2022学年高考化学模拟题汇编专题11水溶液中的离子平衡【含答案】

专题11水溶液中的离子平衡/r/n1．（河北沧州市/r/n高三/r/n三模）/r/n常温下，向/r/n溶液中滴加/r/n的/r/n溶液时，/r/n(/r/n、/r/n或/r/n)/r/n随/r/nV(/r/n溶液/r/n)/r/n变化的曲线如图。下列叙述不正确的是/r/nA．/r/n常温下，/r/n的第一步电离平衡常数/r/nB．/r/n当/r/nV(/r/n溶液/r/n)/r/n时，/r/nC．V(/r/n溶液/r/n)/r/n时，水的电离程度最大/r/nD．/r/n常温下，当/r/n时，有/r/nB/r/nA/r/n．当/r/n时，/r/n，则/r/n，/r/nA/r/n项正确；/r/nB/r/n．当/r/n时，根据电荷守恒得/r/n，根据物料守恒得/r/n，综上可得/r/n，/r/nB/r/n项错误；/r/nC/r/n．当/r/nV(/r/n溶液/r/n)/r/n时，溶液中溶质为/r/n，/r/n是强碱弱酸盐，在水中水解，促进水的电离，所以当/r/nV(/r/n溶液/r/n)/r/n时，水的电离程度最大，/r/nC/r/n项正确；/r/nD/r/n．当/r/n时，/r/n，根据电荷守恒得/r/n，/r/n，此时/r/n，/r/n，/r/nD/r/n项正确；/r/n答案选/r/nB/r/n。/r/n2．（广东珠海市/r/n高三/r/n二模）/r/n常温下，向/r/n的/r/nHA/r/n溶液中逐滴加入/r/n的氨水，溶液的/r/npH/r/n及导电能力变化趋势如图。下列说法错误的是/r/nA．b/r/n点溶液中，水的电离程度最大/r/nB．/r/n溶液中的阳离子浓度总和关系是：/r/nb>c>a/r/nC．/r/n常温下，/r/nHA/r/n的电离平衡常数约为/r/n10/r/n-5/r/nD．/r/n氨水的电离常数小于/r/nHA/r/n的电离平衡常数/r/nD/r/nA/r/n．/r/nb/r/n点（滴入/r/n10mL/r/n氨水时），/r/nHA/r/n和/r/nNH/r/n3/r/n•/r/nH/r/n2/r/nO/r/n恰好完全反应，此时溶液中的溶质仅有/r/nNH/r/n4/r/nA/r/n；/r/nb/r/n点之前，溶液中的溶质有/r/nHA/r/n和/r/nNH/r/n4/r/nA/r/n，/r/nb/r/n点之后，溶液中的溶质有/r/nNH/r/n3/r/n•/r/nH/r/n2/r/nO/r/n和/r/nNH/r/n4/r/nA/r/n，/r/nHA/r/n和/r/nNH/r/n3/r/n•/r/nH/r/n2/r/nO/r/n都会抑制水的电离，/r/nNH/r/n4/r/nA/r/n的水解会促进水的电离，故/r/nb/r/n点溶液中，水的电离程度最大，/r/nA/r/n正确；/r/nB/r/n．溶液中始终存在电荷守恒：/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(A/r/n-/r/n)/r/n；由于溶液的导电能力与溶液中离子浓度呈正比，导电能力：/r/nb/r/n＞/r/nc/r/n＞/r/na/r/n，故溶液中的阳离子浓度综合关系为：/r/nb/r/n＞/r/nc/r/n＞/r/na/r/n，/r/nB/r/n正确；/r/nC/r/n．常温下，/r/n0.1mol/r/n•/r/nL/r/n-1/r/nHA/r/n溶液的/r/npH/r/n约为/r/n3/r/n，则该溶液中/r/nc/r/n(H/r/n+/r/n)/r/n、/r/nc/r/n(A/r/n-/r/n)/r/n的浓度约为/r/n0.001mol/r/n•/r/nL/r/n-1/r/n，则/r/nK/r/n(HA)=/r/n=/r/n=1.0/r/n/r/n10/r/n-5/r/n，/r/nC/r/n正确；/r/nD/r/n．/r/nb/r/n点，溶液的溶质仅有/r/nNH/r/n4/r/nA/r/n，溶液的/r/npH/r/n为/r/n7/r/n，则说明/r/n和/r/nA/r/n-/r/n的水解程度相当，即氨水、/r/nHA/r/n的电离平衡常数相同，/r/nD/r/n错误；/r/n故选/r/nD/r/n。/r/n3．（江西南昌市/r/n高三/r/n三模）/r/n某些难溶性铅盐可用作涂料，如秦俑彩绘中使用的铅白/r/n(PbCO/r/n3/r/n)/r/n和黄金雨中黄色的/r/nPbI/r/n2/r/n。室温下/r/nPbCO/r/n3/r/n和/r/nPbI/r/n2/r/n在不同的溶液中分别达到溶解平衡时/r/n-lgc(Pb/r/n2+/r/n)/r/n与/r/n-lgc(/r/n)/r/n或/r/n-lgc(I/r/n-/r/n)/r/n的关系如图所示。下列说法正确的是/r/nA．K/r/nsp/r/n(PbCO/r/n3/r/n)/r/n的数量级为/r/n10/r/n-14/r/nB．p/r/n点对应的是/r/nPbCO/r/n3/r/n不饱和溶液/r/nC．L/r/n2/r/n对应的是/r/n-lgc(Pb/r/n2+/r/n)/r/n与/r/n-lgc(I/r/n-/r/n)/r/n的关系变化/r/nD．/r/n向浓度相同的/r/nNa/r/n2/r/nCO/r/n3/r/n、/r/nNal/r/n混合溶液中滴入/r/nPb(NO/r/n3/r/n)/r/n2/r/n溶液先产生黄色沉淀/r/nA/r/n【分析】/r/n，则/r/n，同理/r/n，则图像/r/nL/r/n1/r/n/r/n对应的是/r/n-lgc(Pb/r/n2+/r/n)/r/n与/r/n-lgc(I/r/n-/r/n)/r/n的关系变化，以此分析。/r/nA/r/n．/r/n/r/n由上述分析可知图像/r/nL/r/n2/r/n/r/n对应的是/r/n-lgc(Pb/r/n2+/r/n)/r/n与/r/n-lg/r/n的关系变化，则/r/n数量级为/r/n10/r/n-14/r/n，故/r/nA/r/n正确；/r/nB/r/n．离子浓度的负对数越小，则离子浓度越大，/r/np/r/n点位于/r/nL/r/n2/r/n的左下方，则对应/r/nPbCO/r/n3/r/n的过饱和溶液，故/r/nB/r/n错误；/r/nC/r/n．/r/nL/r/n2/r/n/r/n对应的是/r/n-lgc(Pb/r/n2+/r/n)/r/n与/r/n-lgc(I/r/n-/r/n)/r/n的关系变化，故/r/nC/r/n正确；/r/nD/r/n．/r/n/r/n，/r/n，/r/n，则/r/np/r/n点溶液中加入/r/nNa/r/n2/r/nCO/r/n3/r/n/r/n浓溶液，先得白色沉淀，故/r/nD/r/n错误；/r/n故答案选：/r/nA/r/n。/r/n4．（天津高三三模）/r/n电位滴定法是根据滴定过程中指示电极电位的变化来确定滴定终点的一种滴定分析方法。在化学计量点附近，被测离子浓度发生突跃，指示电极电位/r/n(ERC)/r/n也产生了突跃，进而确定滴定终点的位置。