高中化学溶液中水解与电离守恒

高中化学溶液中水解与电离守恒 高中化学 电离与水解 电解质溶液中有关离子浓度的判断是近年高考的重要 题 快递公司问题件快递公司问题件货款处理关于圆的周长面积重点题型关于解方程组的题及答案关于南海问题 型之一。解此类型题的关键是掌握“两平衡、两原理”，即弱电解质的电离平衡、盐的水解平衡和电解质溶液中的电荷守恒、物料守恒原理。首先，我们先来研究一下解决这类问题的理论基础。 一、电离平衡理论和水解平衡理论 1.电离理论: ?弱电解质的电离是微弱的，电离消耗的电解质及产生的微粒都是少量的，同时注意考虑水的电离的存在;?多元弱酸的电离是分步的，主要以第一步电离为主; 2.水解理论: 从盐类的水解的特征 分析 定性数据统计分析pdf销售业绩分析模板建筑结构震害分析销售进度分析表京东商城竞争战略分析 :水解程度是微弱的(一般不超过2‰)。例如:NaHCO溶液中，3 ――c(HCO),,c(HCO)或c(OH ) 323 理清溶液中的平衡关系并分清主次: +-?弱酸的阴离子和弱碱的阳离子因水解而损耗;如NaHCO溶液中有:c(Na) , c(HCO)。33?弱酸的阴离子和弱碱的阳离子的水解是微量的(双水解除外)，因此水解生成的弱电解质 +-及产生H的(或OH)也是微量，但由于水的电离平衡和盐类水解平衡的存在，所以水解后 +-的酸性溶液中c(H)(或碱性溶液中的c(OH))总是大于水解产生的弱电解质的浓度;?一 +-般来说“谁弱谁水解，谁强显谁性”，如水解呈酸性的溶液中c(H),c(OH)，水解呈碱性 -+的溶液中c(OH),c(H);?多元弱酸的酸根离子的水解是分步进行的，主要以第一步水解为主。 。 二、电解质溶液中的守恒关系 1、电荷守恒:电解质溶液中的阴离子的负电荷总数等于阳离子的正电荷总数， 电荷守恒的重要应用是依据电荷守恒列出等式，比较或计算离子的物质的量或物质的量 ，,，―+，-―浓度。如(1)在只含有A、M、H、OH四种离子的溶液中c(A)+c(H)==c(M)+c(OH), ，―+-若c(H),c(OH)，则必然有c(A),c(M)。 例如，在NaHCO溶液中，有如下关系: 3 ，，――2―C(Na)+c(H)==c(HCO)+c(OH)+2c(CO) 33 engineering construction and acceptance specification for 5 GB50275-98 compressor, fan and pump installation engineering construction and acceptance specification for lifting 6 HG20201-2000 construction installation engineering construction standard 7, HG ... Pressure gauges, using an installed spark arrestor for acetylene cylinders, illegal carrying, using gas cylinders, each operator fined 20 Yuan. Gas bottle without the hot sun exposure measures, responsibilities of team a fine of 50 Yuan. (7) at height and the opening and provisional protective measures have been taken, the responsible unit fined 200 Yuan, construction person in charge a fine of 50 Yuan. (8) no permit to work and the job in accordance with the provisions, on the job unit fined 300 yuan, the unit will charge a fine of 100 Yuan. 7.5.3 accident penalties (1) injuries fatalities directly punished 1~2 million. Direct responsibility for the accident and have the corresponding responsibility of leadership, such as concerning administrative sanctions, should be brought to the company or the relevant Department. (2) personal injury accident occurred, the direct punishment 0.5-10,000 yuan, responsible for the direct responsibility for the accident and the responsibility of leadership, such as concerning administrative sanctions should be dealt with by the personnel Department of the company. (3) personal injury accident occurs, the direct punishment 500-1000, who is directly responsible for the accident responsibility, give notice of criticism and 50-100 economic sanctions against them. (4) to conceal the accident, reported without undue delay or false, to inform the administrative leadership of the criticism, resulting in serious consequences, the pursuit of leadership, along with 500-1000 punishment. (5) significant near miss should be 书写电荷守恒式必须?准确的判断溶液中离子的种类;?弄清离子浓度和电荷浓度的关系。 