2021年高考化学真题和模拟题分类汇编专题12弱电解质的电离平衡【含答案】

专题12弱电解质的电离平衡/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下列物质属于弱电解质的是/r/nA．CO/r/n2/r/n B．H/r/n2/r/nO/r/n C．HNO/r/n3/r/n D．NaOH/r/n【KS5U答案】/r/nB/r/n【分析】/r/n在水溶液中或熔融状态下不能够完全电离的电解质叫做若电解质。/r/n【KS5U解析】/r/nA/r/n．/r/nCO/r/n2/r/n在水溶液中或熔融状态下不能够电离，为非电解质，/r/nA/r/n不符合题意；/r/nB/r/n．/r/nH/r/n2/r/nO/r/n在水溶液中或熔融状态下能够部分电离，为弱电解质，/r/nB/r/n符合题意；/r/nC/r/n．/r/nHNO/r/n3/r/n为一种强酸，在水溶液中或熔融状态下能够完全电离，为强电解质，/r/nC/r/n不符合题意；/r/nD/r/n．/r/nNaOH/r/n为一种强碱，在水溶液中或熔融状态下能够完全电离，为强电解质，/r/nD/r/n不符合题意；/r/n故答案选/r/nB/r/n。/r/n3．（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/n4．（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/r/n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/r/n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/n5．（浙江高考真题试卷）/r/n下列物质属于强电解质的是/r/nA．KOH/r/n B．H/r/n3/r/nPO/r/n4/r/n C．SO/r/n3/r/n D．CH/r/n3/r/nCHO/r/n【KS5U答案】/r/nA/r/n【分析】/r/n在水溶液中或熔融状态下能够完全电离的化合物为强电解质。/r/n【KS5U解析】/r/nA/r/n．/r/nKOH/r/n在水溶液中或熔融状态下能够完全电离出/r/nK/r/n+/r/n和/r/nOH/r/n-/r/n，/r/nKOH/r/n为强电解质，/r/nA/r/n符合题意；/r/nB/r/n．/r/nH/r/n3/r/nPO/r/n4/r/n在水溶液中或熔融状态下能不够完全电离，/r/nH/r/n3/r/nPO/r/n4/r/n为弱电解质，/r/nB/r/n不符合题意；/r/nC/r/n．/r/nSO/r/n3/r/n在水溶液中或熔融状态下不能电离，/r/nSO/r/n3/r/n属于非电解质，/r/nC/r/n不符合题意；/r/nD/r/n．/r/nCH/r/n3/r/nCHO/r/n在水溶液中或熔融状态下不能电离，/r/nCH/r/n3/r/nCHO/r/n属于非电解质，/r/nD/r/n不符合题意；/r/n故答案选/r/nA/r/n。/r/n二、多选题/r/n6．（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/r/n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/n2021年化学高考模拟题/r/n一、单选题/r/n1．（福建省南安高三二模）/r/n常温下，有体积相同的四种溶液：①/r/npH=2/r/n的/r/nCH/r/n3/r/nCOOH/r/n溶液；②/r/npH=2/r/n的盐酸；③/r/n0.01mol/L/r/n的醋酸溶液；④/r/n0.01mol/L/r/n的盐酸。下列说法正确的是/r/nA．/r/n把四种溶液稀释到相同/r/npH/r/n，所需水的体积：①/r/n>/r/n②/r/n=/r/n④/r/n=/r/n③/r/nB．/r/n四种溶液中和氢氧化钠的能力：①/r/n=/r/n②/r/n=/r/n③/r/n=/r/n④/r/nC．/r/n与镁条反应的起始速率的大小关系为：①/r/n>/r/n②/r/n=/r/n④/r/n>/r/n③/r/nD．/r/n与镁条反应生成相同体积的氢气所需的时间为：①/r/n</r/n②/r/n=/r/n④/r/n</r/n③/r/n【KS5U答案】/r/nD/r/n【KS5U解析】/r/nA/r/n．醋酸为弱酸，在溶液中存在电离平衡，盐酸是强酸，在溶液中完全电离，则等浓度的盐酸和醋酸溶液稀释到相同/r/npH/r/n，所需水的体积为/r/n④>③/r/n，故/r/nA/r/n错误；/r/nB/r/n．醋酸为弱酸，在溶液中存在电离平衡，盐酸是强酸，在溶液中完全电离，/r/npH=2/r/n的醋酸溶液的浓度大于/r/n0.01mol/L/r/n，中和氢氧化钠的能力大于其他三种溶液，故/r/nB/r/n错误；/r/nC/r/n．/r/npH=2/r/n的醋酸溶液和/r/npH=2/r/n的盐酸中氢离子浓度相同，与镁条反应的起始速率相等，故/r/nC/r/n错/r/n误；/r/nD/r/n．醋酸为弱酸，在溶液中存在电离平衡，盐酸是强酸，在溶液中完全电离，/r/npH=2/r/n的醋酸溶液的浓度大于/r/n0.01mol/L/r/n，与镁条反应过程中，氢离子浓度减小，电离平衡向右移动，氢离子浓度变化小，生成氢气的反应速率最快，/r/npH=2/r/n的盐酸和/r/n0.01mol/L/r/n的盐酸中氢离子浓度小于/r/npH=2/r/n的醋酸溶液，与镁条反应过程中，两种溶液生成氢气的反应速率相等，但慢于/r/npH=2/r/n的醋酸溶液，/r/n0.01mol/L/r/n的醋酸溶液中氢离子浓度最小，与镁条反应过程中，生成氢气的反应速率最慢，则生成相同体积的氢气所需的时间为/r/n①<②=④<③/r/n，故/r/nD/r/n正确；/r/n故选/r/nD/r/n。/r/n2．（青海高三三模）/r/n向/r/n1L/r/n浓度均为/r/n0.1mol·L/r/n-1/r/n的/r/nH/r/n3/r/nAsO/r/n3/r/n和/r/nH/r/n3/r/nAsO/r/n4/r/n水溶液中逐滴加入/r/n0.1mol·L/r/n-1/r/nNaOH/r/n溶液，含砷的各种微粒的分布分数/r/n(/r/n平衡时某物种的浓度占各物种浓度之和的分数/r/n)/r/n与/r/npH/r/n的关系分别如图所示，下列说法错误的是/r/nA．