现利用盐酸滴定某溶液中碳酸钠含量，其电位滴定曲线如图所示。下列说法错误的是/r/nA．b/r/n点溶液呈酸性/r/nB．a/r/n到/r/nb/r/n之间存在：/r/nc(Na/r/n+/r/n)>c(Cl/r/n-/r/n)/r/nC．/r/n该滴定过程中不需任何指示剂/r/nD．a/r/n点溶液中存在：/r/nc(Na/r/n+/r/n)=c(HCO/r/n)+c(CO/r/n)+c(H/r/n2/r/nCO/r/n3/r/n)/r/nD/r/n【分析】/r/n由图可知，/r/na/r/n点为第一个计量点，发生的反应为碳酸钠与盐酸反应生成氯化钠和碳酸氢钠，/r/nb/r/n点为第二个计量点，发生的反应为碳酸氢钠溶液与盐酸反应生成氯化钠、二氧化碳和水。/r/nA/r/n．由分析可知，/r/nb/r/n点溶液为氯化钠和碳酸的饱和溶液，碳酸在溶液中部分电离使溶液呈酸性，故/r/nA/r/n正确；/r/nB/r/n．从/r/na/r/n点到/r/nb/r/n点，溶液由氯化钠、碳酸氢钠混合液变为氯化钠、碳酸混合溶液，溶液中/r/nc/r/n(Na/r/n+/r/n)>/r/nc/r/n(Cl/r/n—/r/n)/r/n故/r/nB/r/n正确；/r/nC/r/n．电位滴定在化学计量点附近，被测离子浓度发生突跃，指示电极电位也产生了突跃，进而确定滴定终点的位置，则滴定时不需指示剂，故/r/nC/r/n正确；/r/nD/r/n．．由分析可知，/r/na/r/n点为氯化钠和碳酸氢钠的混合溶液，也电荷守恒可知溶液中存在/r/nc/r/n(Na/r/n+/r/n)>/r/nc/r/n(HCO/r/n)+/r/nc/r/n(CO/r/n)+/r/nc/r/n(H/r/n2/r/nCO/r/n3/r/n)/r/n，故/r/nD/r/n错误；/r/n故选/r/nD/r/n。/r/n5．（江苏徐州市/r/n高三模拟/r/n）25℃/r/n时，改变/r/n0.1mol·L/r/n-1/r/n弱酸/r/nRCOOH/r/n溶液的/r/npH/r/n，溶液中/r/nRCOOH/r/n分子的物质的量分数/r/nδ(RCOOH)/r/n随之改变/r/n[/r/n已知/r/nδ(RCOOH)=/r/n]/r/n，甲酸/r/n(HCOOH)/r/n与丙酸/r/n(CH/r/n3/r/nCH/r/n2/r/nCOOH)/r/n溶液中/r/nδ(RCOOH)/r/n与/r/npH/r/n的关系如图。下列说法正确的是/r/nA．pH=4/r/n的丙酸和甲酸溶液稀释/r/n100/r/n倍后对应溶液中的/r/nc(H/r/n+/r/n)/r/n：甲酸/r/n>/r/n丙酸/r/nB．25℃/r/n时，等浓度甲酸与丙酸混合后，丙酸的电离平衡常数/r/nK>10/r/n-4.88/r/nC．/r/n浓度均为/r/n0.1mol·L/r/n-1/r/n的/r/nHCOONa/r/n和/r/nCH/r/n3/r/nCH/r/n2/r/nCOONa/r/n的混合溶液中：/r/nc(H/r/n＋/r/n)+c(CH/r/n3/r/nCH/r/n2/r/nCOOH)+c(HCOOH)=c(OH/r/n-/r/n)+c(HCOO/r/n-/r/n)+c(CH/r/n3/r/nCH/r/n2/r/nCOO/r/n-/r/n)/r/nD．/r/n将/r/n0.1mol·L/r/n-1/r/n的/r/nHCOOH/r/n溶液与/r/n0.1mol·L/r/n-1/r/n的/r/nHCOONa/r/n溶液等体积混合，所得溶液中：/r/nc(HCOO/r/n-/r/n)>c(Na/r/n＋/r/n)>c(HCOOH)>c(H/r/n＋/r/n)>c(OH/r/n-/r/n)/r/nD/r/nA/r/n．根据酸溶液中分子的物质的量分数可知，/r/npH=4/r/n的丙酸和甲酸溶液中酸分子的物质的量分数前者大，则丙酸浓度大于甲酸，稀释/r/n100/r/n倍后丙酸电离出的/r/nH/r/n+/r/n更多，故/r/nc(H/r/n+/r/n)/r/n：丙酸/r/n>/r/n甲酸，选项/r/nA/r/n错误；/r/nB/r/n．弱酸的电离平衡常数只与温度有关，与浓度无关，根据/r/nN/r/n点可知，丙酸的分子的物质的量分数为/r/n50%/r/n，则可得/r/nK=c(H/r/n+/r/n)=10/r/n-4.88/r/n，选项/r/nB/r/n错误；/r/nC/r/n．浓度均为/r/n0.1mol·L/r/n-1/r/n的/r/nHCOONa/r/n和/r/nCH/r/n3/r/nCH/r/n2/r/nCOONa/r/n的混合溶液中存在电荷守恒有/r/nc(H/r/n＋/r/n)+c(Na/r/n＋/r/n)=c(OH/r/n-/r/n)+c(HCOO/r/n-/r/n)+c(CH/r/n3/r/nCH/r/n2/r/nCOO/r/n-/r/n)/r/n，但/r/nc(Na/r/n＋/r/n)/r/nc(CH/r/n3/r/nCH/r/n2/r/nCOOH)+c(HCOOH)/r/n，故/r/nc(H/r/n＋/r/n)+c(CH/r/n3/r/nCH/r/n2/r/nCOOH)+c(HCOOH)/r/nc(OH/r/n-/r/n)+c(HCOO/r/n-/r/n)+c(CH/r/n3/r/nCH/r/n2/r/nCOO/r/n-/r/n)/r/n，选项/r/nC/r/n错误；/r/nD/r/n．将/r/n0.1mol·L/r/n-1/r/n的/r/nHCOOH/r/n溶液与/r/n0.1mol·L/r/n-1/r/n的/r/nHCOONa/r/n溶液等体积混合，所得溶液呈酸性，/r/nHCOOH/r/n电离大于/r/nHCOONa/r/n水解，故存在：/r/nc(HCOO/r/n-/r/n)>c(Na/r/n＋/r/n)>c(HCOOH)>c(H/r/n＋/r/n)>c(OH/r/n-/r/n)/r/n，选项/r/nD/r/n正确；/r/n答案选/r/nD/r/n。/r/n6．（辽宁朝阳市/r/n高三模拟/r/n）/r/n常温时，采用甲基橙和酚酞双指示剂，用盐酸滴定/r/nNa/r/n2/r/nCO/r/n3/r/n溶液，溶液中/r/nlg/r/nc/r/n(H/r/n2/r/nCO/r/n3/r/n)/r/n、/r/nlg/r/nc/r/n(HCO/r/n)/r/n、/r/nlg/r/nc/r/n(CO/r/n)/r/n、/r/nlg/r/nc/r/n(H/r/n+/r/n)/r/n、/r/nlg/r/nc/r/n(OH/r/n-/r/n)/r/n随溶液/r/npH/r/n的变化及滴定曲线如图所示：/r/n下列说法不正确的是/r/nA．/r/n整个滴定过程中可先用酚酞再用甲基橙作指示剂/r/nB．n/r/n点的/r/npH/r/n为/r/nm/r/n点和/r/nq/r/n点/r/npH/r/n的平均值/r/nC．r/r/n点溶液中：/r/nc/r/n(OH/r/n-/r/n)=/r/nc/r/n(H/r/n+/r/n)+/r/nc/r/n(HCO/r/n)+/r/nc/r/n(H/r/n2/r/nCO/r/n3/r/n)/r/nD．