2、物料守恒:就电解质溶液而言，物料守恒是指电解质发生变化(反应或电离)前某元素的原子(或离子)的物质的量等于电解质变化后溶液中所有含该元素的原子(或离子)的物质的量之和。 实质上，物料守恒属于原子个数守恒和质量守恒。 2――在NaS溶液中存在着S的水解、HS的电离和水解、水的电离，粒子间有如下关系 2 2――，，2―c(S)+c(HS)+c(HS)==1/2c(Na) ( Na,S守恒) 2 ―2――C(HS)+2c(S)+c(H)==c(OH) (H、O原子守恒) ―在NaHS溶液中存在着HS的水解和电离及水的电离。 ―――，2―，―HS，HOHS，OH HSH，S HOH，OH 222 ―2―，从物料守恒的角度分析，有如下等式:c(HS)+C(S)+c(HS)==c(Na);从电荷守恒的2 ―2――，，角度分析，有如下等式:c(HS)+2(S)+c(OH)==c(Na)+c(H);将以上两式相加，有: 2――，c(S)+c(OH)==c(HS)+c(H) 2 得出的式子被称为质子守恒 3、质子守恒:无论溶液中结合氢离子还是失去氢离子，但氢原子总数始终为定值，也就是说结合的氢离子的量和失去氢离子的量相等。 现将此类题的解题 方法 快递客服问题件处理详细方法山木方法pdf计算方法pdf华与华方法下载八字理论方法下载 作如下总结。 二、典型题――溶质单一型 1、弱酸溶液中离子浓度的大小判断 解此类题的关键是紧抓弱酸的电离平衡 ,点击试题,0.1mol/L 的HS溶液中所存在离子的浓度由大到小的排列顺序是2 _________________ ，，2―――解析:在HS溶液中有下列平衡:HSH+HS,HSH+S 。已知多元弱酸的22 ，电离以第一步为主～第二步电离较第一步弱得多～但两步电离都产生H～因此答案应为: ，2―――c(H)>c(HS)>c(S)>c(OH) 弱酸溶液中离子浓度大小的一般关系是:C(显性离子) > C(一级电离离子) > C(二级电离离子) > C(水电离出的另一离子) 同样的思考方式可以解决弱碱溶液的问题 engineering construction and acceptance specification for 5 GB50275-98 compressor, fan and pump installation engineering construction and acceptance specification for lifting 6 HG20201-2000 construction installation engineering construction standard 7, HG ... Pressure gauges, using an installed spark arrestor for acetylene cylinders, illegal carrying, using gas cylinders, each operator fined 20 Yuan. Gas bottle without the hot sun exposure measures, responsibilities of team a fine of 50 Yuan. (7) at height and the opening and provisional protective measures have been taken, the responsible unit fined 200 Yuan, construction person in charge a fine of 50 Yuan. (8) no permit to work and the job in accordance with the provisions, on the job unit fined 300 yuan, the unit will charge a fine of 100 Yuan. 7.5.3 accident penalties (1) injuries fatalities directly punished 1~2 million. Direct responsibility for the accident and have the corresponding responsibility of leadership, such as concerning administrative sanctions, should be brought to the company or the relevant Department. (2) personal injury accident occurred, the direct punishment 0.5-10,000 yuan, responsible for the direct responsibility for the accident and the responsibility of leadership, such as concerning administrative sanctions should be dealt with by the personnel Department of the company. (3) personal injury accident occurs, the direct punishment 500-1000, who is directly responsible for the accident responsibility, give notice of criticism and 50-100 economic sanctions against them. (4) to conceal the accident, reported without undue delay or false, to inform the administrative leadership of the criticism, resulting in serious consequences, the pursuit of leadership, along with 500-1000 punishment. (5) significant near miss should be 2、弱碱溶液 ,点击试题,室温下，0.1mol/L的氨水溶液中，下列关系式中不正确的是 -+A. c(OH),c(H) +B.c(NH?HO)+c(NH)=0.1mol/L 324 -++C.c(NH),c(NH?HO),c(OH),c(H) 432 -++D.c(OH)=c(NH)+c(H) 