n)/r/n，因此常温下加入的/r/n溶液呈碱性；/r/n(6)/r/n由在高温条件下，/r/n、葡萄糖/r/n(/r/n)/r/n和/r/n可制备电极材料/r/n，同时生成/r/n和/r/n可知，该反应中/r/nFe/r/n价态降低，/r/nC/r/n价态升高，结合得失电子守恒、原子守恒可知该反应的化学方程式为/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年化学高考模拟题/r/n一、单选题/r/n1．（九龙坡区·重庆市育才中学高三三模）/r/n室温下，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二甲胺在水中的电离与一水合氨相似/r/n)/r/n滴定/r/n10.00mL0.1mol/L/r/n的盐酸溶液。溶液/r/npH/r/n随加入二甲胺溶液体积变化曲线如图所示/r/n(/r/n忽略溶液混合时的体积变化/r/n)/r/n。下列说法正确的是/r/nA．/r/n本实验应该选择酚酞作指示剂/r/nB．/r/n室温下，/r/nC．a/r/n点溶液中水的电离程度最大/r/nD．b/r/n点溶液中存在：/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为一元弱碱，二甲胺在水中的电离与一水合氨相似，二甲胺/r/n[(CH/r/n3/r/n)/r/n2/r/nNH]/r/n电离方程式为/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，电离平衡常数/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是强酸弱碱盐，其水溶液呈酸性，结合电荷守恒关系分析解答。/r/n【KS5U解析】/r/nA/r/n．用二甲胺溶液滴定盐酸溶液达到滴定终点时生成强酸弱碱盐/r/n(CH/r/n3/r/n)/r/n2/r/nNH/r/n2/r/nCl/r/n，溶液呈酸性，应该选择甲基橙作指示剂，故/r/nA/r/n错误；/r/nB/r/n．/r/n(CH/r/n3/r/n)/r/n2/r/nNH/r/n2/r/nCl/r/n是强酸弱碱盐，其水溶液呈酸性，电荷关系为/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，图中/r/na/r/n点时溶液呈中性，/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、并且二甲胺溶液过量、/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的浓度/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，电离方程式为/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，则电离平衡常数/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正确；/r/nC/r/n．强酸弱碱盐能够促进水的电离，并且其浓度越大、促进作用越强，二者恰好完全反应时生成/r/n(CH/r/n3/r/n)/r/n2/r/nNH/r/n2/r/nCl/r/n，此时溶液呈酸性，图中/r/na/r/n点呈中性、二甲胺过量，所以/r/na/r/n点以前的某点溶液中水的电离程度最大，故/r/nC/r/n错误；/r/nD/r/n．图中/r/nb/r/n点溶液的/r/npH=8/r/n、呈碱性，/r/nc/r/n(OH/r/n-/r/n)/r/n＞/r/nc/r/n(H/r/n+/r/n)/r/n，溶液中主要离子为/r/n(CH/r/n3/r/n)/r/n2/r/nNH/r/n和/r/nCl/r/n-/r/n，电荷关系为/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错误；/r/n故选/r/nB/r/n。/r/n2．（福建省南安高三二模）/r/n亚砷酸/r/n(H/r/n3/r/nAsO/r/n3/r/n)/r/n可以用于治疗白血病，其在溶液中存在多种微粒形态，将/r/nKOH/r/n溶液滴入亚砷酸溶液，各种微粒物质的量分数与溶液的/r/npH/r/n关系如图所示。下列说法不正确的是/r/nA．/r/n人体血液的/r/npH/r/n在/r/n7.35-7.45/r/n之间，患者用药后人体中含/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之间，随/r/npH/r/n增大/r/nHAsO/r/n水解程度减小/r/nC．/r/n通常情况下，/r/nH/r/n2/r/nAsO/r/n电离程度大于水解程度/r/nD．/r/n交点/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．由图可知，当溶液/r/npH/r/n在/r/n7.35-7.45/r/n之间时，砷元素的主要存在形式为/r/nH/r/n3/r/nAsO/r/n3/r/n，故/r/nA/r/n正确；/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之间时，溶液/r/npH/r/n增大，氢氧根离子浓度增大，平衡向逆反应方向移动，/r/nHAsO/r/n水解程度减小，故/r/nB/r/n正确；/r/nC/r/n．由图可知，当溶液中砷元素的主要存在形式为/r/nH/r/n2/r/nAsO/r/n时，溶液/r/npH/r/n约为/r/n11/r/n，溶液呈碱性，说明/r/nH/r/n2/r/nAsO/r/n的水解程度大于电离程度，故/r/nC/r/n错误；/r/nD/r/n．