H/r/n3/r/nAsO/r/n4/r/n第二步电离方程式为/r/nH/r/n2/r/nAsO/r/nHAsO/r/n＋/r/nH/r/n＋/r/nB．/r/n往/r/nH/r/n3/r/nAsO/r/n3/r/n溶液中加/r/nNaOH/r/n溶液至/r/npH=10.0/r/n时，主要存在/r/nH/r/n2/r/nAsO/r/nC．pH=11/r/n时，对应的溶液中：/r/nc(H/r/n2/r/nAsO/r/n)/r/n＋/r/nc(HAsO/r/n)/r/n＋/r/nc(AsO/r/n)/r/n＋/r/nc(H/r/n3/r/nAsO/r/n3/r/n)=0.1mol·L/r/n-1/r/nD．H/r/n3/r/nAsO/r/n4/r/n第三步电离的电离常数为/r/nK/r/na3/r/n，若/r/na3/r/n=-lgK/r/na3/r/n，则/r/na3/r/n=11.5/r/n【KS5U答案】/r/nC/r/n【KS5U解析】/r/nA/r/n．/r/n/r/n多元弱酸是分步电离的，/r/nH/r/n3/r/nAsO/r/n4/r/n第二步电离方程式为/r/nH/r/n2/r/nAsO/r/nHAsO/r/n＋/r/nH/r/n＋/r/n，故/r/nA/r/n正确；/r/nB/r/n．/r/n/r/n由分步图，往/r/nH/r/n3/r/nAsO/r/n3/r/n溶液中加/r/nNaOH/r/n溶液至/r/npH=10.0/r/n时，/r/nH/r/n3/r/nAsO/r/n3/r/n减少，/r/nH/r/n2/r/nAsO/r/n增大，约占/r/n80%/r/n，主要存在/r/nH/r/n2/r/nAsO/r/n，故/r/nB/r/n正确；/r/nC/r/n．/r/n/r/n根据物料守恒，/r/npH=11/r/n时，对应的溶液中：/r/nn(H/r/n2/r/nAsO/r/n)/r/n＋/r/nn(HAsO/r/n)/r/n＋/r/nn(AsO/r/n)/r/n＋/r/nn(H/r/n3/r/nAsO/r/n3/r/n)=0.1/r/nmol·L/r/n-1/r/n×1L/r/n，由于加入的/r/nNaOH/r/n的体积末知，无法计算含砷粒子的总浓度，故/r/nC/r/n错误；/r/nD/r/n．/r/nH/r/n3/r/nAsO/r/n4/r/n第三步电离的电离常数为/r/nK/r/na3/r/n=/r/n/r/n，/r/npH=11.5/r/n时，/r/n=/r/n，若/r/na3/r/n=-lgK/r/na3/r/n，则/r/na3/r/n=-lgK/r/na3/r/n=-lg/r/n=11.5/r/n，故/r/nD/r/n正确；/r/n故选/r/nC/r/n。/r/n3．（安徽高三一模）25℃/r/n，向/r/n20mL0.1mol•L/r/n-1/r/n的弱碱/r/nBOH/r/n溶液/r/n(/r/nK/r/nb/r/n=1.0×10/r/n-5/r/n)/r/n中逐滴加入/r/n0.1mol•L/r/n-1/r/n盐酸，/r/npH/r/n～/r/nV/r/n曲线如图所示，下列说法正确的是/r/nA．a/r/n点到/r/nb/r/n点，水的电离程度先减小后增大/r/nB．a/r/n点时，/r/nc(Cl/r/n-/r/n)+c(OH/r/n-/r/n)=c(H/r/n+/r/n)+c(BOH)/r/nC．b/r/n点时，/r/nc(Cl/r/n-/r/n)=c(BOH)+c(B/r/n+/r/n)+c(H/r/n+/r/n)/r/nD．V=20mL/r/n时，/r/nc(Cl/r/n-/r/n)>c(B/r/n+/r/n)>c(OH/r/n-/r/n)>c(H/r/n+/r/n)/r/n【KS5U答案】/r/nB/r/n【KS5U解析】/r/nA/r/n．当/r/nBOH/r/n与盐酸恰好完全反应时，溶液中的溶质为/r/n0.05mol/LBCl/r/n，此时溶液中存在/r/nB/r/n+/r/n的水解，/r/nBOH/r/n的电离平衡常数/r/nK/r/nb/r/n=1.0×10/r/n-5/r/n，则/r/nB/r/n+/r/n的水解平衡常数为/r/nK/r/nh/r/n=/r/n=10/r/n-9/r/n；则此时溶液中满足/r/nK/r/nh/r/n=c(H/r/n+/r/n)c(BOH)/c(B/r/n+/r/n)≈c(H/r/n+/r/n)c(BOH)/0.05mol·L/r/n−1/r/n，/r/nc(H/r/n+/r/n)=c(BOH)/r/n，解得/r/nc(H/r/n+/r/n)=/r/n×10/r/n-5.5/r/nmol/L/r/n，所以此时溶液的/r/npH/r/n＞/r/n5/r/n，即/r/nb/r/n点/r/nBOH/r/n已经完全反应，/r/na/r/n点到/r/nb/r/n点，/r/nBOH/r/n逐渐和盐酸完全反应，然后盐酸过量，则水的电离程度先增大后减小，/r/nA/r/n错误；/r/nB/r/n．/r/na/r/n点时，溶液中存在电荷守恒/r/nc(Cl/r/n-/r/n)+c(OH/r/n-/r/n)=c(H/r/n+/r/n)+c(B/r/n+/r/n)/r/n，此时/r/npH=9/r/n，则/r/nc(OH/r/n-/r/n)=10/r/n-5/r/nmol/L/r/n，/r/nBOH/r/n的电离平衡常数/r/nK/r/nb/r/n=/r/n=1.0×10/r/n-5/r/n，所以此时/r/nc(BOH)=c(B/r/n+/r/n)/r/n，则/r/nc(Cl/r/n-/r/n)+c(OH/r/n-/r/n)=c(H/r/n+/r/n)+c(BOH)/r/n，/r/nB/r/n正确；/r/nC/r/n．