4 下面我们以弱酸强碱盐为例，来介绍一下能发生水解的盐溶液中离子浓度大小比较的解题方法 3、能发生水解的盐溶液中离子浓度大小比较---弱酸强碱型 解此类题型的关键是抓住盐溶液中水解的离子 在CHCOONa 溶液中各离子的浓度由大到小排列顺序正确的是( ) 3 ，――，A、 c(Na)>c(CHCOO)>c(OH)>c(H) 3 ―，―，B、 c(CHCOO)>c(Na)>c(OH)>c(H) 3 ，―，―C、 c(Na)>c(CHCOO)>c(H)>c(OH) 3 ，――，D、 c(Na)>c(OH)>c(CHCOO)>c(H) 3 ，―――解析:在CHCOONa溶液中: CHCOONaNa+CHCOO ～CHCOO+HO CHCOOH+OH ,333323 ―，――，而使c(CHCOO)降低且溶液呈现碱性～则c(Na)>c(CHCOO)～c(OH)>c(H)～又因一般盐33 ――的水解程度较小～则c(CHCOO)>c(OH),因此A选项正确。 3 一元弱酸盐溶液中离子浓度的一般关系是:C(不水解离子) > C(水解离子)>C(显性离子)>C(水电离出的另外一种离子) ,点击试题,在NaCO溶液中各离子的浓度由小到大的排列顺序是______ 23 ，2―2――――解析:在NaCO溶液中～NaCO==2Na+CO ～CO+HOHCO+OH ～HCO232333233 ――，2―2―+HOHCO+OH 。CO水解使溶液呈现碱性～则C(OH)>C(H)～由于CO少部分水解～22333 ―2――――则C(CO)>C(HCO)～HCO又发生第二步水解～则C(OH)>C(HCO)～第二步水解较第一步3333 ――，――水解弱得多～则C(HCO)与C(OH)相关不大～但C(H)比C(OH)小得多～因此C(HCO) > C(H33，，―，―2―)。此题的答案为:C(H) C(水解离子)>C(显性离子)>C(二级水解离子)>C(水电离出的另一离子) ,随堂练习,在NaS溶液中下列关系不正确的是 2 +,2,A,c(Na) =2c(HS) +2c(S) +c(HS) 2 ++,,2,B,c(Na) +c(H)=c(OH)+c(HS)+2c(S) +2,,,C,c(Na),c(S),c(OH),c(HS) engineering construction and acceptance specification for 5 GB50275-98 compressor, fan and pump installation engineering construction and acceptance specification for lifting 6 HG20201-2000 construction installation engineering construction standard 7, HG ... Pressure gauges, using an installed spark arrestor for acetylene cylinders, illegal carrying, using gas cylinders, each operator fined 20 Yuan. Gas bottle without the hot sun exposure measures, responsibilities of team a fine of 50 Yuan. (7) at height and the opening and provisional protective measures have been taken, the responsible unit fined 200 Yuan, construction person in charge a fine of 50 Yuan. (8) no permit to work and the job in accordance with the provisions, on the job unit fined 300 yuan, the unit will charge a fine of 100 Yuan. 7.5.3 accident penalties (1) injuries fatalities directly punished 1~2 million. Direct responsibility for the accident and have the corresponding responsibility of leadership, such as concerning administrative sanctions, should be brought to the company or the relevant Department. (2) personal injury accident occurred, the direct punishment 0.5-10,000 yuan, responsible for the direct responsibility for the accident and the responsibility of leadership, such as concerning administrative sanctions should be dealt with by the personnel Department of the company. (3) personal injury accident occurs, the direct punishment 500-1000, who is directly responsible for the accident responsibility, give notice of criticism and 50-100 economic sanctions against them. (4) to conceal the accident, reported without undue delay or false, to inform the administrative leadership of the criticism, resulting in serious consequences, the pursuit of leadership, along with 500-1000 punishment. (5) significant near miss should be ,,+D,c(OH)=c(HS)+c(H)+c(HS) 2 ,点击试题,判断0.1mol/L 的NaHCO溶液中离子浓度的大小关系 3 ，―，―――2――解析:因NaHCO==Na+HCO～HCO+HOHCO+OH～HCO+CO 。