r/r/n点到/r/nk/r/n点对应的变化过程中，溶液中水的电离程度始终减小/r/nC/r/nA/r/n．/r/n/r/n已知：酚酞变色的/r/npH/r/n范围为/r/n8.2～10/r/n，/r/npH/r/n＜/r/n8.2/r/n时酚酞溶液变无色，甲基橙变色的/r/npH/r/n范围为/r/n3.1～4.4/r/n，/r/npH/r/n＜/r/n4.4/r/n时甲基橙变橙色，由图可知：盐酸溶液滴定/r/n溶液时，先用酚酞再用甲基橙作指示剂，/r/nA/r/n项正确；/r/nB/r/n．/r/n/r/n设/r/nm/r/n点的/r/npH=a/r/n，/r/nm/r/n点：①/r/n，/r/n，/r/n；设/r/nq/r/n点的/r/npH=b/r/n，/r/nq/r/n点：②/r/n，/r/n，/r/n，/r/nn/r/n点，/r/n，/r/n，则/r/nn/r/n点，/r/n，/r/nB/r/n项正确；/r/nC/r/n．/r/nr/r/n点由/r/n水解生成的/r/n和/r/n的量是相等的可知，此时溶质为/r/n，由质子守恒得/r/n，/r/nC/r/n项错误；/r/nD/r/n．/r/nr/r/n点溶质为/r/n，此时水的电离程度最大，/r/nr/r/n点到/r/nk/r/n点滴加盐酸，水的电离程度始终减小，/r/nD/r/n项正确；/r/n故选/r/nC/r/n。/r/n7．（河北石家庄市/r/n高三/r/n二模）/r/n用/r/n0.1mol/LNaOH/r/n标准溶液滴定/r/n20mL/r/n等浓度的三元酸/r/nH/r/n3/r/nA/r/n，得到/r/npH/r/n与/r/nV(NaOH)/r/n、/r/nlgX[X=/r/n或/r/n或/r/n]/r/n的关系分别如图/r/n1/r/n、图/r/n2/r/n所示。/r/n下列说法错误的是/r/nA．/r/n曲线/r/nI/r/n表示/r/npH/r/n与/r/nlg/r/n的变化关系/r/nB．/r/n由图/r/n1/r/n知中和/r/nNa/r/n2/r/nHA/r/n溶液时，未出现/r/npH/r/n突跃/r/nC．/r/n当体系中溶液的/r/npH=9/r/n时，/r/n=1.8/r/nD．/r/n当滴入/r/n20mLNaOH/r/n溶液时，/r/nc(Na/r/n+/r/n)/r/n＞/r/nc(H/r/n2/r/nA/r/n-/r/n)/r/n＞/r/nc(H/r/n+/r/n)/r/n＞/r/nc(HA/r/n2-/r/n)/r/n＞/r/nc(OH/r/n-/r/n)/r/nC/r/n【分析】/r/n将/r/nNaOH/r/n溶液滴入/r/nH/r/n3/r/nA/r/n溶液中，有①/r/nH/r/n3/r/nA+NaOH=NaH/r/n2/r/nA+H/r/n2/r/nO/r/n；②/r/nNaH/r/n2/r/nA+NaOH=Na/r/n2/r/nHA+H/r/n2/r/nO/r/n；③/r/nNa/r/n2/r/nHA+NaOH=Na/r/n3/r/nA+H/r/n2/r/nO/r/n，二者浓度均为/r/n0.1mol·L/r/n−1/r/n，三元酸/r/nH/r/n3/r/nA/r/n溶液体积为/r/n20mL/r/n，则反应①②③完全对应/r/nNaOH/r/n溶液体积分别为/r/n20mL/r/n、/r/n40mL/r/n、/r/n60mL/r/n。/r/nA/r/n．加入/r/nNaOH/r/n过程中，/r/npH/r/n增大，当/r/nlgX=0/r/n时，直线对应/r/npH/r/n最大，曲线/r/nI/r/n表示/r/npH/r/n与/r/nlg/r/n的变化关系，选项/r/nA/r/n正确；/r/nB/r/n．中和/r/nNa/r/n2/r/nHA/r/n为反应③，/r/nNaOH/r/n溶液体积从/r/n40mL/r/n到/r/n60mL/r/n，从图中分析/r/n40mLNaOH/r/n溶液加入时，溶液碱性已经很强，当/r/n60mLNaOH/r/n溶液加入时，从图中看，/r/npH=12/r/n时，未出现/r/npH/r/n突跃，选项/r/nB/r/n正确；/r/nC/r/n．曲线Ⅱ表示/r/npH/r/n与/r/nlg/r/n的变化关系，从图中看，根据点（/r/n10/r/n，/r/n2.8/r/n）可知/r/npH=10/r/n时，/r/nlgX=2.8/r/n，则/r/n=10/r/n2.8/r/n，/r/nK/r/n2/r/n=/r/n=10/r/n-10/r/n×10/r/n2.8/r/n=/r/n10/r/n-7.2/r/n，当/r/npH=9/r/n时，即/r/nc(H/r/n+/r/n)=10/r/n-9/r/n，/r/n=/r/n，选项/r/nC/r/n错误；/r/nD/r/n．当加入/r/n20mLNaOH/r/n溶液时，溶质为/r/nNaH/r/n2/r/nA/r/n，从图/r/n1/r/n看溶液显酸性，则/r/nH/r/n2/r/nA/r/n-/r/n的电离程度大于/r/nH/r/n2/r/nA/r/n-/r/n的水解程度，离子浓度大小顺序为/r/nc(Na/r/n+/r/n)/r/n＞/r/nc(H/r/n2/r/nA/r/n−/r/n)/r/n＞/r/nc(H/r/n+/r/n)/r/n＞/r/nc(HA/r/n2−/r/n)/r/n＞/r/nc(OH/r/n−/r/n)/r/n，选项/r/nD/r/n正确；/r/n答案选/r/nC/r/n。/r/n8．（四川成都市·成都七中高三/r/n模拟/r/n）/r/n甲胺/r/n(CH/r/n3/r/nNH/r/n2/r/n)/r/n的性质与氨相似，/r/n25/r/n℃时向/r/n10mL0.1mol·L/r/n-1/r/n的甲胺中逐滴滴入/r/n0.1mol·L/r/n-1/r/n稀盐酸。溶液中水电离出的氢离子浓度的负对数/r/n[/r/n用/r/npC/r/n表示，/r/npC=-1g/r/nc/r/n水/r/n(H/r/n+/r/n)]/r/n与加入盐酸体积的关系如图所示。下列叙述错误的是/r/nA．/r/nK/r/nb/r/n(CH/r/n3/r/nNH/r/n2/r/n)/r/n的数量级为/r/n10/r/n-4/r/nB．/r/n甲胺的电离方程式：/r/nCH/r/n3/r/nNH/r/n2/r/n+H/r/n2/r/nO/r/n+OH/r/n-/r/nC．b/r/n点溶液的/r/npH/r/n＜/r/n7/r/nD．e/r/n点溶液：/r/nc/r/n(Cl/r/n-/r/n)/r/n＞/r/nc/r/n(/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/nC/r/nA/r/n．