当/r/nBOH/r/n和盐酸恰好完全反应时，溶液中存在物料守恒：/r/nc(Cl/r/n-/r/n)=c(BOH)+c(B/r/n+/r/n)/r/n，/r/nb/r/n点时盐酸过量，所加入的盐酸中存在/r/nc′(Cl/r/n-/r/n)=c′(H/r/n+/r/n)/r/n，若加入到混合溶液中后，/r/nB/r/n+/r/n的水解不受影响，则/r/nc(Cl/r/n-/r/n)=c(BOH)+c(B/r/n+/r/n)+c(H/r/n+/r/n)/r/n，但盐酸的电离会抑制/r/nB/r/n+/r/n，所以/r/nc(Cl/r/n-/r/n)/r/n＞/r/nc(BOH)+c(B/r/n+/r/n)+c(H/r/n+/r/n)/r/n，/r/nC/r/n错误；/r/nD/r/n．/r/nV=20mL/r/n时，溶液中溶质为/r/nBCl/r/n，溶液存在/r/nB/r/n+/r/n的水解，溶液显酸性，但水解是微弱的，所以/r/nc(Cl/r/n-/r/n)/r/n＞/r/nc(B/r/n+/r/n)/r/n＞/r/nc(H/r/n+/r/n)/r/n＞/r/nc(OH/r/n-/r/n)/r/n，/r/nD/r/n错误；/r/n综上所述答案为/r/nB/r/n。/r/n4．（浙江高三其他模拟）25/r/n℃时，下列说法正确的是/r/nA．H/r/n2/r/nA/r/n溶液与/r/nNaOH/r/n溶液按物质的量/r/n1/r/n：/r/n1/r/n恰好完全反应时，溶液酸碱性无法判断/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对水的电离没有影响/r/nC．/r/n醋酸的电离度：/r/npH=3/r/n的醋酸溶液大于/r/npH=4/r/n的醋酸溶液/r/nD．pH=2/r/n的/r/nHCl/r/n和/r/npH=12/r/n的/r/nBa(OH)/r/n2/r/n溶液等体积混合后，溶液显碱性/r/n【KS5U答案】/r/nA/r/n【KS5U解析】/r/nA/r/n．/r/nH/r/n2/r/nA/r/n溶液与/r/nNaOH/r/n溶液按物质的量/r/n1/r/n：/r/n1/r/n恰好完全反应后，生成的是/r/nNaHA/r/n，但是/r/nH/r/n2/r/nA/r/n的电离常数不知道，所以该溶液酸碱性无法判断，/r/nA/r/n正确；/r/nB/r/n．可溶性正盐/r/nBA/r/n溶液呈中性，只能说明HA和BOH的强弱相同，但是如果都是弱酸，且电离常数相同，则/r/nBA/r/n的阴阳离子水解程度相同，促进水的电离，/r/nB/r/n错误；/r/nC/r/n．醋酸是弱电解质，稀释促进电离，稀释后水的PH增大，/r/nPH/r/n越大说明越稀，水的电离程度越大，/r/nC/r/n错误；/r/nD/r/n．/r/npH=2/r/n的/r/nHCl/r/n和/r/npH=12/r/n的/r/nBa(OH)/r/n2/r/n当中，氢离子和氢氧根离子浓度相同，溶液等体积混合后，溶液显中性，D错误；/r/n故选/r/nA/r/n。/r/n5．（天津）/r/n草酸/r/n(H/r/n2/r/nC/r/n2/r/nO/r/n4/r/n)/r/n是一种二元弱酸。实验室配制了/r/n0.0100mol/r/n·/r/nL/r/n-1/r/nNa/r/n2/r/nC/r/n2/r/nO/r/n4/r/n标准溶液，现对/r/n25/r/n℃时该溶液的性质进行探究，下列所得结论正确的是/r/nA．/r/n测得/r/n0.0100mol/r/n·/r/nL/r/n-1/r/nNa/r/n2/r/nC/r/n2/r/nO/r/n4/r/n溶液/r/npH/r/n为/r/n8.6/r/n，此时溶液中存在：/r/nc(Na/r/n+/r/n)/r/n＞/r/nc(HC/r/n2/r/nO/r/n)/r/n＞/r/nc(C/r/n2/r/nO/r/n)/r/n＞/r/nc(H/r/n+/r/n)/r/nB．/r/n向该溶液中滴加稀盐酸至溶液/r/npH=7/r/n，/r/n/r/n此时溶液中存在，/r/nc(Na/r/n+/r/n)=c(HC/r/n2/r/nO/r/n)+2c(C/r/n2/r/nO/r/n)/r/nC．/r/n已知/r/n25/r/n℃时/r/nK/r/nsp/r/n(CaC/r/n2/r/nO/r/n4/r/n)=2.5/r/n×/r/n10/r/n-9/r/n向该溶液中加入等体积/r/n0.0200mol/r/n·/r/nL/r/n-1/r/nCaCl/r/n2/r/n溶液，所得/r/n上层清液中/r/nc(C/r/n2/r/nO/r/n)=5.00/r/n×/r/n10/r/n-7/r/nmol/r/n·/r/nL/r/n-1/r/nD．/r/n向该溶液中加入足量稀硫酸酸化后，再滴加/r/nKMnO/r/n4/r/n溶液，发生反应的离子方程式为/r/nC/r/n2/r/nO/r/n+4MnO/r/n+14H/r/n+/r/n=2CO/r/n2/r/n↑+4Mn/r/n2+/r/n+7H/r/n2/r/nO/r/n【KS5U答案】/r/nC/r/n【KS5U解析】/r/nA/r/n．/r/nNa/r/n2/r/nC/r/n2/r/nO/r/n4/r/n溶液/r/npH/r/n为/r/n8.6/r/n，/r/nC/r/n2/r/nO/r/n水解导致溶液显碱性，但其水解程度较小，故/r/nc(HC/r/n2/r/nO/r/n)/r/n＜/r/nc(C/r/n2/r/nO/r/n)/r/n，Ａ项错误；/r/nＢ．向该溶液中滴加稀盐酸至溶液/r/npH=7/r/n，则溶液呈中性，则/r/nc(H/r/n+/r/n)=c(OH/r/n-/r/n)/r/n，溶液中电荷守恒，/r/nc(H/r/n+/r/n)+c(Na/r/n+/r/n)=c(HC/r/n2/r/nO/r/n)+2c(C/r/n2/r/nO/r/n)+c(OH/r/n-/r/n)+c(Cl/r/n-/r/n)/r/n，即/r/nc(Na/r/n+/r/n)=c(HC/r/n2/r/nO/r/n)+2c(C/r/n2/r/nO/r/n)+c(Cl/r/n-/r/n)/r/n，/r/nB/r/n项错误；/r/nC/r/n．