HCO的水解H333223333 ―2―程度大于电离程度～因此溶液呈碱性～且C(OH) > C(CO)。由于少部分水解和电离～则3 ，―，―2―C(Na)>C(HCO)>C(OH)>C(H) > C(CO)。 33 二元弱酸的酸式盐溶液中离子浓度大小的一般关系是:C(不水解离子)>C(水解离子)>C(显性离子)>C(水电离出的另一离子)>C(电离得到的酸根离子) ,随堂练习,草酸是二元弱酸，草酸氢钾溶液呈酸性，在0.1mol/LKHCO溶液中，下列24关系正确的是,CD, ++--2-A,c(K)+c(H)=c(HCO)+c(OH)+ c(CO) 2424 -2-B,c(HCO)+ c(CO)=0.1mol/L 2424 2-C,c(CO),c(HCO) 24224 +-2-D,c(K)= c(HCO)+ c(HCO)+ c(CO) 2242424 下面再让我们利用上述规律来解决一下强酸弱碱盐的问题 ,点击试题,在氯化铵溶液中，下列关系正确的是( ) -++-+-+-A.c(Cl),c(NH),c(H),c(OH) B.c(NH),c(Cl),c(H),c(OH) 44 -+--+-++C.c(NH),c(Cl),c(H),c(OH) D.c(Cl),c(NH),c(H),c(OH) 44 三、典型题----两种电解质溶液相混合型的离子浓度的判断 解此类题的关键是抓住两溶液混合后生成的盐的水解情况以及混合时弱电解质有无剩余，若有剩余，则应讨论弱电解质的电离。下面以一元酸、一元碱和一元酸的盐为例进行分析。 1、强酸与弱碱混合 ,点击试题,PH=13的NH?HO和PH=1的盐酸等体积混合后所得溶液中各离子浓度由大到32 小的排列顺序是____________ ，―解析:PH==1的HCl～C(H)==0.1 mol/L ～PH=13的NH〃HO～C(OH)== 0.1 mol/L ～则32 NH 〃HO 的浓度远大于0.1 mol/L ～因此～两溶液混合时生成NHCl为强酸弱碱盐～氨水324过量～且C(NH 〃HO)>C(NHCl)～则溶液的酸碱性应由氨水决定。即NH〃HO的电离大于32432 ，，，――NH的水解～所以溶液中的离子浓度由大到小的顺序为:C(NH)>C(Cl)>C(OH)>C(H)。 44 需要我们注意的是，强酸弱碱盐溶液中加入一定量的弱碱，解题方法与此题相同。 2、强碱与弱酸混合 ,点击试题,PH=X的NaOH溶液与PH=Y的CHCOOH溶液，已知X+Y=14，且Y<3。将上述两3 溶液等体积混合后，所得溶液中各离子浓度由大到小的顺序正确的是( ) ，――，A、 C(Na)>C(CHCOO)>C(OH)>C(H) 3 ―，，―B、 C(CHCOO)>C(Na)>C(H)>C(OH) 3 engineering construction and acceptance specification for 5 GB50275-98 compressor, fan and pump installation engineering construction and acceptance specification for lifting 6 HG20201-2000 construction installation engineering construction standard 7, HG ... Pressure gauges, using an installed spark arrestor for acetylene cylinders, illegal carrying, using gas cylinders, each operator fined 20 Yuan. Gas bottle without the hot sun exposure measures, responsibilities of team a fine of 50 Yuan. (7) at height and the opening and provisional protective measures have been taken, the responsible unit fined 200 Yuan, construction person in charge a fine of 50 Yuan. (8) no permit to work and the job in accordance with the provisions, on the job unit fined 300 yuan, the unit will charge a fine of 100 Yuan. 7.5.3 accident penalties (1) injuries fatalities directly punished 1~2 million. Direct responsibility for the accident and have the corresponding responsibility of leadership, such as concerning administrative sanctions, should be brought to the company or the relevant Department. (2) personal injury accident occurred, the direct punishment 0.5-10,000 yuan, responsible for the direct responsibility for the accident and the responsibility of leadership, such as concerning administrative sanctions should be dealt with by the personnel Department of the company. (3) personal injury accident occurs, the direct punishment 500-1000, who is directly responsible for the accident responsibility, give notice of criticism and 50-100 economic sanctions against them. (4) to conceal the