在/r/na/r/n点，/r/npC=11.8/r/n，则溶液中的/r/nc/r/n(OH/r/n-/r/n)=10/r/n-(14-11.8)/r/n=10/r/n-2.2/r/n，/r/nK/r/nb/r/n(CH/r/n3/r/nNH/r/n2/r/n)=/r/n=10/r/n-3/r/n./r/n4/r/n，由此可知/r/nK/r/nb/r/n(CH/r/n3/r/nNH/r/n2/r/n)/r/n的数量级为/r/n10/r/n-4/r/n，/r/nA/r/n正确；/r/nB/r/n．氨的电离方程式为/r/nNH/r/n3/r/n+H/r/n2/r/nO/r/n+OH/r/n-/r/n，采用类推法，则甲胺的电离方程式：/r/nCH/r/n3/r/nNH/r/n2/r/n+H/r/n2/r/nO/r/n+OH/r/n-/r/n，/r/nB/r/n正确；/r/nC/r/n．在/r/nb/r/n点，/r/n10mL0.1mol·L/r/n-1/r/n的甲胺中逐滴滴入/r/n5mL0.1mol·L/r/n-1/r/n稀盐酸，所得溶液中/r/nc/r/n(/r/n)/r/n≈/r/nc/r/n(CH/r/n3/r/nNH/r/n2/r/n)/r/n，/r/nK/r/nb/r/n(CH/r/n3/r/nNH/r/n2/r/n)=/r/n=10/r/n-3/r/n./r/n4/r/n，/r/nc/r/n(OH/r/n-/r/n)/r/n≈/r/n10/r/n-3/r/n./r/n4/r/n，溶液的/r/npH/r/n＞/r/n7/r/n，/r/nC/r/n错误；/r/nD/r/n．在/r/ne/r/n点时，/r/nCH/r/n3/r/nNH/r/n2/r/n与/r/nHCl/r/n刚好完全反应生成/r/n和水，此时发生水解反应/r/n+H/r/n2/r/nO/r/nCH/r/n3/r/nNH/r/n2/r/n∙H/r/n2/r/nO+H/r/n+/r/n，从而得出溶液中/r/nc/r/n(Cl/r/n-/r/n)/r/n＞/r/nc/r/n(/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，/r/nD/r/n正确；/r/n故选/r/nC/r/n。/r/n9．（广东广州市/r/n高三/r/n三模）/r/n已知常温下/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=1.8×10/r/n-5/r/n。常温下向/r/n1L0.1mol•L/r/n-1/r/n的/r/nNH/r/n4/r/nCl/r/n溶液中不断加入/r/nNaOH/r/n固体，溶液中/r/nNH/r/n和/r/nNH/r/n3/r/n•H/r/n2/r/nO/r/n的物质的量浓度变化曲线如图所示/r/n(/r/n忽略溶液体积变化/r/n)/r/n。下列说法正确的是/r/nA．M/r/n点时，加入的/r/nNaOH/r/n少于/r/n0.05mol/r/nB．M/r/n点时，溶液/r/npH=7/r/nC．M/r/n点时，/r/nc(OH/r/n-/r/n)+0.05=a+c(H/r/n+/r/n)/r/nD．n(NaOH)=0.05mol/r/n时，/r/nc(Cl/r/n-/r/n)>c(NH/r/n)>c(H/r/n+/r/n)>c(OH/r/n-/r/n)/r/nC/r/nA/r/n．/r/n/r/n向/r/n1L0.1mol•L/r/n-1/r/n的/r/nNH/r/n4/r/nCl/r/n溶液中不断加入/r/nNaOH/r/n固体，/r/na=0.05mol/r/n时，得到物质的量均为/r/n0.05mol/r/n的氯化铵和一水合氨/r/n,/r/n,/r/n即一水合氨的电离程度大于铵根离子的水解程度，所以/r/n，所以要想满足/r/n/r/n，加入的/r/nNaOH/r/n要大于/r/n0.05mol/r/n，故/r/nA/r/n错误；/r/nB/r/n．由/r/nA/r/n分析知，当/r/na=0.05mol/r/n时，一水合氨的电离程度大于铵根离子的水解程度，溶液显碱性，/r/nM/r/n点/r/na/r/n大于/r/n0.05mol,/r/n碱性更强，溶液/r/npH>7,/r/n故/r/nB/r/n错误；/r/nC/r/n．/r/nM/r/n点时存在电荷守恒/r/n，且/r/nc(Na/r/n+/r/n)=amol/L/r/n，/r/nc(Cl/r/n-/r/n)=0.1mol/L,/r/n,/r/n则/r/n整理可得/r/nc(OH/r/n-/r/n)+0.05=a+c(H/r/n+/r/n),/r/n故/r/nC/r/n正确；/r/nD/r/n．/r/nn(NaOH)=0.05mol/r/n时，一水合氨的电离程度大于铵根离子的水解程度，所以/r/n，所以/r/nc(OH/r/n-/r/n)>c(H/r/n+/r/n)/r/n，所以/r/nc(Cl/r/n-/r/n)>c(NH/r/n)>c(H/r/n+/r/n)>c(OH/r/n-/r/n)/r/n不成立，故/r/nD/r/n错误；/r/n故/r/nC。/r/n10．（辽宁铁岭市/r/n高三/r/n二模）25℃/r/n时，向/r/n20mL0.1mol/L/r/n的/r/n溶液中滴入盐酸，溶液的/r/npH/r/n与离子浓度变化关系如图所示。已知：/r/n是二元弱酸，/r/nY/r/n表示/r/n或/r/n，/r/n。下列叙述正确的是/r/nA．/r/nB．NaHX/r/n溶液中/r/nC．/r/n曲线/r/nn/r/n表示/r/npH/r/n与/r/n的变化关系/r/nD．/r/n当溶液呈中性时，/r/n0.2mol/L=/r/nC/r/nA/r/n．由/r/nN/r/n点可求/r/n，由/r/nM/r/n点可求/r/n，所以/r/n，故/r/nA/r/n错误；/r/nB/r/n．根据计算，/r/nNaHX/r/n溶液中/r/n电离常数小于水解常数，溶液呈碱性，溶液中离子浓度大小关系为：/r/n，故/r/nB/r/n错误；/r/nC/r/n．由图可知，当/r/npH/r/n相等时，/r/n，则曲线/r/nn/r/n表示/r/npH/r/n与/r/n的变化关系，故/r/nC/r/n正确；/r/nD/r/n．当溶液呈中性时，根据电荷守恒/r/n，因为该实验中是在/r/n25℃/r/n时，向/r/n的/r/n溶液中滴入盐酸，加入盐酸的过程相当于稀释了溶液，/r/n，所以/r/n，故/r/nD/r/n错误。/r/n故选/r/nC/r/n。/r/n11．（江苏南京师大附中高三/r/n模拟/r/n）/r/n工业生产上用过量烧碱溶液处理某矿物/r/n(/r/n含/r/nAl/r/n2/r/nO/r/n3/r/n、/r/nMgO)/r/n，过滤后得到的滤液用/r/nNaHCO/r/n3/r/n溶液处理，测得溶液/r/npH/r/n和/r/nAl(OH)/r/n3/r/n生成的量随加入/r/nNaHCO/r/n3/r/n溶液体积变化的曲线如图，下列说法不正确的是/r/nA．/r/n生成沉淀的离子方程式为/r/nHCO/r/n＋/r/nAlO/r/n＋/r/nH/r/n2/r/nO=Al(OH)/r/n3/r/n↓＋/r/nCO/r/nB．a/r/n点水的电离程度大于/r/nb/r/n点水的电离程度/r/nC．