向溶液中加入等体积/r/n0.0200mol·L/r/n-1/r/nCaCl/r/n2/r/n溶液，二者以/r/n1:1/r/n反应，则/r/nCaCl/r/n2/r/n剩余，混合溶液中/r/nc(Ca/r/n2+/r/n)=/r/nmol·L/r/n-1/r/n=0.005mol·L/r/n-1/r/n，/r/nc(C/r/n2/r/nO/r/n)=/r/nmol·L/r/n-1/r/n=5.00/r/n×/r/n10/r/n-7/r/nmol·L/r/n-1/r/n＜/r/n5.00/r/n×/r/n10/r/n-5/r/nmol·L/r/n-1/r/n，/r/nC/r/n项正确；/r/nD/r/n．草酸具有还原性，酸性高锰酸钾溶液具有强氧化性，二者发生氧化还原反应生成二氧化碳、锰离子和水，草酸是弱酸写化学式，故发生反应的离子方程式为/r/n5H/r/n2/r/nC/r/n2/r/nO/r/n4/r/n+2MnO/r/n+6H/r/n+/r/n=10CO/r/n2/r/n↑+2Mn/r/n2+/r/n+8H/r/n2/r/nO/r/n，/r/nD/r/n项错误；/r/n答案选/r/nC/r/n。/r/n6．（阜新市第二高三其他模拟）/r/n常温下，下列说法正确的是/r/nA．/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/n增大/r/nB．/r/n向冰醋酸中逐滴加水，溶液的导电性、醋酸的电离程度均先增大后减小/r/nC．0.1mol•L/r/n-1/r/nNa/r/n2/r/nCO/r/n3/r/n溶液和/r/n0.1mol•L/r/n-1/r/nNaHCO/r/n3/r/n溶液等体积混合：/r/nc(CO/r/n)+2c(OH/r/n-/r/n)=c(HCO/r/n)+3c(H/r/n2/r/nCO/r/n3/r/n)+2c(H/r/n+/r/n)/r/nD．/r/n将/r/nCH/r/n3/r/nCOONa/r/n、/r/nHCl/r/n两溶液混合后，溶液呈中性，则溶液中/r/nc(Na/r/n+/r/n)<c(Cl/r/n-/r/n)/r/n【KS5U答案】/r/nC/r/n【KS5U解析】/r/nA/r/n．向/r/n0.1mol•L/r/n-1/r/n氨水中加入少量硫酸铵固体，铵根离子浓度增大，根据一水合氨的电离平衡常数可知所求等式/r/n=/r/n，而温度不变时/r/nK/r/nb/r/n(NH/r/n3/r/n•H/r/n2/r/nO)/r/n大小不变，则该比值减小，故/r/nA/r/n错误；/r/nB/r/n．冰醋酸/r/n(/r/n不含自由移动的离子/r/n)/r/n不导电，往冰醋酸中加水，醋酸的电离平衡右移，开始时离子浓度增大，后来减小，故溶液的导电性开始增大后来减小；因醋酸的电离平衡一直右移，故电离程度始终增大，故/r/nB/r/n错误；/r/nC/r/n．/r/n0.1mol•L/r/n-1/r/nNa/r/n2/r/nCO/r/n3/r/n溶液与/r/n0.1mol•L/r/n-1/r/nNaHCO/r/n3/r/n溶液等体积混合，根据电荷守恒和物料守恒可得/r/nc(CO/r/n3/r/n2-/r/n)+2c(OH/r/n-/r/n)=c(HCO/r/n3/r/n-/r/n)+3c(H/r/n2/r/nCO/r/n3/r/n)+2c(H/r/n+/r/n)/r/n，故/r/nC/r/n正确；/r/nD/r/n．根据电荷守恒，/r/nc(Na/r/n+/r/n)+c(H/r/n+/r/n)=c(OH/r/n-/r/n)+c(Cl/r/n-/r/n)+c(CH/r/n3/r/nCOO/r/n-/r/n)/r/n，溶液呈中/r/n/r/n性，即/r/nc(H/r/n+/r/n)=c(OH/r/n-/r/n)/r/n，则/r/nc(Na/r/n+/r/n)=c(Cl/r/n-/r/n)+c(CH/r/n3/r/nCOO/r/n-/r/n)/r/n，溶液中/r/nc(Na/r/n+/r/n)>c(Cl/r/n-/r/n)/r/n，故/r/nD/r/n错误；/r/n选/r/nC/r/n。/r/n7．（天津高三三模）/r/n设/r/nN/r/nA/r/n为阿伏加德罗常数的值，下列说法正确的是/r/nA．0.1molCl/r/n2/r/n与足量的/r/nH/r/n2/r/nO/r/n反应，转移的电子数为/r/n0.2N/r/nA/r/nB．/r/n电解/r/nAgNO/r/n3/r/n溶液当阳极产生标况下/r/n2.24L/r/n气体时转移的电子数为/r/n0.4N/r/nA/r/nC．25/r/n℃、/r/n101kPa/r/n下，/r/n1mol/r/n乙炔和甲醛的混合气体含有的共用电子对数为/r/n3N/r/nA/r/nD．25/r/n℃，/r/n0.1mol/L/r/n的某酸/r/nHA/r/n溶液中，/r/nc(A/r/n-/r/n)=0.1mol/L/r/n【KS5U答案】/r/nB/r/n【KS5U解析】/r/nA/r/n．氯气和水的反应为可逆反应，不能进行彻底，转移的电子数小于/r/n0.2N/r/nA/r/n，/r/nA/r/n错误；/r/nB/r/n．电解/r/nAgNO/r/n3/r/n溶液，阳极生成的气体为氧气，物质的量/r/nn=V/Vm=2.24L÷22.4L/mol=0.1mol/r/n，电极反应式为/r/n4OH/r/n-/r/n-4e/r/n-/r/n=2H/r/n2/r/nO+O/r/n2/r/n↑/r/n，则转移的电子数为/r/n0.4N/r/nA/r/n个，/r/nB/r/n正确；/r/nC/r/n．/r/n1mol/r/n乙炔共用电子对数为/r/n5N/r/nA/r/n，/r/n1mol/r/n甲醛共用电子对数为/r/n4N/r/nA/r/n，故/r/n1mol/r/n乙炔和甲醛的混合气体含有的共用电子对数无法计算，/r/nC/r/n错误；/r/nD/r/n．未知/r/nHA/r/n是强酸还是弱酸，无法求算/r/nc(A/r/n-/r/n)/r/n，/r/nD/r/n错误；/r/n故选：/r/nB/r/n。/r/n8．（天津高三三模）/r/n下列说法错误的是/r/nA．/r/n将/r/n0.1mol/r/n·/r/nL/r/n-1/r/n的/r/nCH/r/n3/r/nCOOH/r/n溶液加水稀释，/r/n变大/r/nB．