accident, reported without undue delay or false, to inform the administrative leadership of the criticism, resulting in serious consequences, the pursuit of leadership, along with 500-1000 punishment. (5) significant near miss should be ―，―，C、 C(CHCOO)>C(Na)>C(OH)>C(H) 3 ，―，―D、 C(Na)>C(CHCOO)>C(H)>C(OH) 3 -(14-X)―解析:同上～PH==X的NaOH溶液中～C(OH)==10 mol/L～PH==Y的CHCOOH 溶液中～C(H3，-Y，―)==10 mol/L～因为X+Y==14～NaOH溶液中C(OH)等于CHCOOH溶液中C(H)。因此C(CHCOOH)33 -Y 远大于10mol/L～CHCOOH过量～因此选项B正确。 3 上述两题的特点是PH+PH==14，且等体积混合。其溶液中各离子浓度的关系的特点是 12 C(弱电解质的离子)>C(强电解质的离子)>C(显性离子) > C (水电离出的另一离子) 3、强碱弱酸盐与强酸混合和强酸弱碱盐与强碱混合 ,点击试题,0.2 mol/L的CHCOOK与0.1 mol/L的盐酸等体积混合后，溶液中下列粒子的3 物质的量关系正确的是( ) ――，A、 C(CHCOO)==C(Cl)==C(H)>C(CHCOOH) 33 ――，B、 C(CHCOO)==C(Cl)>C(CHCOOH)>C(H) 33 ――，C、 C(CHCOO)>C(Cl)>C(H)>C(CHCOOH) 33 ――，D、 C(CHCOO)>C(Cl)>C(CHCOOH)>C(H) 33 KCl+CHCOOH～又知CHCOOK过量～反应后溶液中解析:两溶液混合后CHCOOK+HCl 333 ―CHCOOK、CHCOOH和KCl物质的量相等。由于CHCOOH的电离和CHCOO的水解程度均很小～3333 ，―――且CHCOOH的电离占主导地位～因此～C(CHCOO)>C(H)>C(OH)。又知C(Cl)==0.05 mol/L～33 C(CHCOOH)<0.05 mol/L。因此～选项中D是正确的。 3 4、酸碱中和型 (1) 恰好中和型 -1 ,点击试题,在10ml 0.1mol?LNaOH溶液中加入同体积、同浓度HAc 溶液，反应后溶液中各微粒的浓度关系错误的是( )。 +-+-+--+A(c(Na),c(Ac),c(H),c(OH) B(c(Na),c(Ac),c(OH),c(H) +-++--C(c(Na),c(Ac)，c(HAC) D(c(Na)，c(H),c(Ac)，c(OH) -3×10解析:由于混合的NaOH与HAc物质的量都为1mol～两者恰好反应生成NaAc～等同于 ---。+单一溶质～故与题型?方法相同。由于少量Ac发生水解:Ac + HOHAc+ OH故有c(Na)2 --+,c(Ac),c(OH),c(H)～根据物料守恒C正确～根据电荷守恒D正确～A错误。故该题选项为A。 (2) pH等于7型 ,点击试题,常温下，将甲酸和氢氧化钠溶液混合，所得溶液pH,7，则此溶液中( )。 -+-+A(c(HCOO),c(Na) B(c(HCOO),c(Na) -+-+C(c(HCOO),c(Na) D(无法确定c(HCOO)与c(Na)的关系 解析:本题绝不能理解为恰好反应～因完全反应生成甲酸钠为强碱弱酸盐～溶液呈碱性～而engineering construction and acceptance specification for 5 GB50275-98 compressor, fan and pump installation engineering construction and acceptance specification for lifting 6 HG20201-2000 construction installation engineering construction standard 7, HG ... Pressure gauges, using an installed spark arrestor for acetylene cylinders, illegal carrying, using gas cylinders, each operator fined 20 Yuan. Gas bottle without the hot sun exposure measures, responsibilities of team a fine of 50 Yuan. (7) at height and the opening and provisional protective measures have been taken, the responsible unit fined 200 Yuan, construction person in charge a fine of 50 Yuan. (8) no permit to work and the job in accordance with the provisions, on the job unit fined 300 yuan, the unit will charge a fine of 100 Yuan. 7.5.3 accident penalties (1) injuries fatalities directly punished 1~2 million. Direct responsibility for the accident and have the corresponding responsibility of leadership, such as concerning administrative sanctions, should be brought to the company or the relevant Department. (2) personal injury accident occurred, the direct punishment 0.5-10,000 yuan, responsible for the direct responsibility for the accident and the responsibility of leadership, such as concerning