25℃时，0.1mol?L-1的HA溶液中cH+cOH?=1010，0.01mol?L-1的BOH溶液pH=12．请回答下列问题：（1）HA是 25℃时，浓度均为0.1的HA溶液和BOH溶液，pH分别是1...
\uff081\uff0925\u2103\u65f6\uff0c0.1mol?L-1\u7684HA\u6eb6\u6db2\u4e2dc\uff08OH-\uff09=10-10 mol/L\uff0c0.01mol?L-1\u7684BOH\u6eb6\u6db2pH=12\uff0e\u8bf7\u56de\u7b54\u4e0b\u5217\u95ee\u9898\uff1a\uff081\uff09\u2460\uff09\u2460c\uff08OH-\uff09=10-10 mol/L\uff0c\u52190.1mol?L-1\u7684HA\u6eb6\u6db2\u4e2dc\uff08H+\uff09=10?1410?10mol/L\uff0c\u6240\u4ee5HA\u5f31\u9178\uff0c\u6240\u4ee5\u7535\u79bb\u65b9\u7a0b\u5f0f\u4e3a\uff1aHA?H++A-\uff0c\u6545\u7b54\u6848\u4e3a\uff1aHA?H++A-\uff1b \u2461A\uff0e\u52a0\u6c34\u7a00\u91ca\u4fc3\u8fdb\u9178\u7535\u79bb\uff0c\u6c22\u79bb\u5b50\u6d53\u5ea6\u3001\u9178\u6d53\u5ea6\u3001\u9178\u6839\u79bb\u5b50\u6d53\u5ea6\u90fd\u964d\u4f4e\uff0c\u4f46\u6c22\u79bb\u5b50\u6d53\u5ea6\u51cf\u5c0f\u7684\u91cf\u5c0f\u4e8e\u9178\u5206\u5b50\u51cf\u5c0f\u7684\u91cf\uff0c\u6240\u4ee5c(H+)c(HA)\u589e\u5927\uff0c\u6545A\u9519\u8bef\uff1bB\uff0e\u52a0\u6c34\u7a00\u91ca\u4fc3\u8fdb\u9178\u7535\u79bb\uff0c\u9178\u6d53\u5ea6\u3001\u9178\u6839\u79bb\u5b50\u6d53\u5ea6\u90fd\u964d\u4f4e\uff0c\u4f46\u9178\u6839\u79bb\u5b50\u6d53\u5ea6\u51cf\u5c0f\u7684\u91cf\u5c0f\u4e8e\u9178\u5206\u5b50\u51cf\u5c0f\u7684\u91cf\uff0c\u6240\u4ee5\u6c22\u6c27\u6839\u79bb\u5b50\u6d53\u5ea6\u589e\u5927\uff0c\u5219c(HA)c(A?)\u51cf\u5c0f\uff0c\u6545B\u6b63\u786e\uff1bC\uff0e\u6e29\u5ea6\u4e0d\u53d8\uff0c\u6c34\u7684\u79bb\u5b50\u79ef\u5e38\u6570\u4e0d\u53d8\uff0c\u6545C\u9519\u8bef\uff1bD\uff0e\u52a0\u6c34\u7a00\u91ca\u4fc3\u8fdb\u9178\u7535\u79bb\uff0c\u6c22\u79bb\u5b50\u6d53\u5ea6\u964d\u4f4e\uff0c\u4f46\u6c22\u6c27\u6839\u79bb\u5b50\u6d53\u5ea6\u589e\u5927\uff0c\u6545D\u9519\u8bef\uff1b\u6545\u9009B\uff1b \uff083\uff09\u7b49pH\u7684HA\u6eb6\u6db2\u548c\u76d0\u9178\uff0cHA\u662f\u5f31\u9178\uff0c\u76d0\u9178\u662f\u5f3a\u9178\uff0c\u6240\u4ee5HA\u7684\u7269\u8d28\u7684\u91cf\u6d53\u5ea6\u5927\u4e8e\u76d0\u9178\uff0c\u7b49\u4f53\u79ef\u7684HA\u548c\u76d0\u9178\uff0cHA\u7684\u7269\u8d28\u7684\u91cf\u5927\u4e8e\u76d0\u9178\uff0c\u6240\u4ee5\u5411\u7b49\u4f53\u79ef\u3001\u7b49pH\u7684HA\u6eb6\u6db2\u548c\u76d0\u9178\u4e2d\u5206\u522b\u52a0\u5165\u8db3\u91cfZn\uff0c\u4ea7\u751f\u7684H2\u662fHA\u591a\uff0c\u6545\u7b54\u6848\u4e3a\uff1a\u591a\uff1b \uff082\uff09\u2460\u82e5a=7\u6eb6\u6db2\u5448\u4e2d\u6027\uff0c\u662f\u5f3a\u78b1\u5f3a\u9178\u76d0\uff0c\u5219HB\u662f\u5f3a\u9178\uff0c\u6545\u7b54\u6848\u4e3a\uff1a\u5f3a\uff1b\u2461\u5f53a=10\u65f6\uff0c\u8be5\u76d0\u6eb6\u6db2\u5448\u78b1\u6027\uff0c\u8bf4\u660e\u662f\u5f3a\u78b1\u5f31\u9178\u76d0\uff0c\u6c34\u89e3\u5448\u78b1\u6027\uff0c\u65b9\u7a0b\u5f0f\u4e3a\uff1aB-+H2O?HB+OH-\uff0c\u6545\u7b54\u6848\u4e3a\uff1aB-+H2O?HB+OH-\uff1b\u2462NaB\u662f\u5f3a\u78b1\u5f31\u9178\u76d0\uff0c\u6eb6\u6db2\u5448\u78b1\u6027\uff0c\u4e0d\u6c34\u89e3\u7684\u6eb6\u6db2\uff1e\u6c34\u89e3\u7684\u79bb\u5b50\uff1e\u663e\u6027\u79bb\u5b50\uff1e\u9690\u6027\u79bb\u5b50\uff0c\u6240\u4ee5\u79bb\u5b50\u6d53\u5ea6\u5927\u5c0f\u4e3a\uff1ac\uff08Na+\uff09\uff1ec\uff08B-\uff09\uff1ec\uff08OH-\uff09\uff1ec\uff08H+\uff09\uff0c\u7535\u8377\u5b88\u6052\u5f97\uff1ac\uff08Na+ \uff09+c\uff08H+ \uff09=c\uff08B- \uff09+c\uff08OH- \uff09\uff0c\u7269\u6599\u5b88\u6052\u53ef\u77e5\uff1ac\uff08Na+ \uff09=c\uff08B- \uff09+c\uff08HB \uff09\uff0c\u6240\u4ee5c\uff08OH-\uff09-c\uff08HB\uff09=c\uff08H+ \uff09=10-10 mol/L\uff0c\u6545\u7b54\u6848\u4e3a\uff1ac\uff08Na+\uff09\uff1ec\uff08B-\uff09\uff1ec\uff08OH-\uff09\uff1ec\uff08H+\uff09\uff1bc\uff08Na+ \uff09+c\uff08H+ \uff09=c\uff08B- \uff09+c\uff08OH- \uff09\uff1b10-10 mol/L\uff0e
0.1mol/LHA\u6eb6\u6db2\u7684PH=1\uff0c\u8bf4\u660eHA\u5b8c\u5168\u7535\u79bb\uff0c\u4e3a\u5f3a\u7535\u89e3\u8d28\uff0c0.1mol/L\u7684BOH\u6eb6\u6db2\u7684pH=11\uff0c\u8bf4\u660e\u78b1\u4e0d\u5b8c\u5168\u7535\u79bb\uff0c\u4e3a\u5f31\u7535\u89e3\u8d28\uff0cA\uff0e\u4efb\u4f55\u7535\u89e3\u8d28\u6eb6\u6db2\u4e2d\u90fd\u5b58\u5728\u7535\u8377\u5b88\u6052\uff0c\u6839\u636e\u7535\u8377\u5b88\u6052\u5f97c\uff08B+\uff09+c\uff08H+\uff09=c\uff08A-\uff09+c\uff08OH-\uff09\uff0c\u6839\u636e\u7269\u6599\u5b88\u6052\u5f97c\uff08BOH\uff09+c\uff08B+\uff09=c\uff08A?