/r/n向/r/n0.1mol/r/n·/r/nL/r/n-1/r/n的氨水中通氨气，则/r/nNH/r/n3/r/n·/r/nH/r/n2/r/nO/r/n的电离程度增大/r/nC．/r/n已知/r/n[Cu(H/r/n2/r/nO)/r/n4/r/n]/r/n2+/r/n(/r/n蓝色/r/n)+4Cl/r/n-/r/n[CuCl/r/n4/r/n]/r/n2-/r/n(/r/n黄色/r/n)+4H/r/n2/r/nO/r/n，加水，溶液颜色变蓝/r/nD．/r/n分别向甲容器/r/n(/r/n恒温恒容/r/n)/r/n中充入/r/nlmolPCl/r/n5/r/n，乙容器/r/n(/r/n绝热恒容/r/n)/r/n充入/r/nPCl/r/n3/r/n和/r/nCl/r/n2/r/n各/r/nlmol/r/n，发生反应/r/nPCl/r/n5/r/nPCl/r/n3/r/n(g)+Cl/r/n2/r/n(g)/r/n△/r/nH/r/n>0/r/n，平衡时/r/nK/r/n(/r/n甲/r/n)</r/nK/r/n(/r/n乙/r/n)/r/n【KS5U答案】/r/nB/r/n【KS5U解析】/r/nA/r/n．/r/nCH/r/n3/r/nCOOH/r/n溶液加水稀释，平衡向正反应方向移动，/r/nc(CH/r/n3/r/nCOO/r/n-/r/n)/r/n浓度减小，/r/n变大，/r/nA/r/n正确；/r/nB/r/n．向氨水中通入/r/nNH/r/n3/r/n，一水合氨浓度增大，一水合氨的电离程度减小，/r/nB/r/n错误；/r/nC/r/n．加水即增大生成物浓度，平衡逆向移动，/r/n[Cu(H/r/n2/r/nO)/r/n4/r/n]/r/n2+/r/n浓度增大，溶液颜色变蓝，/r/nC/r/n正确；/r/nD/r/n．乙容器充入/r/nPCl/r/n3/r/n和/r/nCl/r/n2/r/n即为/r/nPCl/r/n3/r/n(g)+Cl/r/n2/r/n(g)/r/nPCl/r/n5/r/n△/r/nH/r/n<0/r/n，该反应是放热反应，在绝热恒容条件下进行，升温使平衡逆向移动，则平衡时乙容器中/r/nPCl/r/n3/r/n和/r/nCl/r/n2/r/n浓度大于甲容器，/r/nK/r/n(/r/n甲/r/n)</r/nK/r/n(/r/n乙/r/n)/r/n，/r/nD/r/n正确。/r/n故选/r/nB/r/n。/r/n9．（天津高三一模）/r/n下列说法正确的是/r/nA．/r/n常温下，由水电离产生的/r/nc(H/r/n+/r/n)=10/r/n-12/r/nmol/L/r/n的溶液中：/r/nK/r/n+/r/n、/r/nNa/r/n+/r/n、/r/nClO/r/n-/r/n、/r/nI/r/n-/r/n大量共存/r/nB．/r/n苯酚溶液中：/r/nCl/r/n-/r/n、/r/nNH/r/n、/r/nCO/r/n可以大量共存/r/nC．/r/n次氯酸钠溶液中通少量/r/nCO/r/n2/r/n：/r/nClO/r/n-/r/n+CO/r/n2/r/n+H/r/n2/r/nO=HClO+HCO/r/nD．/r/n用银氨溶液检验醛基的离子方程式：/r/nRCHO+2Ag(NH/r/n3/r/n)/r/n2/r/nOH/r/nRCOO/r/n-/r/n+NH/r/n+2Ag↓+3NH/r/n3/r/n+H/r/n2/r/nO/r/n【KS5U答案】/r/nC/r/n【KS5U解析】/r/nA/r/n．/r/nClO/r/n-/r/n、/r/nI/r/n-/r/n发生氧化还原反应而不能大量共存；常温下，由水电离产生的/r/nc(H/r/n+/r/n)=10/r/n-12/r/nmol/L/r/n的溶液，说明水的电离受到了抑制，溶液可能因溶质电离显酸性，也可能因溶质电离显碱性，酸性溶液中/r/nH/r/n+/r/n能与/r/nClO/r/n-/r/n、发生反应，不能大量共存，/r/nA/r/n错误；/r/nB/r/n．/r/n/r/n苯酚的酸性介于碳酸的一级电离与二级电离之间，则苯酚能与/r/nCO/r/n反应生成/r/nHCO/r/n，不可以大量共存，/r/nB/r/n错误；/r/nC/r/n．/r/n/r/n次氯酸的电离平衡常数介于碳酸的一级电离平衡常数与二级电离平衡常数之间，则次氯酸钠溶液中通少量/r/nCO/r/n2/r/n：/r/nClO/r/n-/r/n+CO/r/n2/r/n+H/r/n2/r/nO=HClO+HCO/r/n，/r/nC/r/n正确；/r/nD/r/n．/r/n/r/n醛基能被银氨溶液氧化，则用银氨溶液检验醛基的离子方程式：/r/nRCHO+2Ag(NH/r/n3/r/n)/r/n+2OH/r/n-/r/nRCOO/r/n-/r/n+NH/r/n+2Ag↓+3NH/r/n3/r/n+H/r/n2/r/nO/r/n，/r/nD/r/n错误。/r/n答案选/r/nC/r/n。/r/n10．（天津高三二模）/r/n下列说法不正确的是/r/nA．/r/n某离子被沉淀完全是指该离子在溶液中的浓度为/r/n0/r/nB．/r/n将/r/nKCl/r/n溶液从常温加热至/r/n100/r/n℃，溶液的/r/npH/r/n变小但仍保持中性/r/nC．/r/n常温下，/r/nNaCN/r/n溶液呈碱性，说明/r/nHCN/r/n是弱电解质/r/nD．/r/n常温下，/r/npH/r/n为/r/n3/r/n的醋酸溶液中加入醋酸钠固体，溶液/r/npH/r/n增大/r/n【KS5U答案】/r/nA/r/n【KS5U解析】/r/nA/r/n．某离子被沉淀完全是指该离子在溶液中的浓度小于/r/n，/r/nA/r/n说法错误；/r/nＢ/r/n．将/r/nKCl/r/n溶液从常温加热至/r/n100℃/r/n，水电离产生的氢离子浓度增大，则溶液的/r/npH/r/n变小但仍保持中性，/r/nB/r/n说法正确；/r/nＣ/r/n．常温下，/r/nNaCN/r/n溶液呈碱性，说明/r/nCN/r/n-/r/n水解产生/r/nHCN/r/n和氢氧根离子，则/r/nHCN/r/n是弱电解质，/r/nC/r/n说法正确；/r/nＤ/r/n．常温下，/r/npH/r/n为/r/n3/r/n的醋酸溶液中加入醋酸钠固体，导致溶液中醋酸根离子浓度增大，醋酸的电离平衡逆向移动，氢离子浓度减小，则溶液/r/npH/r/n增大，/r/nD/r/n说法正确。