由图可知，交点/r/nb/r/n的溶液为碱性溶液，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，由电荷守恒关系/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正确；/r/n故选/r/nC/r/n。/r/n3．（福建省南安高三二模）/r/n常温下，下列说法正确的是/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，充分反应后，四种离子的浓度不变的是/r/n(/r/n忽略反应前后溶液体积的变化/r/n)/r/nB．/r/n水电离的/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，下列离子能大量共存：/r/n、/r/nNa/r/n+/r/n、/r/n、/r/nC．/r/n氢氧化铁溶于/r/nHI/r/n溶液中的离子方程式为：/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，下列离子能大量共存：/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．向溶液中加入具有强氧化性的过氧化钠固体，过氧化钠能将溶液中的亚硫酸根离子氧化为硫酸根离子，溶液中硫酸根离子浓度会增大，故/r/nA/r/n错误；/r/nB/r/n．水电离的/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的溶液可能为酸溶液，也可能为碱溶液，酸溶液中，氢离子与碳酸氢根离子反应，不能大量共存，碱溶液中，氢氧根离子与铵根离子、碳酸氢根离子反应，不能大量共存，故/r/nB/r/n错误；/r/nC/r/n．氢氧化铁溶于氢碘酸溶液的反应为氢氧化铁与氢碘酸溶液反应生成碘化亚铁、碘和水，反应的离子方程式为/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正确；/r/nD/r/n．在硫氢化钠溶液中，硫氢根离子与铝离子会发生双水解反应生成氢氧化铝沉淀和硫化氢气体，不能大量共存，故/r/nD/r/n错误；/r/n故选/r/nC/r/n。/r/n4．（重庆市第十一中学校高三二模）/r/n常温下，向/r/n20mL/r/n浓度均为/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，测得溶液的电阻率/r/n(/r/n溶液的电阻率越大，导电能力越弱/r/n)/r/n与加入氨水的体积/r/n(V)/r/n的关系如图。/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下列说法正确的是/r/nA．/r/n同浓度的/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过程，水的电离程度逐渐减小/r/nC．c/r/n点时，/r/nD．d/r/n点时，/r/n【KS5U答案】/r/nC/r/n【分析】/r/n电阻率与离子浓度成反比，即/r/na→b/r/n过程中溶液的导电性减弱，向混合溶液中加入等物质的量浓度的/r/nNH/r/n3/r/n•H/r/n2/r/nO/r/n溶液时，发生反应先后顺序为：/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溶液中电阻率增大、导电性减弱，/r/nb/r/n点最小，原因是溶液体积增大导致/r/nb/r/n点离子浓度减小，/r/nb/r/n点溶液中溶质为/r/nNH/r/n4/r/nX/r/n、/r/nCH/r/n3/r/nCOOH/r/n；继续加入/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是弱电解质，生成的/r/nCH/r/n3/r/nCOONH/r/n4/r/n是强电解质，导致溶液中离子浓度增大，溶液的电导性增大，/r/nc/r/n点时醋酸和一水合氨恰好完全反应生成醋酸铵，/r/nc/r/n点溶液中溶质为/r/nNH/r/n4/r/nX/r/n、/r/nCH/r/n3/r/nCOONH/r/n4/r/n，且二者的物质的量相等；/r/nd/r/n点溶液中溶质为等物质的量浓度的/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，据此分析解答。/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，则随着/r/nNH/r/n3/r/n•H/r/n2/r/nO/r/n的加入，酸碱反应生成盐，溶液导电性将增强、电阻率将减小，但图象上随着/r/nNH/r/n3/r/n•H/r/n2/r/nO/r/n的加入溶液电阻率增大、导电性反而减弱，说明原混合溶液中离子浓度更大，即/r/nHX/r/n为强电解质，同浓度的/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错误；/r/n

B/r/n．酸或碱都抑制水的电离，滴加/r/nNH/r/n3/r/n•H/r/n2/r/nO/r/n溶液的过程：/r/na→c/r/n为/r/nHX/r/n和/r/nCH/r/n3/r/nCOOH/r/n转化为/r/nNH/r/n4/r/nX/r/n、/r/nCH/r/n3/r/nCOONH/r/n4/r/n的过程，溶液的酸性减弱，水的电离程度增大，故/r/nB/r/n错误；/r/n