administrative sanctions should be dealt with by the personnel Department of the company. (3) personal injury accident occurs, the direct punishment 500-1000, who is directly responsible for the accident responsibility, give notice of criticism and 50-100 economic sanctions against them. (4) to conceal the accident, reported without undue delay or false, to inform the administrative leadership of the criticism, resulting in serious consequences, the pursuit of leadership, along with 500-1000 punishment. (5) significant near miss should be ++--现在Ph=7～故酸略为过量。根据溶液中电荷守恒:c(Na)+ c(H)= c(HCOO)，c(OH) 因 +-+-pH=7～故c(H)= c(OH)～所以有c(Na)= c(HCOO)～答案为C。 (3) 反应过量型 ,点击试题,常温下将稀NaOH溶液与稀CHCOOH溶液混合，不可能出现的结果是 3 —++—A(pH,7，且 (OH) , (Na) , (H) ,(CHCOO) ccc c3 ++——B(pH,7，且 c(Na) + c(H) = c(CHCOO) + c(OH) 3 —++—C(pH,7，且c(CHCOO) , c(H) ,c(Na), c(OH) 3 —++—D(pH,7，且c(CHCOO) , c(Na) ,c(H) = c(OH) 3 ,随堂练习,1、将 标准 excel标准偏差excel标准偏差函数exl标准差函数国标检验抽样标准表免费下载红头文件格式标准下载 状况下的2.24LCO通入150ml1mol/LnaOH溶液中，下列说2 法正确的是,, -2-A,c(HCO)略大于c(CO) 33 -2- B,c(HCO)等于c(CO) 33 +2--C,c(Na)等于c(CO)与c(HCO)之和 33 -2-D,c(HCO)略小于c(CO) 33 ,12、向0.1mol?LNaOH溶液中通入过量CO后，溶液中存在的主要离子是, , 2 ，，2,, A Na、CO B Na、HCO 33 ，,2,,C HCO、CO D Na、OH 33 四、守恒问题在电解质溶液中的应用 解此类题的关键是抓住溶液呈中性(即阴阳离子所带电荷总数相等)及变化前后原子的个数 守恒两大特点。若题中所给选项为阴阳离子的浓度关系，则应考虑电荷守恒，若所给选项等 式关系中包含了弱电解质的分子浓度在内，则应考虑物料守恒。 ,点击试题, 表 关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf 示0.1 mol/L NaHCO溶液中有关粒子浓度的关系正确的是( ) 3 ，，――2― A、C(Na)>C(HCO)>C(CO)>C(H)>C(OH) 33 ，，―2―― B、C(Na)+C(H)==C(HCO)+C(CO)+C(OH) 33 ，，―2―― C、C(Na)+C(H)==C(HCO)+2C(CO)+C(OH) 33 ，―2― D、C(Na)==C(HCO)+C(CO)+C(HCO) 3323 ―，解析:A、NaHCO溶液因为水解大于电离而呈碱性～因此C(OH)>C(H)。 3 2―B、应考虑电荷守恒～C(CO)前应乘以2,C、电荷守恒符合题意,D、含弱电解质分子应考3 ，，――2―虑物料守恒～在NaHCO溶液中存在下列关系:NaHCO==Na+HCO ,HCOH+CO ,33333 ―，――2―HCO+HO HCO+OH 则C(Na)==C(HCO)+C(CO)+C(HCO)符合题意。故选CD 32233323 1、两种物质混合不反应: ,点击试题,用物质的量都是0.1 mol的CHCOOH和CHCOONa配制成1L混合溶液，已知其33 -+中C(CHCOO),C(Na)，对该混合溶液的下列判断正确的是( ) 3 engineering construction and acceptance specification for 5 GB50275-98 compressor, fan and pump installation engineering construction and acceptance specification for lifting 6 HG20201-2000 construction installation engineering construction standard 7, HG ... Pressure gauges, using an installed spark arrestor for acetylene cylinders, illegal carrying, using gas cylinders, each operator fined 20 Yuan. Gas bottle without the hot sun exposure measures, responsibilities of team a fine of 50 Yuan. (7) at height and the opening and provisional protective measures have been taken, the responsible unit fined 200 Yuan, construction person in charge a fine of 50 Yuan. (8) no permit to work and the job in accordance with the provisions, on the job unit fined 300 yuan, the unit will charge a fine of 100 Yuan. 7.5.3 accident penalties (1) injuries fatalities directly punished 1~2 million. Direct responsibility for the accident and have