/r/n故选/r/nA/r/n。/r/n11．（广东广州市·华南师大附中高三三模）/r/n常温下，用/r/n0.1mol·L/r/n-1/r/n氨水滴定/r/n10mL/r/n浓度均为/r/n0.1mol·L/r/n-1/r/n的/r/nHCl/r/n和/r/nCH/r/n3/r/nCOOH/r/n的混合液，已知醋酸的电离常数为/r/nK/r/na/r/n=1.8×10/r/n-5/r/n，下列说法错误的是/r/nA．/r/n在氨水滴定前，混合溶液/r/nc(Cl/r/n-/r/n)/r/n＞/r/nc(CH/r/n3/r/nCOOH)/r/nB．/r/n在氨水滴定前，混合溶液/r/nc(CH/r/n3/r/nCOO/r/n-/r/n)≈Ka/r/nC．/r/n当滴入氨水/r/n10mL/r/n时，/r/nc(NH/r/n)/r/n＋/r/nc(NH/r/n3/r/n·H/r/n2/r/nO)=c(CH/r/n3/r/nCOOH)+c(CH/r/n3/r/nCOO/r/n-/r/n)/r/nD．/r/n当溶液呈中性时，氨水滴入量等于/r/n20mL/r/n，且/r/nc(NH/r/n)/r/n＜/r/nc(Cl/r/n-/r/n)/r/n【KS5U答案】/r/nD/r/n【KS5U解析】/r/nA/r/n．/r/nHCl/r/n完全电离，故/r/nc/r/n(Cl/r/n-/r/n)=/r/nc/r/n(HCl)/r/n，/r/nCH/r/n3/r/nCOOH/r/n微弱电离损耗，故/r/nc/r/n(Cl/r/n-/r/n)/r/n＞/r/nc/r/n(CH/r/n3/r/nCOOH)/r/n，/r/nA/r/n正确；/r/nB/r/n．由电离方程式/r/nCH/r/n3/r/nCOOH/r/nCH/r/n3/r/nCOO/r/n-/r/n+H/r/n+/r/n，知/r/nK/r/na/r/n=/r/n，由于/r/nCH/r/n3/r/nCOOH/r/n电离是微弱的，故溶液中/r/nc/r/n(H/r/n+/r/n)/r/n≈/r/nc/r/n(HCl)/r/n≈/r/nc/r/n(CH/r/n3/r/nCOOH)/r/n，故/r/nK/r/na/r/n≈/r/nc/r/n(CH/r/n3/r/nCOO/r/n/r/n)/r/n，/r/nB/r/n正确；/r/nC/r/n．当加入氨水/r/n10mL/r/n时，/r/nHCl/r/n恰好反应，故反应后溶液组成为：/r/nNH/r/n4/r/nCl/r/n、/r/nCH/r/n3/r/nCOOH/r/n，根据物料守恒知，/r/nc/r/n(/r/n)+/r/nc/r/n(NH/r/n3/r/n·/r/nH/r/n2/r/nO)=/r/nc/r/n(CH/r/n3/r/nCOOH)+/r/nc/r/n(CH/r/n3/r/nCOO/r/n/r/n)/r/n，/r/nC/r/n正确；/r/nD/r/n．当氨水加入/r/n20mL/r/n时，/r/nHCl/r/n、/r/nCH/r/n3/r/nCOOH/r/n恰好反应，反应后溶液组成为：/r/nNH/r/n4/r/nCl/r/n、/r/nCH/r/n3/r/nCOONH/r/n4/r/n，/r/nCH/r/n3/r/nCOONH/r/n4/r/n使溶液显中性，/r/nNH/r/n4/r/nCl/r/n使溶液显酸性，故此时溶液不可能显中性，要使溶液显中性，加入氨水体积应大于/r/n20mL/r/n（或依据溶液显中性时，根据电荷守恒：/r/nc/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(Cl/r/n-/r/n)+/r/nc/r/n(CH/r/n3/r/nCOO/r/n/r/n)/r/n，知/r/nc/r/n(/r/n)=/r/nc/r/n(Cl/r/n-/r/n)+/r/nc/r/n(CH/r/n3/r/nCOO/r/n/r/n)/r/n，则/r/nc/r/n(/r/n)/r/n＞/r/nc/r/n(Cl/r/n-/r/n)/r/n），/r/nD/r/n错误；/r/n故答案选/r/nD/r/n。/r/n12．（四川广元市·高三三模）/r/n对下列反应的推断或解释正确的是/r/n操作/r/n实验现象/r/n推断或解释/r/nA/r/n将少量饱和硼酸溶液滴加到碳酸钠溶液中/r/n无气泡/r/n酸性：/r/nH/r/n2/r/nCO/r/n3/r/n＞/r/nH/r/n3/r/nBO/r/n3/r/nB/r/n将/r/nC/r/n2/r/nH/r/n4/r/n通入溴的四氯化碳溶液中/r/n溴的四氯化碳溶液褪色/r/nC/r/n2/r/nH/r/n4/r/n与溴发生了加成反应/r/nC/r/n同温同压下用/r/npH/r/n试纸测定相同浓度的碳酸钠和乙酸铵溶液的酸碱性/r/n碳酸钠溶液显碱性，乙酸铵溶液显中性/r/n碳酸钠溶液发生了水解，乙酸铵溶液没有水解/r/nD/r/n向均盛有/r/n2mL5%H/r/n2/r/nO/r/n2/r/n溶液的两支试管中分别滴入/r/n0.3mol/LFeCl/r/n3/r/n和/r/n0.2mol/LCuCl/r/n2/r/n溶液各/r/n1mL/r/n前者生成气泡的速率更快/r/n催化效果：/r/nFe/r/n3+/r/n＞/r/nCu/r/n2+/r/nA．A/r/n B．B/r/n C．C/r/n D．D/r/n【KS5U答案】/r/nB/r/n【KS5U解析】/r/nA/r/n．没有气泡，说明无二氧化碳生成，但由于硼酸是少量的，故可能生成碳酸氢钠，或不反应，不能比较碳酸和硼酸的酸性强弱，故/r/nA/r/n错误；/r/nB/r/n．将/r/nC/r/n2/r/nH/r/n4/r/n通入溴的四氯化碳溶液中，溴的四氯化碳溶液褪色，/r/nC/r/n2/r/nH/r/n4/r/n与溴发生了加成反应，故/r/nB/r/n正确；/r/nC/r/n．乙酸铵溶液显中性是因为醋酸根离子的水解程度与铵根离子的水解程度相等，故/r/nC/r/n错误；/r/nD/r/n．