/r/nC/r/n．根据分析可知，/r/nc/r/n点溶液中溶质为/r/nNH/r/n4/r/nX/r/n、/r/nCH/r/n3/r/nCOONH/r/n4/r/n，且二者的物质的量相等，溶液为酸性，则/r/nc(H/r/n+/r/n)/r/n＞/r/nc(OH/r/n-/r/n)/r/n，根据电荷守恒/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正确；/r/n

D/r/n．/r/nd/r/n点溶液中溶质为等物质的量浓度的/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时，/r/nNH/r/n3/r/n•H/r/n2/r/nO/r/n转化为/r/nNH/r/n，/r/n40/r/n～/r/n60mL/r/n时，/r/nNH/r/n3/r/n•H/r/n2/r/nO/r/n过量，/r/nd/r/n点时，溶液体积共为/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错误；/r/n

/r/n故选：/r/nC/r/n。/r/n5．（青海高三三模）/r/n用下列实验装置/r/n(/r/n部分夹持装置略去/r/n)/r/n，能达到实验目的的是/r/nA．/r/n加热装置/r/nI/r/n中的烧杯，分离/r/nI/r/n2/r/n和高锰酸钾固体/r/nB．/r/n利用装置/r/nII/r/n除去/r/nCO/r/n中的/r/nCO/r/n2/r/nC．/r/n利用装置/r/nIII/r/n制备/r/nFe(OH)/r/n3/r/n胶体/r/nD．/r/n利用装置/r/nIV/r/n蒸干/r/nAlCl/r/n3/r/n溶液制无水/r/nAlCl/r/n3/r/n固体/r/n【KS5U答案】/r/nB/r/n【KS5U解析】/r/nA/r/n．/r/n/r/n加热装置/r/nI/r/n中的烧杯，高锰酸钾固体会分解，故/r/nA/r/n错误；/r/nB/r/n．/r/n/r/n利用装置/r/nII/r/n除去/r/nCO/r/n中的/r/nCO/r/n2/r/n，只有二氧化碳能与/r/nNaOH/r/n溶液反应，故/r/nB/r/n正确；/r/nC/r/n．/r/n/r/n利用装置/r/nIII/r/n制备/r/nFe(OH)/r/n3/r/n胶体，应使用酒精灯的外焰加热，加热至液体呈红褐色即可，故/r/nC/r/n错误；/r/nD/r/n．/r/n/r/n利用装置/r/nIV/r/n蒸干/r/nAlCl/r/n3/r/n溶液制无水/r/nAlCl/r/n3/r/n固体，氯化铝会水解，应在氯化氢气流中加热，故/r/nD/r/n错误；/r/n故选/r/nB/r/n。/r/n6．（四川成都市·成都七中高二零模）/r/n常温下，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，碳酸根离子浓度与氯化钙溶液体积的关系如图所示。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下列说法正确的是/r/n

/r/nA．/r/n图像中/r/nV/r/n0/r/n=20/r/n，/r/nm=5/r/nB．a/r/n点溶液：/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溶液的浓度变为/r/n0.05mol·L/r/n-1/r/n，则/r/nn/r/n点向/r/nc/r/n点方向迁移/r/nD．/r/n若用/r/nCdCl/r/n2/r/n溶液替代/r/nCaCl/r/n2/r/n溶液，则/r/nn/r/n点向/r/nb/r/n点方向迁移/r/n【KS5U答案】/r/nC/r/n【KS5U解析】/r/nA/r/n．/r/n/r/n图像中/r/nV/r/n0/r/n=20/r/n，/r/nNa/r/n2/r/nCO/r/n3/r/n溶液与/r/n/r/nCaCl/r/n2/r/n溶液恰好完全反应/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错误；/r/nB/r/n．/r/na/r/n点溶液：溶质为/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，溶液中存在电荷守恒/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错误；/r/nC/r/n．/r/n/r/n若/r/nNa/r/n2/r/nCO/r/n3/r/n溶液的浓度变为/r/n0.05mol·L/r/n-1/r/n，用的/r/n0.1mol·L/r/n-1/r/nCaCl/r/n2/r/n溶液体积减小，则/r/nn/r/n点向/r/nc/r/n点方向迁移，故/r/nC/r/n正确；/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，反应后，碳酸根离子浓度减小，/r/npC=-lgc(CO/r/n)/r/n增大，则/r/nn/r/n点向/r/nd/r/n点方向迁移，故/r/nD/r/n错误；/r/n故选/r/nC/r/n。/r/n7．（四川成都市·成都七中高二零模）/r/n下列有关电解质溶液的说法正确的是/r/nA．/r/n加水稀释，/r/nNa/r/n2/r/nS/r/n溶液中离子浓度均减小/r/nB．0.1mol/LNaOH/r/n溶液中滴加等体积等浓度醋酸溶液，溶液的导电性增强/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三种溶液的/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铵固体，则溶液中/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溶液呈碱性，加水稀释，/r/nNa/r/n2/r/nS/r/n溶液中氢离子浓度增大，故/r/nA/r/n错误；/r/nB/r/n．