the corresponding responsibility of leadership, such as concerning administrative sanctions, should be brought to the company or the relevant Department. (2) personal injury accident occurred, the direct punishment 0.5-10,000 yuan, responsible for the direct responsibility for the accident and the responsibility of leadership, such as concerning administrative sanctions should be dealt with by the personnel Department of the company. (3) personal injury accident occurs, the direct punishment 500-1000, who is directly responsible for the accident responsibility, give notice of criticism and 50-100 economic sanctions against them. (4) to conceal the accident, reported without undue delay or false, to inform the administrative leadership of the criticism, resulting in serious consequences, the pursuit of leadership, along with 500-1000 punishment. (5) significant near miss should be +-- A.C(H),C(OH) B.C(CHCOOH)，C(CHCOO),0.2 mol/L 33 --- C.C(CHCOOH),C(CHCOO) D.C(CHCOO)，C(OH),0.2 mol/L 333 2、两种物质恰好完全反应 -1,点击试题,在10ml 0.1mol?LNaOH溶液中加入同体积、同浓度HAc溶液，反应后溶液中各微粒的浓度关系错误的是( )。 +-+-A(c(Na),c(Ac),c(H),c(OH) +--+B(c(Na),c(Ac),c(OH),c(H) +-C(c(Na),c(Ac)，c(HAC) ++--D(c(Na)，c(H),c(Ac)，c(OH) 3、两种物质反应，其中一种有剩余: (1)酸与碱反应型 关注所给物质的量是物质的量浓度还是pH。 在审题时，要关注所给物质的量是“物质的量浓度”还是“pH”，否则会很容易判断错误。(解答此类题目时应抓住两溶液混合后剩余的弱酸或弱碱的电离程度和生成盐的水解程度的相对大小。) -1-1,点击试题,把0.02 mol?L HAc溶液与0.01 mol?LNaOH溶液等体积混合，则混合液中微粒浓度关系正确的是( ) -+- A、c(Ac),c(Na) B、c(HAc),c(Ac) +---1 C、2c(H),c(Ac)-c(HAc) D、c(HAc)+c(Ac),0.01 mol?L(2)盐与碱(酸)反应型 ,讲,解答此类题目时应抓住两溶液混合后生成的弱酸或弱碱的电离程度和剩余盐的水解程度的相对大小。 -1-1将0.1mol?L 醋酸钠溶液20mL与0.1mol?L盐酸10mL混合后，溶液显酸性，则溶液中有关粒子浓度关系正确的是( )。 --+A(c(CHCOO),c(Cl),c(H),c(CHCOOH) 33 --+B(c(CHCOO),c(Cl)，c(CHCOOH),c(H) 33 --+C(c(CHCOO),c(Cl),c(H),c(CHCOOH) 33 ++---D(c(Na)，c(H),c(CHCOO)，c(Cl)，c(OH) 3 4、不同物质同种离子浓度比较型: +,点击试题,物质的量浓度相同的下列溶液中，NH浓度最大的是( )。 4 A(NHCl B(NHHSO C(NHCOONH D(NHHCO 4443443 +++ 解析:NH在溶液中存在下列平衡:NH + HO NH〃HO + HB中NHHSO电离出大4423244 +++--量H～使平衡向左移动～故B中c(NH)大于A中的c(NH)～C项的CHCOO和D项的HCO4433 ++水解均呈碱性～使平衡向右移动～故C、D中c(NH)小于A中c(NH)～正确答案为B。 44 engineering construction and acceptance specification for 5 GB50275-98 compressor, fan and pump installation engineering construction and acceptance specification for lifting 6 HG20201-2000 construction installation engineering construction standard 7, HG ... Pressure gauges, using an installed spark arrestor for acetylene cylinders, illegal carrying, using gas cylinders, each operator fined 20 Yuan. Gas bottle without the hot sun exposure measures, responsibilities of team a fine of 50 Yuan. (7) at height and the opening and provisional protective measures have been taken, the responsible unit fined 200 Yuan, construction person in charge a fine of 50 Yuan. (8) no permit to work and the job in accordance with the provisions, on the job unit fined 300 yuan, the unit will charge a fine of 100 Yuan. 7.5.3 accident