/r/nFeCl/r/n3/r/n、/r/nCuCl/r/n2/r/n溶液的浓度不同，可为浓度的影响，应加入浓度相同的/r/nFeCl/r/n3/r/n、/r/nCuCl/r/n2/r/n，当研究某一因素对反应速率的影响时，应用控制变量法，保证其他因素相同，故/r/nD/r/n错误。/r/n故选/r/nB/r/n。/r/n13．（黑龙江哈尔滨市校高三其他模拟）/r/n根据下列各图曲线表征的信息，得出的结论错误的是/r/nA．/r/n图/r/n1/r/n表示常温下向体积为/r/n10mL0.1mol·L/r/n-1/r/n的/r/nNaOH/r/n溶液中逐滴加入/r/n0.1mol·L/r/n-1/r/nCH/r/n3/r/nCOOH/r/n溶液后溶液的/r/npH/r/n变化曲线，则/r/nc/r/n点处有/r/nc(CH/r/n3/r/nCOOH)/r/n＋/r/n2c(H/r/n＋/r/n)=2c(OH/r/n-/r/n)/r/n＋/r/nc(CH/r/n3/r/nCOO/r/n-/r/n)/r/nB．/r/n图/r/n2/r/n表示用水稀释/r/npH/r/n相同的盐酸和醋酸时溶液的/r/npH/r/n变化曲线，其中/r/nⅠ/r/n表示醋酸，/r/nⅡ/r/n表示盐酸，且溶液导电性：/r/nc/r/n＞/r/nb/r/n＞/r/na/r/nC．/r/n图/r/n3/r/n表示/r/nH/r/n2/r/n与/r/nO/r/n2/r/n发生反应过程中的能量变化，表示/r/nH/r/n2/r/n燃烧热的/r/nΔH=-285.8kJ·mol/r/n-1/r/nD．/r/n结合图/r/n4/r/n分析可知，向/r/n100mL/r/n含/r/nCu/r/n2+/r/n、/r/nMn/r/n2+/r/n、/r/nFe/r/n2+/r/n、/r/nZn/r/n2+/r/n均为/r/n10/r/n-5/r/nmol/L/r/n的混合溶液中逐滴加入/r/n1×10/r/n-4/r/nmol/LNa/r/n2/r/nS/r/n溶液，/r/nCu/r/n2+/r/n先沉淀/r/n【KS5U答案】/r/nB/r/n【KS5U解析】/r/nA/r/n．图/r/n1/r/n中/r/nc/r/n点加入/r/n0.1mol/L20mLCH/r/n3/r/nCOOH/r/n溶液与/r/n10mL0.1mol/LNaOH/r/n充分反应后得到等物质的量浓度的/r/nCH/r/n3/r/nCOONa/r/n和/r/nCH/r/n3/r/nCOOH/r/n的混合液，溶液中的电荷守恒为/r/nc/r/n(Na/r/n+/r/n)+/r/nc/r/n(H/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/n，物料守恒为/r/n2/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(CH/r/n3/r/nCOOH)/r/n，两式整理消去/r/nc/r/n(Na/r/n+/r/n)/r/n得/r/nc/r/n(CH/r/n3/r/nCOOH)/r/n＋/r/n2/r/nc/r/n(H/r/n＋/r/n)=2/r/nc/r/n(OH/r/n-/r/n)/r/n＋/r/nc/r/n(CH/r/n3/r/nCOO/r/n-/r/n)/r/n，/r/nA/r/n正确；/r/nB/r/n．盐酸属于强酸溶液，醋酸溶液属于弱酸溶液，加水稀释促进醋酸的电离，将/r/npH/r/n相同的盐酸和醋酸稀释相同的倍数后，/r/npH/r/n较大的为盐酸、/r/npH/r/n较小的为醋酸，则图/r/n2/r/n中/r/nI/r/n表示盐酸，/r/nII/r/n表示醋酸，/r/na/r/n、/r/nb/r/n、/r/nc/r/n三点/r/npH/r/n由大到小的顺序为/r/nc/r/n＞/r/nb/r/n＞/r/na/r/n，三点溶液中离子物质的量浓度由大到小的顺序为/r/na>b>c/r/n，溶液的导电性：/r/na>b>c/r/n，/r/nB/r/n错误；/r/nC/r/n．根据图/r/n3/r/n可写出热化学方程式为/r/n2H/r/n2/r/n(g)+O/r/n2/r/n(g)=2H/r/n2/r/nO(l)∆/r/nH/r/n=-571.6kJ/mol/r/n，则/r/n1molH/r/n2/r/n(g)/r/n完全燃烧生成/r/nH/r/n2/r/nO(l)/r/n放出/r/n285.8kJ/r/n的热量，表示/r/nH/r/n2/r/n燃烧热的/r/nΔ/r/nH/r/n=-285.8kJ·mol/r/n-1/r/n，/r/nC/r/n正确；/r/nD/r/n．根据题给图像知，/r/nKsp(CuS)/r/n＜/r/nKsp(FeS)/r/n＜/r/nKsp(ZnS)/r/n，向/r/n100mL/r/n含/r/nCu/r/n2+/r/n、/r/nMn/r/n2+/r/n、/r/nFe/r/n2+/r/n、/r/nZn/r/n2+/r/n均为/r/n10/r/n-5/r/nmol/L/r/n的混合溶液中逐滴加入/r/n1×10/r/n-4/r/nmol/LNa/r/n2/r/nS/r/n溶液，/r/nD/r/n正确；/r/n答案选/r/nB/r/n。/r/n14．（浙江高三其他模拟）25/r/n℃时，下列说法正确的是/r/nA．/r/n分别取/r/n20.00mL0.1000mol/L/r/n的盐酸和醋酸溶液，以酚酞作指示剂，用/r/n0.1000mol/LNaOH/r/n标准溶液滴定至终点时，两者消耗的/r/nNaOH/r/n溶液体积相等/r/nB．/r/n将/r/npH=3/r/n的醋酸溶液稀释到原体积的/r/n10/r/n倍后，溶液的/r/npH=4/r/nC．/r/n均为/r/n0.1mol/L/r/n的/r/nNa/r/n2/r/nSO/r/n3/r/n、/r/nNa/r/n2/r/nCO/r/n3/r/n、/r/nH/r/n2/r/nSO/r/n4/r/n溶液中阴离子的浓度依次减小/r/nD．/r/n常温下/r/npH=11/r/n的碱溶液中水电离产生的/r/nc/r/n(H/r/n+/r/n)/r/n是纯水电离产生的/r/nc/r/n(H/r/n+/r/n)/r/n的/r/n10/r/n4/r/n倍/r/n【KS5U答案】/r/nA/r/n【KS5U解析】/r/nA/r/n．