/r/n0.1mol/LNaOH/r/n溶液中滴加等体积等浓度醋酸溶液，产物为/r/n0.05mol/L/r/n的/r/nCH/r/n3/r/nCOONa/r/n溶液，离子浓度减小，溶液的导电性减小，故/r/nB/r/n错误；/r/nC/r/n．/r/nCH/r/n3/r/nCOO/r/n-/r/n、/r/n、/r/nClO/r/n-/r/n水解程度依次增强，/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三种溶液的浓度/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正确；/r/nD/r/n．向/r/n0.1mol·L/r/n-1/r/n的氨水中加入少量硫酸铵固体，铵根离子浓度增大，氨水电离平衡逆向移动，/r/nc(OH/r/n﹣/r/n)/r/n减小、/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减小，故/r/nD/r/n错误；/r/n选/r/nC/r/n。/r/n8．（浙江高三其他模拟）/r/n已知：/r/np/r/n=-lg/r/n。室温下向/r/nHX/r/n溶液中滴加等物质的量浓度的/r/nNaOH/r/n溶液，溶液/r/npH/r/n随/r/np/r/n变化关系如图所示。下列说法正确的是/r/nA．a/r/n点溶液中：/r/n10c(Na/r/n+/r/n)=c(HX)/r/nB．/r/n溶液中由水电离出的/r/nc(H/r/n+/r/n)/r/n：/r/na>b>c/r/nC．b/r/n点溶液中：/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当溶液呈中性时：/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点溶液中存在电荷守恒：/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，此时/r/np/r/n=-1/r/n，则/r/nc(X/r/n-/r/n)=10c(HX)/r/n，代入电荷守恒得/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，选项/r/nA/r/n错误；/r/nB/r/n．根据图示可知，/r/na/r/n、/r/nb/r/n、/r/nc/r/n均为酸性溶液，则溶质为/r/nHX/r/n和/r/nNaX/r/n，/r/npH<7/r/n的溶液中，/r/nHX/r/n的电离程度大于/r/nX/r/n-/r/n的水解程度，可只考虑/r/nH/r/n+/r/n对水的电离的抑制，溶液/r/npH/r/n越大氢离子浓度越小，水的电离程度越大，则溶液中水的电离程度：/r/na<b<c/r/n，选项/r/nB/r/n错误；/r/nC/r/n．/r/nb/r/n点溶液中存在电荷守恒：/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，此时/r/np/r/n=0/r/n，则/r/nc(X/r/n-/r/n)=c(HX)/r/n，代入电荷守恒有/r/nc(Na/r/n+/r/n)+c(H/r/n+/r/n)=c(HX)+c(OH/r/n-/r/n)/r/n，选项/r/nC/r/n正确；/r/nD/r/n．溶液中存在电荷守恒：/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，当溶液呈中性时/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，则/r/nc(X/r/n-/r/n)=c(HX)/r/n，此时/r/np/r/n=0/r/n，即为/r/nb/r/n点，但/r/nb/r/n点溶液呈酸性，不符合，选项/r/nD/r/n错误；/r/n答案选/r/nC/r/n。/r/n9．（河南新乡市·新乡县一中高三其他模拟）/r/n为测定某二元弱酸/r/nH/r/n2/r/nA/r/n与/r/nNaOH/r/n溶液反应过程中溶液/r/npH/r/n与粒子关系，在/r/n25℃/r/n时进行实验，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随溶液/r/npH/r/n的变化关系如图所示。下列说法正确的是/r/nA．/r/n直线/r/nII/r/n中/r/nX/r/n表示的是/r/nB．/r/n当/r/npH=3.81/r/n时，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当/r/npH=6.91/r/n时，对应的溶液中，/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．当/r/npH=0/r/n时，/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，故直线/r/nⅡ/r/n中/r/nX/r/n表示/r/n，/r/nA/r/n错误；/r/nB/r/n．当/r/nlgX=0/r/n时溶液的/r/npH=1.81/r/n，带入/r/nK/r/na1/r/n计算式中可求出/r/nK/r/na1/r/n=1×10/r/n-1.81/r/n，当/r/npH=3.81/r/n时，/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错误；/r/nC/r/n．