penalties (1) injuries fatalities directly punished 1~2 million. Direct responsibility for the accident and have the corresponding responsibility of leadership, such as concerning administrative sanctions, should be brought to the company or the relevant Department. (2) personal injury accident occurred, the direct punishment 0.5-10,000 yuan, responsible for the direct responsibility for the accident and the responsibility of leadership, such as concerning administrative sanctions should be dealt with by the personnel Department of the company. (3) personal injury accident occurs, the direct punishment 500-1000, who is directly responsible for the accident responsibility, give notice of criticism and 50-100 economic sanctions against them. (4) to conceal the accident, reported without undue delay or false, to inform the administrative leadership of the criticism, resulting in serious consequences, the pursuit of leadership, along with 500-1000 punishment. (5) significant near miss should be engineering construction and acceptance specification for 5 GB50275-98 compressor, fan and pump installation engineering construction and acceptance specification for lifting 6 HG20201-2000 construction installation engineering construction standard 7, HG ... Pressure gauges, using an installed spark arrestor for acetylene cylinders, illegal carrying, using gas cylinders, each operator fined 20 Yuan. Gas bottle without the hot sun exposure measures, responsibilities of team a fine of 50 Yuan. (7) at height and the opening and provisional protective measures have been taken, the responsible unit fined 200 Yuan, construction person in charge a fine of 50 Yuan. (8) no permit to work and the job in accordance with the provisions, on the job unit fined 300 yuan, the unit will charge a fine of 100 Yuan. 7.5.3 accident penalties (1) injuries fatalities directly punished 1~2 million. Direct responsibility for the accident and have the corresponding responsibility of leadership, such as concerning administrative sanctions, should be brought to the company or the relevant Department. (2) personal injury accident occurred, the direct punishment 0.5-10,000 yuan, responsible for the direct responsibility for the accident and the responsibility of leadership, such as concerning administrative sanctions should be dealt with by the personnel Department of the company. (3) personal injury accident occurs, the direct punishment 500-1000, who is directly responsible for the accident responsibility, give notice of criticism and 50-100 economic sanctions against them. (4) to conceal the accident, reported without undue delay or false, to inform the administrative leadership of the criticism, resulting in serious consequences, the pursuit of leadership, along with 500-1000 punishment. (5) significant near miss should be