盐酸和醋酸均是一元酸，在二者量相等以及选用相同指示剂的条件下，用/r/n0.1000mol/LNaOH/r/n标准溶液滴定至终点时，两者消耗的/r/nNaOH/r/n溶液体积相等，故/r/nA/r/n正确；/r/nB/r/n．醋酸是弱酸，稀释促进电离，将/r/npH=3/r/n的醋酸溶液稀释到原体积的/r/n10/r/n倍后，溶液的/r/n3/r/n＜/r/npH/r/n＜/r/n4/r/n，故/r/nB/r/n错误；/r/nC/r/n．硫酸是强酸，只电离不水解，亚硫酸根离子的水解程度小于碳酸根离子的水解程度，水解程度越大，阴离子浓度越大，则均为/r/n0.1mol/L/r/n的/r/nNa/r/n2/r/nCO/r/n3/r/n、/r/nNa/r/n2/r/nSO/r/n3/r/n、/r/nH/r/n2/r/nSO/r/n4/r/n溶液中阴离子的浓度依次减小，故/r/nC/r/n错误；/r/nD/r/n．碱抑制水的电离，常温下/r/npH=11/r/n的碱溶液中水电离产生的/r/nc/r/n(H/r/n+/r/n)/r/n小于纯水电离产生的/r/nc/r/n(H/r/n+/r/n)/r/n，故/r/nD/r/n错误。/r/n答案选/r/nA/r/n。/r/n15．（江西抚州市·临川一中高三其他模拟）/r/n锂金属电池的电解液在很大程度上制约着锂电池的发展，某种商业化锂电池的电解质的结构如图所示。已知短周期主族元素/r/nX/r/n、/r/nY/r/n、/r/nZ/r/n、/r/nM/r/n和/r/nW/r/n的原子序数依次增大，且/r/n0.1mol•L/r/n-1/r/nXW/r/n的水溶液呈酸性且/r/npH>1/r/n。下列说法正确的是/r/nA．/r/n最高价含氧酸的酸性：/r/nY>Z/r/nB．/r/n氢化物的沸点：/r/nM>W>Z/r/nC．W/r/n的单质可与/r/nM/r/n的氢化物发生置换反应/r/nD．YW/r/n3/r/n分子中各原子最外层都达到/r/n8/r/n电子稳定结构/r/n【KS5U答案】/r/nC/r/n【分析】/r/n短周期主族元素/r/nX/r/n、/r/nY/r/n、/r/nZ/r/n、/r/nM/r/n和/r/nW/r/n的原子序数依次增大，/r/n0.1mol•L/r/n-1/r/nXW/r/n的水溶液呈酸性且/r/npH>1/r/n，则/r/nXW/r/n为一元弱酸，/r/nX/r/n为/r/nH/r/n，/r/nW/r/n为/r/nF/r/n；由结构可知负一价的阴离子中/r/n4/r/n个/r/nZ/r/n都形成/r/n4/r/n个共价键，则/r/nX/r/n为/r/nC/r/n；/r/n2/r/n个/r/nM/r/n均形成/r/n2/r/n对共用电子对，则/r/nM/r/n为/r/nO/r/n；负一价的阴离子中一个/r/nY/r/n形成/r/n4/r/n对共用电子对，则/r/nY/r/n的最外层电子数为/r/n3/r/n，/r/nY/r/n为/r/nB/r/n；综上所述，/r/nX/r/n、/r/nY/r/n、/r/nZ/r/n、/r/nM/r/n和/r/nW/r/n分别为/r/nH/r/n、/r/nB/r/n、/r/nC/r/n、/r/nO/r/n、/r/nF/r/n。/r/n【KS5U解析】/r/nA/r/n．/r/nY/r/n为/r/nB/r/n，/r/nZ/r/n为/r/nC/r/n，元素的非金属性/r/n，因此最高价含氧酸的酸性：/r/nC>B/r/n，选项/r/nA/r/n错误；/r/nB/r/n．/r/nC/r/n的氢化物很多，沸点有高有低，/r/nH/r/n2/r/nO/r/n、/r/nHF/r/n分子间存在氢键，且/r/nH/r/n2/r/nO/r/n常温呈液态，/r/nHF/r/n呈气态，则氢化物的沸点/r/n，/r/nZ/r/n的氢化物的沸点可能比/r/nM/r/n的高，也可能比/r/nW/r/n的还低，还可能介于/r/nM/r/n和/r/nW/r/n之间，选项/r/nB/r/n错误；/r/nC/r/n．/r/nW/r/n为/r/nF/r/n，/r/nM/r/n为/r/nO/r/n，/r/n2F/r/n2/r/n+2H/r/n2/r/nO=4HF+O/r/n2/r/n，即/r/nW/r/n的单质可与/r/nM/r/n的氢化物发生置换反应，选项/r/nC/r/n正确；/r/nD/r/n．/r/nY/r/n为/r/nB/r/n，/r/nW/r/n为/r/nF/r/n，/r/nBF/r/n3/r/n中/r/nB/r/n的最外层电子为/r/n6/r/n，未达到/r/n8/r/n电子稳定结构，选项/r/nD/r/n错误；/r/n答案选/r/nC/r/n。/r/n16．（江苏高三其他模拟）/r/n室温下，通过实验研究亚硫酸盐的性质，已知/r/nK/r/na1/r/n(H/r/n2/r/nSO/r/n3/r/n)=1.54×10/r/n-2/r/n，/r/nK/r/na2/r/n(H/r/n2/r/nSO/r/n3/r/n)=1.02×10/r/n-7/r/n实验/r/n实验操作和现象/r/n1/r/n把/r/nSO/r/n2/r/n通入氨水，测得溶液/r/npH=7/r/n2/r/n向/r/n0.1mol/LNa/r/n2/r/nSO/r/n3/r/n溶液中加入过量/r/n0.2mol/LCaCl/r/n2/r/n溶液，产生白色沉淀/r/n3/r/n向/r/n0.1mol/LNaHSO/r/n3/r/n溶液中滴入一定量/r/nNaOH/r/n溶液至/r/npH=7/r/n4/r/n把少量氯气通入/r/nNa/r/n2/r/nSO/r/n3/r/n溶液中，测得/r/npH/r/n变小/r/n下列说法错误的是/r/nA．/r/n实验/r/n1/r/n中，可算出/r/n=3.04/r/nB．/r/n实验/r/n2/r/n反应静置后的上层清液中有/r/nc/r/n(Ca/r/n2+/r/n)·/r/nc/r/n(SO/r/n)=/r/nK/r/nsp/r/n(CaSO/r/n3/r/n)/r/nC．/r/n实验/r/n3/r/n中：/r/nc/r/n(Na/r/n+/r/n)/r/n＞/r/n2[/r/nc/r/n(HSO/r/n)+/r/nc/r/n(