a/r/n点溶液中存在：/r/nc(Na/r/n＋/r/n)/r/n＋/r/nc(H/r/n＋/r/n)=c(AlO/r/n)/r/n＋/r/nc(OH/r/n-/r/n)/r/nD．/r/n原/r/nNaHCO/r/n3/r/n液中/r/nc(HCO/r/n)/r/n＋/r/nc(CO/r/n)/r/n＋/r/nc(H/r/n2/r/nCO/r/n3/r/n)=1.0mol·L/r/n-1/r/nB/r/nA/r/n．/r/nAlO/r/n2/r/n-/r/n促进/r/nHCO/r/n3/r/n-/r/n的电离，偏铝酸钠和碳酸氢钠发生生成氢氧化铝沉淀和碳酸钠，反应的离子方程式为/r/nHCO/r/n＋/r/nAlO/r/n＋/r/nH/r/n2/r/nO=Al(OH)/r/n3/r/n↓/r/n＋/r/nCO/r/n，故/r/nA/r/n正确；/r/nB/r/n．加入前/r/n8mLNaHCO/r/n3/r/n溶液时，没有沉淀生成，说明/r/na/r/n点的溶液中含有氢氧化钠，氢氧化钠抑制水电离，/r/nb/r/n点溶液含有偏铝酸钠、碳酸钠，偏铝酸钠、碳酸钠水解促进水电离，所以/r/na/r/n点水的电离程度小于/r/nb/r/n点水的电离程度，故/r/nB/r/n错误；/r/nC/r/n．/r/na/r/n点溶液中含有偏铝酸钠、氢氧化钠，根据电荷守，溶液中存在：/r/nc(Na/r/n＋/r/n)/r/n＋/r/nc(H/r/n＋/r/n)=c(AlO/r/n)/r/n＋/r/nc(OH/r/n-/r/n)/r/n，故/r/nC/r/n正确；/r/nD/r/n．由图象可知，加入前/r/n8mL/r/n碳酸氢钠溶液不生成沉淀，生成/r/n0.032mol/r/n氢氧化铝沉淀消耗/r/n32mLNaHCO/r/n3/r/n溶液，根据/r/nHCO/r/n3/r/n-/r/n+AlO/r/n2/r/n-/r/n+H/r/n2/r/nO═Al/r/n（/r/nOH/r/n）/r/n3/r/n↓+CO/r/n3/r/n2-/r/n可知，/r/nc/r/n（/r/nNaHCO/r/n3/r/n）/r/n=/r/n1.0mol/L/r/n，原/r/nNaHCO/r/n3/r/n液中/r/nc(HCO/r/n)/r/n＋/r/nc(CO/r/n)/r/n＋/r/nc(H/r/n2/r/nCO/r/n3/r/n)=1.0mol·L/r/n-1/r/n，原/r/nNaHCO/r/n3/r/n液中/r/nc(HCO/r/n)/r/n＋/r/nc(CO/r/n)/r/n＋/r/nc(H/r/n2/r/nCO/r/n3/r/n)=1.0mol·L/r/n-1/r/n，故/r/nD/r/n正确；/r/n故选/r/nB/r/n。/r/n12．（天津高三二模）/r/n常温下，用/r/n溶液滴定/r/n溶液的滴定曲线如图所示/r/n(d/r/n点为恰好反应点/r/n)/r/n。下列说法错误的是/r/nA．b→c/r/n过程中，/r/n不断增大/r/nB．a→d/r/n过程中/r/nd/r/n点时水的电离程度最大/r/nC．c/r/n点时溶液的导电能力最强/r/nD．d/r/n点混合液中离子浓度大小关系为：/r/nC/r/nd/r/n点为恰好反应点，此时溶质为/r/nKA/r/n，溶液显碱性，则说明/r/nKA/r/n为强碱弱酸盐，/r/nHA/r/n为弱酸，/r/nA/r/n．/r/nb→c/r/n过程中发生反应：/r/nKOH+HA=KA+H/r/n2/r/nO/r/n，氢离子被消耗，/r/nHA/r/n的电离平衡正向移动，则/r/n不断增大，故/r/nA/r/n正确；/r/nB/r/n．/r/na→d/r/n过程中，/r/nd/r/n点为恰好反应点，此时溶质为/r/nKA/r/n，对水的电离只有促进作用，则/r/nd/r/n点时水的电离程度最大，故/r/nB/r/n正确；/r/nC/r/n．将/r/nKOH/r/n溶液加入弱酸/r/nHA/r/n溶液中，发生反应：/r/nKOH+HA=KA+H/r/n2/r/nO/r/n，逐渐将弱电解质溶液转化为强电解质溶液，溶液导电性逐渐增强，当恰好反应后，继续加入/r/nKOH/r/n溶液，溶液导电性继续增强，故/r/nC/r/n错误；/r/nD/r/n．/r/nd/r/n点为恰好反应点，此时溶质为/r/nKA/r/n，弱酸根/r/nA/r/n-/r/n离子水解使溶液显碱性，且水解是微弱的，则/r/nd/r/n点混合液中离子浓度大小关系为：/r/n，故/r/nD/r/n正确；/r/n故选/r/nC/r/n。/r/n13．（福建厦门外国语学校高三/r/n模拟/r/n）25/r/n℃时，/r/n0.10L/r/n某二元弱酸/r/nH/r/n2/r/nA/r/n用/r/n1.00mol·L/r/n-1/r/nNaOH/r/n溶液调节其/r/npH/r/n，溶液中/r/nH/r/n2/r/nA/r/n、/r/nHA/r/n-/r/n及/r/nA/r/n2-/r/n的物质的量浓度变化如图所示：/r/n下列说法正确的是/r/nA．H/r/n2/r/nA/r/n的/r/nK/r/na/r/n2/r/n=1×10/r/n-11/r/nB．/r/n在/r/nZ/r/n点，由水电离出的氢离子浓度约/r/n10/r/n-10/r/nmol/L/r/nC．/r/n在/r/nY/r/n点时，/r/nc/r/n(Na/r/n+/r/n)/r/n＞/r/n3/r/nc/r/n(A/r/n2-/r/n)/r/nD．0.1mol·L/r/n-1/r/nNaHA/r/n溶液中：/r/nc/r/n(Na/r/n+/r/n)/r/n＞/r/nc/r/n(HA/r/n-/r/n)/r/n＞/r/nc/r/n(H/r/n2/r/nA)/r/n＞/r/nc/r/n(A/r/n2-/r/n)/r/nC/r/n图中虚线/r/nX-Y-Z/r/n表示溶液/r/npH/r/n，加入的/r/nV(NaOH)/r/n越大，溶液中/r/nc/r/n(A/r/n2-/r/n)/r/n越大、/r/nc/r/n(H/r/n2/r/nA)/r/n越小。/r/nA/r/n．/r/nY/r/n点/r/nc/r/n(HA/r/n-/r/n)=/r/nc/r/n(A/r/n2-/r/n)/r/n，/r/nH/r/n2/r/nA/r/n的/r/nK/r/na2/r/n=/r/n=/r/nc/r/n(H/r/n+/r/n)=10/r/n-8/r/nmol/L/r/n，/r/nA/r/n错误；/r/nB/r/n．在/r/nZ/r/n点时溶液/r/npH=10/r/n，说明溶液中/r/nc(/r/nH/r/n+/r/n)=10/r/n-10/r/nmol/L/r/n，则根据水的离子积常数可知由水电离产生的氢离子浓度/r/nc(/r/nH/r/n+/r/n)/r/n水/r/n=/r/nc/r/n(OH/r/n-/r/n)/r/n水/r/n=/r/n，/r/nB/r/n错误；/r/nC/r/n．