与/r/nB/r/n项同理，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的电离能力强于其水解能力，电离生成的/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错误；/r/nD/r/n．当/r/npH=6.91/r/n时，对应的溶液中/r/nc/r/n(HA/r/n-/r/n)=/r/nc/r/n(A/r/n2-/r/n)/r/n，又因电荷守恒/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正确；/r/n综上所述答案为/r/nD/r/n。/r/n10．（安徽高三其他模拟）/r/n常温下，将/r/nHCl/r/n气体通入/r/n0.1mol/L/r/n氨水中，混合溶液中/r/npH/r/n与微粒浓度的对数值/r/n(lgc)/r/n和反应物物质的量之比/r/nX[X=/r/n]/r/n的关系如图所示/r/n(/r/n忽略溶液体积的变化/r/n)/r/n，下列说法正确的是/r/nA．NH/r/n3/r/n·H/r/n2/r/nO/r/n的电离平衡常数为/r/n10/r/n-9.25/r/nB．P/r/n2/r/n点由水电离出的/r/nc(H/r/n+/r/n)=1.0×10/r/n-7/r/nmol/L/r/nC．P/r/n3/r/n为恰好完全反应点，/r/nc(Cl/r/n-/r/n)+c(NH/r/n)=0.2mol/L/r/nD．P/r/n3/r/n之后，水的电离程度一直减小/r/n【KS5U答案】/r/nB/r/n【KS5U解析】/r/nA/r/n．一水合氨电离平衡状态下，溶液中铵根离子和氢氧根离子浓度相同时，氨水浓度为/r/n0.1mol/L/r/n，图象分析可知，/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的电离平衡常数/r/nK/r/nb/r/n=/r/n=/r/n=10/r/n-5/r/n，/r/nA/r/n错误；/r/nB/r/n．由图可知，/r/nP/r/n2/r/n点对应的溶液/r/npH=7/r/n，故由水电离出的/r/nc(H/r/n+/r/n)=1.0×10/r/n-7/r/nmol/L/r/n，/r/nB/r/n正确；/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错误；/r/nD/r/n．/r/nP/r/n3/r/n点为恰好完全反应，溶质为/r/nNH/r/n4/r/nCl/r/n，故之后，加入的/r/nHCl/r/n越来越多，由于/r/nH+/r/n对水解的抑制作用，水的电离程度减小，当/r/nHCl/r/n达到饱和溶液时，水的电离程度将不再改变，故不是一直减小，/r/nD/r/n错误；/r/n故/r/nB/r/n。/r/n11．（陕西宝鸡市·高三其他模拟）/r/n常温下，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，溶液中由水电离出的/r/nc/r/n水/r/n(OH/r/n-/r/n)/r/n的对数随加入/r/nCH/r/n3/r/nCOOH/r/n溶液体积的变化如图所示，下列说法正确的是/r/nA．H/r/n、/r/nF/r/n点溶液显中性/r/nB．E/r/n点溶液中由水电离的/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点溶液中离子浓度关系为/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点溶液中各离子浓度关系为/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氢氧化钠在溶液中抑制水的电离，向氢氧化钠溶液中加入醋酸，对水的电离的抑制作用逐渐减弱，当溶液为醋酸钠溶液时，水的电离程度最大，则/r/nG/r/n点为醋酸钠溶液；从/r/nE/r/n点到/r/nG/r/n点的反应过程中，所得溶液为氢氧化钠和醋酸钠的混合溶液，溶液为碱性；/r/nH/r/n点为醋酸和醋酸钠混合溶液，溶液呈中性。/r/n【KS5U解析】/r/nA/r/n．由分析可知，/r/nH/r/n点为醋酸和醋酸钠混合溶液，溶液呈中性，/r/nF/r/n点为氢氧化钠和醋酸钠的混合溶液，溶液为碱性，故/r/nA/r/n错误；/r/nB/r/n．氢氧化钠在溶液中抑制水的电离，/r/n0.01mol/r/n·/r/nL/r/n-1/r/n的氢氧化钠溶液中氢氧根离子的浓度为/r/n0.01mol/r/n·/r/nL/r/n-1/r/n，则溶液中水电离的氢离子浓度为/r/n10/r/n-12/r/nmol/r/n·/r/nL/r/n-1/r/n，故/r/nB/r/n错误；/r/nC/r/n．由分析可知，/r/nH/r/n点为醋酸和醋酸钠混合溶液，溶液呈中性，由电荷守恒关系/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，溶液中离子浓度关系为/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错误；/r/nD/r/n．由分析可知，/r/nG/r/n点为醋酸钠溶液，由电荷守恒关系/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，溶液中各离子浓度关系为/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