/r/nY/r/n点/r/nc/r/n(HA/r/n-/r/n)=/r/nc/r/n(A/r/n2-/r/n)/r/n，溶液中存在电荷守恒：/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(HA/r/n-/r/n)+2/r/nc/r/n(A/r/n2-/r/n)/r/n，该点溶液呈碱性，则/r/nc/r/n(H/r/n+/r/n)/r/n＜/r/nc/r/n(OH/r/n-/r/n)/r/n，所以存在/r/nc/r/n(Na/r/n+/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，由于/r/nc/r/n(HA/r/n-/r/n)=/r/nc/r/n(A/r/n2-/r/n)/r/n，所以即/r/nc/r/n(Na/r/n+/r/n)/r/n＞/r/n3/r/nc/r/n(A/r/n2-/r/n)/r/n，/r/nC/r/n正确；/r/nD/r/n．加入/r/n10mLNaOH/r/n溶液时，酸碱恰好完全反应生成/r/nNaHA/r/n，此时溶液/r/npH/r/n＜/r/n7/r/n，溶液呈酸性，说明/r/nHA/r/n-/r/n电离程度大于其水解程度，则/r/nc/r/n(A/r/n2-/r/n)/r/n＞/r/nc/r/n(H/r/n2/r/nA)/r/n，但其电离和水解程度都较小，主要以/r/nHA/r/n-/r/n形式存在，/r/nHA/r/n-/r/n电离、水解消耗，所以/r/nc/r/n(Na/r/n+/r/n)/r/n＞/r/nc/r/n(HA/r/n-/r/n)/r/n，则该溶液中微粒浓度关系为：/r/nc/r/n(Na/r/n+/r/n)/r/n＞/r/nc/r/n(HA/r/n-/r/n)/r/n＞/r/nc/r/n(A/r/n2-/r/n)/r/n＞/r/nc/r/n(H/r/n2/r/nA)/r/n，/r/nD/r/n错误；/r/n故合理选项是/r/nC/r/n。/r/n14．（浙江高三/r/n模拟/r/n）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/nA/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．（浙江高三/r/n模拟/r/n）/r/n一定温度下，难溶电解质在水中存在沉淀溶解和生成的平衡，常用溶度积/r/nK/r/nsp/r/n来表示溶解程度的大小，如/r/nK/r/nsp/r/n(AgCl)=/r/nc/r/n(Ag/r/n+/r/n)·/r/nc/r/n(Cl/r/n—/r/n)/r/n。常温下用/r/n0.1000mol/L/r/n的/r/nAgNO/r/n3/r/n标准溶液/r/n(pH≈5)/r/n分别滴定/r/n20.00mL0.1000mol/L/r/n的/r/nNaCl/r/n溶液和/r/nNaBr/r/n溶液，混合溶液的/r/npAg[/r/n定义为/r/npAg=/r/n—/r/nlg/r/nc/r/n(Ag/r/n+/r/n)]/r/n与/r/nAgNO/r/n3/r/n溶液体积的变化关系如图所示。/r/n下列说法不正确的是/r/nA．/r/nK/r/nsp/r/n(AgCl)>/r/nK/r/nsp/r/n(AgBr)/r/nB．/r/n当/r/nV/r/n(AgNO/r/n3/r/n溶液/r/n)=20mL/r/n时，/r/nNaBr/r/n溶液中：/r/nc/r/n(Ag/r/n+/r/n)+/r/nc/r/n(H/r/n+/r/n)=/r/nc/r/n(Br/r/n—/r/n)+/r/nc/r/n(OH/r/n—/r/n)/r/nC．/r/n当/r/nV/r/n(AgNO/r/n3/r/n溶液/r/n)=30mL/r/n时，/r/nNaCl/r/n溶液中：/r/nc/r/n(NO/r/n)>/r/nc/r/n(Na/r/n+/r/n)>/r/nc/r/n(Ag/r/n+/r/n)>/r/nc/r/n(Cl/r/n—/r/n)>/r/nc/r/n(H/r/n+/r/n)/r/nD．/r/n相同实验条件下，若改用/r/n10.00mL0.2000mol/LNaCl/r/n溶液，则滴定曲线/r/n(/r/n滴定终点前/r/n)/r/n可能变为/r/na/r/nC/r/nA/r/n．由图可知，/r/nV/r/n(AgNO/r/n3/r/n溶液/r/n)=0mL/r/n时，等浓度的氯化钠溶液中银离子浓度大于溴化钠溶液，则溶度积的关系为/r/nK/r/nsp/r/n(AgCl)>/r/nK/r/nsp/r/n(AgBr)/r/n，故/r/nA/r/n正确；/r/nB/r/n．当/r/nV/r/n(AgNO/r/n3/r/n溶液/r/n)=20mL/r/n时，/r/nNaBr/r/n溶液与硝酸银溶液恰好完全反应，溶液中钠离子浓度等于硝酸根离子浓度，由电荷守恒关系/r/nc/r/n(Ag/r/n+/r/n)+/r/nc/r/n(H/r/n+/r/n)+/r/nc/r/n(Na/r/n+/r/n)=/r/nc/r/n(Br/r/n—/r/n)+/r/nc/r/n(OH/r/n—/r/n)+/r/nc/r/n(NO/r/n)/r/n可得，溶液中/r/nc/r/n(Ag/r/n+/r/n)+/r/nc/r/n(H/r/n+/r/n)=/r/nc/r/n(Br/r/n—/r/n)+/r/nc/r/n(OH/r/n—/r/n)/r/n，故/r/nB/r/n正确；/r/nC/r/n．由图可知，当/r/nV/r/n(AgNO/r/n3/r/n溶液/r/n)=30mL/r/n时，/r/nNaCl/r/n溶液中银离子浓度接近与/r/n0.1mol/L/r/n，由/r/nK/r/nsp/r/n(AgCl)=/r/nc/r/n(Ag/r/n+/r/n)·/r/nc/r/n(Cl/r/n—/r/n)/r/n可知，溶液中氯离子浓度小于氢离子，故/r/nC/r/n错误；/r/nD/r/n．相同实验条件下，若改用/r/n10.00mL0.2000mol/LNaCl/r/n溶液，未加入硝酸银溶液时，银离子浓度减小，消耗硝酸银溶液的体积增大，则滴定终点前的滴定曲线可能变为/r/na/r/n，故/r/nD/r/n正确；/r/n故选/r/nC/r/n。/r/n16．（黑龙江哈尔滨市校高三/r/n模拟/r/n）/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/nB/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/n17．（/r/n河北衡水/r/n高三/r/n模拟/r/n）25℃/r/n时，向/r/n20.00mL0.1000mol·L/r/n-1/r/n的/r/nNa/r/n2/r/nR/r/n溶液中滴入/r/n0.1000mol·L/r/n-1/r/n的盐酸，溶液的/r/npH/r/n与所加盐酸体积的关系如图所示/r/n(/r/n忽略混合过程中溶液体积的变化/r/n)/r/n。下列说法错误的是/r/nA．K/r/na2/r/n(H/r/n2/r/nR)/r/n＞/r/nK/r/nh1/r/n(R/r/n2-/r/n)/r/nB．K/r/na1/r/n(H/r/n2/r/nR)/r/n的数量级为/r/n10/r/n-4/r/nC．/r/n滴定过程中水的电离程度先减小后增大/r/nD．m/r/n点时，溶液中存在/r/nc(Cl/r/n-/r/n)+2c(R/r/n2-/r/n)+c(HR/r/n-/r/n)=c(Na/r/n+/r/n)/r/nC/r/nA/r/n．/r/nm/r/n点时，/r/nNa/r/n2/r/nR/r/n的物质的量等于/r/nNaHR/r/n的物质的量，此时/r/npH=7/r/n，溶液中存在/r/nR/r/n2-/r/n、/r/nHR/r/n-/r/n的水解，以及/r/nHR/r/n-/r/n的电离，所以/r/nK/r/na2/r/n(H/r/n2/r/nR)/r/n＞/r/nK/r/nh1/r/n(R/r/n2-/r/n)/r/n，故/r/nA/r/n正确；/r/n/r/nB/r/n．加入/r/n40ml/r/n盐酸时，得到/r/n60.oomlpH=2.5/r/n的/r/nH/r/n2/r/nR/r/n和/r/nNaCl/r/n的混合溶液，由元素守恒可知，/r/n，/r/n/r/n，/r/n，故/r/nB/r/n正确；/r/nC/r/n．滴定过程中水的电离程度逐渐减小，故/r/nC/r/n错误；/r/nD/r/n．/r/nm/r/n点，由电荷守恒有：/r/n/r/n，此时/r/npH=7/r/n，/r/n/r/n，则有/r/n故/r/nD/r/n正确；/r/n故答案为/r/nC/r/n。/r/n18．（江苏省如皋中学高三三模）/r/n室温下，通过下列实验探究/r/nNaHS/r/n溶液的性质。/r/n实验/r/n实验操作和现象/r/n1/r/n向/r/n0.1mol·L/r/n-1/r/nNaHS/r/n溶液中滴加几滴酚酞试剂，溶液变红/r/n2/r/n向/r/n0.1mol·L/r/n-1/r/nNaHS/r/n溶液中加入等体积/r/n0.1mol·L/r/n-1/r/nNaOH/r/n溶液充分混合/r/n3/r/n向/r/n0.1mol·L/r/n-1/r/nNaHS/r/n溶液中通入过量氯气，无淡黄色沉淀产生/r/n4/r/n向/r/n0.1mol·L/r/n-1/r/nNaHS/r/n溶液中滴加过量/r/nCuCl/r/n2/r/n溶液，产生黑色沉淀/r/n下列有关说法正确的是/r/nA．0.1mol·L/r/n-1/r/nNaHS/r/n溶液中：/r/nc(S/r/n2-/r/n)>c(H/r/n2/r/nS)/r/nB．/r/n实验/r/n2/r/n所得溶液中：/r/nc(Na/r/n＋/r/n)=c(S/r/n2-/r/n)/r/n＋/r/nc(HS/r/n-/r/n)/r/n＋/r/nc(H/r/n2/r/nS)/r/nC．/r/n实验/r/n3/r/n说明/r/nHS/r/n-/r/n不能被氯气氧化/r/nD．/r/n实验/r/n4/r/n反应静置后的上层清液中有/r/nc(Cu/r/n2+/r/n)·c(S/r/n2-/r/n)=K/r/nsp/r/n(CuS)/r/nD/r/nA/r/n．/r/nNaHS/r/n溶液中存在电离平衡/r/nHS/r/n-/r/nH/r/n+/r/n+S/r/n2-/r/n和水解平衡/r/nHS/r/n-/r/n+H/r/n2/r/nO/r/nH/r/n2/r/nS+OH/r/n-/r/n，由实验/r/n1/r/n可知/r/nNaHS/r/n溶液显碱性，则/r/nHS/r/n-/r/n的水解程度大于电离程度，因此/r/nc(S/r/n2-/r/n)<c(H/r/n2/r/nS)/r/n，/r/nA/r/n错误；/r/nB/r/n．实验/r/n2/r/n中/r/n0.1mol·L/r/n-1/r/nNaHS/r/n溶液和/r/n0.1mol·L/r/n-1/r/nNaOH/r/n溶液等体积混合，则所得溶液为/r/nNa/r/n2/r/nS/r/n溶液，存在物料守恒/r/nc(Na/r/n＋/r/n)=2[c(S/r/n2-/r/n)/r/n＋/r/nc(HS/r/n-/r/n)/r/n＋/r/nc(H/r/n2/r/nS)]/r/n，/r/nB/r/n错误；/r/nC/r/n．氯气有强氧化性，/r/nHS/r/n-/r/n有较强还原性，二者能发生氧化还原反应，实验/r/n3/r/n没有淡黄色沉淀产生可能是氯气将/r/nHS/r/n-/r/n氧化为硫酸根，/r/nC/r/n错误；/r/nD/r/n．实验/r/n4/r/n中产生黑色沉淀，说明生成了/r/nCuS/r/n，则静置后的上层清液中有/r/nc(Cu/r/n2+/r/n)·c(S/r/n2-/r/n)=K/r/nsp/r/n(CuS)/r/n，/r/nD/r/n正确；/r/n选/r/nD/r/n。/r/n19．（湖南邵阳市/r/n高三模拟/r/n）/r/n亚砷酸/r/n(H/r/n3/r/nAsO/r/n3/r/n)/r/n可用于白血病的治疗。室温下，配制一组/r/nc(H/r/n3/r/nAsO/r/n3/r/n)/r/n＋/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/n0.100mol·L/r/n－/r/n1/r/n的/r/nH/r/n3/r/nAsO/r/n3/r/n和/r/nNaOH/r/n混合溶液，溶液中部分微粒的物质的量浓度随/r/npH/r/n的变化关系曲线如图所示。下列指定溶液中微粒的物质的量浓度关系不正确的是/r/nA．pH/r/n＝/r/n11/r/n的溶液中：几乎只有/r/nH/r/n2/r/nAsO/r/nB．c(Na/r/n＋/r/n)/r/n＝/r/n0.100mol·L/r/n－/r/n1/r/n的溶液中：/r/nc(H/r/n2/r/nAsO/r/n)>c(H/r/n3/r/nAsO/r/n3/r/n)>c(HAsO/r/n)/r/nC．c(Na/r/n＋/r/n)/r/n＝/r/n0.200mol·L/r/n－/r/n1/r/n的溶液中：/r/nc(H/r/n＋/r/n)/r/n＋/r/n2c(H/r/n3/r/nAsO/r/n3/r/n)/r/n＝/r/nc(OH/r/n－/r/n)/r/n＋/r/nc(AsO/r/n)/r/nD．pH/r/n＝/r/n12.8/r/n的溶液中：/r/nc(H/r/n＋/r/n)/r/n＋/r/nc(Na/r/n＋/r/n)>4c(H/r/n2/r/nAsO/r/n)/r/n＋/r/nc(HAsO/r/n)/r/n＋/r/nc(OH/r/n－/r/n)/r/nC/r/nA/r/n．由图象可知，/r/npH=11/r/n时溶液中主要存在/r/nH/r/n2/r/nAsO/r/n，其它离子的浓度几乎为/r/n0/r/n，/r/npH/r/n＝/r/n11/r/n的溶液中：几乎只有/r/nH/r/n2/r/nAsO/r/n，故/r/nA/r/n正确；/r/nB/r/n．/r/nc(Na/r/n+/r/n)=0.100mol•L/r/n-1/r/n时，/r/nc(H/r/n3/r/nAsO/r/n3/r/n)+c(H/r/n2/r/nAsO/r/n)+c(HAsO/r/n)+c(AsO/r/n)=0.100mol•L/r/n-1/r/n，说明溶质为/r/nNaH/r/n2/r/nAsO/r/n3/r/n，结合图示可知/r/nH/r/n2/r/nAsO/r/n溶液的/r/npH>7/r/n，呈碱性，说明/r/nH/r/n2/r/nAsO/r/n的水解程度大于其电离程度，则/r/nc(H/r/n3/r/nAsO/r/n3/r/n)>c(HAsO/r/n)/r/n，溶液中离子浓度大小为：/r/nc(H/r/n2/r/nAsO/r/n)>c(H/r/n3/r/nAsO/r/n3/r/n)>c(HAsO/r/n)/r/n，故/r/nB/r/n正确；/r/nC/r/n．/r/nc(Na/r/n+/r/n)=0.200mol•L/r/n-1/r/n时，/r/nc(H/r/n3/r/nAsO/r/n3/r/n)+c(H/r/n2/r/nAsO/r/n3/r/n-/r/n)+c(HAsO/r/n)+c(AsO/r/n)=0.100mol•L/r/n-1/r/n，说明溶质为/r/nNa/r/n2/r/nHAsO/r/n3/r/n，结合质子守恒可知：/r/nc(H/r/n+/r/n)+2c(H/r/n3/r/nAsO/r/n3/r/n)+c(H/r/n2/r/nAsO/r/n)=c(OH/r/n-/r/n)+c(AsO/r/n)/r/n，故/r/nC/r/n错误；/r/nD/r/n．/r/n/r/n结合图示可知，/r/npH=12.8/r/n的溶液中/r/nc(AsO/r/n)=c(H/r/n2/r/nAsO/r/n)/r/n，结合电荷守恒/r/nc(H/r/n+/r/n)+c(Na/r/n+/r/n)=3c(AsO/r/n)+c(H/r/n2/r/nAsO/r/n)+2c(HAsO/r/n)+c(OH/r/n-/r/n)=4c(H/r/n2/r/nAsO/r/n)+2c(HAsO/r/n)+c(OH/r/n-/r/n)/r/n，/r/npH/r/n＝/r/n12.8/r/n的溶液中：/r/nc(H/r/n＋/r/n)/r/n＋/r/nc(Na/r/n＋/r/n)>4c(H/r/n2/r/nAsO/r/n)/r/n＋/r/nc(HAsO/r/n)/r/n＋/r/nc(OH/r/n－/r/n)/r/n，故/r/nD/r/n正确；/r/n故选/r/nC/r/n。/r/n20．（广东高三/r/n模拟/r/n）/r/n常温下，向/r/n1L0.1mol/r/n·/r/nL/r/n-1/r/n—酸/r/nHR/r/n溶液中缓慢通氨气/r/n[/r/n已知常温下/r/n]/r/n，假设溶液温度和体积保持不变，混合溶液的/r/npH/r/n与/r/n的变化关系如图所示。下列叙述正确的是/r/nA．/r/n的/r/nHR/r/n溶液的/r/npH/r/n约为/r/n3/r/nB．HR/r/n为弱酸，常温时/r/nC．/r/n当通入/r/n时，溶液中/r/nD．/r/n当溶液呈中性时，/r/nA/r/n【分析】/r/nA/r/n和/r/nB/r/n，根据图像中坐标点（/r/n5/r/n，/r/n0/r/n）可得/r/n时，/r/npH=5/r/n。再结合电离常数的表达式可得/r/nK/r/n；/r/nC/r/n，当通入/r/n时，溶质是/r/nNH/r/n4/r/nR/r/n，再根据越弱越水解判断溶液的酸碱性，进而判断离子/r/nR/r/n-/r/n和/r/nNH/r/n4/r/n+/r/n浓度大小；/r/nD/r/n，当溶液为中性时，结合图像中横坐标为/r/n7/r/n时，纵坐标为/r/n2/r/n，可得出离子/r/nR/r/n-/r/n和/r/nHR/r/n浓度大小。/r/nA/r/n．根据/r/n时，/r/npH=5/r/n，则/r/n，/r/n的/r/nHR/r/n溶液中/r/n，则溶液的/r/npH/r/n约为/r/n3/r/n，/r/nA/r/n正确；/r/nB/r/n．根据/r/n时，/r/npH=5/r/n，则/r/n，/r/nB/r/n错误；/r/nC/r/n．当通入/r/n时，溶质是/r/nNH/r/n4/r/nR/r/n，/r/n、/r/n，水解常数/r/n，所以溶液呈碱性，/r/n，/r/nC/r/n错误；/r/nD/r/n．当溶液中性时，根据电荷守恒/r/n知/r/n，由图可知，/r/n则/r/n，/r/nD/r/n错误；/r/n电解质离子浓度问题关键在于识别/r/nPH/r/n与/r/n之间的定量关系。/r/n2/r/n1/r/n．（安徽师范大学附属中学高三/r/n模拟/r/n）/r/n已知琥珀酸/r/n[(CH/r/n2/r/nCOOH)/r/n2/r/n]/r/n是二元弱酸，以/r/nH/r/n2/r/nA/r/n代表其化学式，且/r/npX/r/n代表/r/n-lg/r/n或/r/n-lg/r/n。常温下，用/r/nNaOH/r/n溶液滴定琥珀酸溶液，混合溶液的/r/npH/r/n与/r/npX/r/n的关系如图所示。下列有关说法正确的是/r/nA．/r/n该溶液中/r/nB．/r/n水的电离程度/r/nd>c/r/nC．/r/n当/r/n被完全中和时，/r/nD．/r/n滴定至/r/nc/r/n点时，/r/nc(Na/r/n+/r/n)/r/n＜/r/nc(A/r/n2-/r/n)+2c(HA/r/n-/r/n)/r/nD/r/n【分析】/r/nH/r/n2/r/nA/r/n为二元弱酸，电离方程式为：/r/n，/r/n，/r/n，/r/n，所以/r/n，/r/n，因为/r/n，所以/r/na/r/n线为代表/r/n，/r/nb/r/n线代表/r/n，据此分析答题。/r/nA/r/n．/r/n，/r/n同一溶液中，/r/n相同，/r/n，所以/r/n，故/r/nA/r/n错误；/r/nB/r/n．/r/n，/r/n，/r/nd/r/n点酸性强于/r/nc/r/n点，溶液酸性越强，对水的电离抑制成都越大，所以水的电离程度/r/nd/r/n＜/r/nc/r/n，故/r/nB/r/n错误；/r/nC/r/n．当/r/n被完全中和时，生成/r/n，列出质子守恒为/r/n，故/r/nC/r/n错误；/r/nD/r/n．/r/n此时/r/n，所以/r/n，列出电荷守恒：/r/n此时/r/nPH/r/n＝/r/n4.2/r/n，所以，/r/n，推知/r/n，故/r/nD/r/n正确；/r/n答案为/r/nD/r/n。/r/n2/r/n2/r/n．（山东潍坊市/r/n高三/r/n三模）/r/n常温下，将/r/n0.1mol·L/r/n-1/r/n的/r/n溶液加水稀释，混合溶液中/r/n与/r/n的关系如图所示。下列说法正确的是/r/n

/r/nA．/r/n的值为/r/n10/r/n-4.7/r/nB．M/r/n、/r/n