\uff09\uff0c\u5219c\uff08H+\uff09=c\uff08BOH\uff09+c\uff08OH-\uff09\uff0c\u6545A\u9519\u8bef\uff1bB\uff0e\u5f31\u7535\u89e3\u8d28BOH\u6eb6\u6db2\u4e2d\u90fd\u5b58\u5728\u7535\u79bb\u5e73\u8861\uff0c\u52a0\u6c34\u7a00\u91ca\u4fc3\u8fdbBOH\u7535\u79bb\uff0c\u6240\u4ee5\u5c060.1mol?L-1BOH\u6eb6\u6db2\u7a00\u91ca\u81f30\uff0eOO1mol?L-1\u5219\u6eb6\u6db2\u7684pH\uff1e9\uff0c\u6545B\u9519\u8bef\uff1bC\uff0e\u6df7\u5408\u6eb6\u6db2pH=7\uff0c\u5219\u6eb6\u6db2\u4e2dc\uff08OH-\uff09=c\uff08H+\uff09\uff0c\u6eb6\u6db2\u4e2d\u5b58\u5728\u7535\u8377\u5b88\u6052c\uff08B+\uff09+c\uff08H+\uff09=c\uff08A-\uff09+c\uff08OH-\uff09\uff0c\u6240\u4ee5\u5f97c\uff08A-\uff09=c\uff08B+\uff09\uff0c\u6545C\u9519\u8bef\uff1bD\uff0e\u5c06\u4e0a\u8ff0\u4e24\u6eb6\u6db2\u6309\u4f53\u79ef\u6bd41\uff1a2\u6df7\u5408\uff0c\u5219\u4e24\u79cd\u6eb6\u6db2\u4e2d\u6eb6\u8d28\u4e3aBA\u548cBOH\uff0cBOH\u7684\u7535\u79bb\u7a0b\u5ea6\u5927\u4e8eB+\u6c34\u89e3\u7a0b\u5ea6\uff0c\u6240\u4ee5\u6eb6\u6db2\u5448\u78b1\u6027\uff0c\u5219c\uff08OH-\uff09\uff1ec\uff08H+\uff09\u3001c\uff08B+\uff09\uff1ec\uff08A-\uff09\uff0c\u6eb6\u6db2\u4e2dBOH\u6d53\u5ea6\u5927\u4e8ec\uff08OH-\uff09\uff0c\u6240\u4ee5\u79bb\u5b50\u6d53\u5ea6\u5927\u5c0f\u987a\u5e8f\u662fc\uff08B+\uff09\uff1ec\uff08A-\uff09\uff1ec\uff08BOH\uff09\uff1ec\uff08OH-\uff09\uff1ec\uff08H+\uff09\uff0c\u6545D\u6b63\u786e\uff1b\u6545\u9009D\uff0e
（1）25℃时，0.1mol?L-1的某酸HA中，如果该酸是强酸，氢离子浓度为0.1mol/L，氢氧根离子浓度为：10-13mol/L，则| c(H+) |
| c(OH?) |
故答案为：弱电解质；强电解质；
（2）弱酸在水溶液里存在电离平衡，其电离方程式为HA?H++A-，故答案为：HA?H++A-；
（3）A．加水稀释促进了弱酸的电离，溶液中氢离子的物质的量增大，HA的物质的量减小，相同溶液中：
| n(H+) |
| n(HA) |
| c(H+) |
| c(HA) |
B．加水稀释促进酸电离，酸浓度、酸根离子浓度都降低，但酸根离子浓度减小的量小于酸分子减小的量，所以氢氧根离子浓度增大，则
| c(HA) |
| c(A?) |
C．c（H+）与c（OH-）的乘积为水的离子积，温度不变，水的离子积常数不变，故C错误；
D．加水稀释促进酸电离，氢离子浓度降低，但氢氧根离子浓度增大，故D错误；
故选B；
（4）等pH的HA溶液和盐酸，HA是弱酸，盐酸是强酸，所以HA的物质的量浓度大于盐酸，等体积等pH的HA和盐酸，HA的物质的量大于盐酸，所以向等体积、等pH的HA溶液和盐酸中分别加入足量Zn，产生的H2是HA多，
故答案为：多．
绛旓細锛1锛夆憼寮辩數瑙ｈ川锛1鍒嗭級 鈶锛2鍒嗭級 鈶 D锛2鍒嗭級锛2锛夆憽>鈶>鈶>鈶紙2鍒嗭級锛3锛-1185 kJ路mol -1 锛2鍒嗭級 锛4锛夆懀锛2鍒嗭級锛5锛塁H 4 鈥8e 鈥 + 2H 2 O锛滳O 2 + 8H + 锛2鍒嗭級 6锛4锛2鍒嗭級 璇曢鍒嗘瀽锛氾紙1锛夆憼25鈩冩椂锛0锛1 mol路L 锛1 鐨凥A婧舵恫...
绛旓細A锛庯紙1锛0.1mol?L-1HCl婧舵恫涓璫锛圚+锛=0.1mol/L锛屾晠pH=-lgc锛圚+锛=-lg0.1=1锛屾晠绛旀涓猴細1锛涳紙2锛塁H3COONa灞炰簬寮虹⒈寮遍吀鐩愶紝閱嬮吀鏍规按瑙H3COO-+H2O?CH3COOH+OH-锛岀牬鍧忔按鐨勭數绂诲钩琛★紝婧舵恫鍛堢⒈鎬э紝婧舵恫鐨刾H锛7锛屾晠绛旀涓猴細锛烇紱CH3COO-+H2O?CH3COOH+OH-锛涳紙3锛夊皢0.1mol?L-1HCl涓0....
绛旓細姘㈡哀鍖栭挔鏄己纰辫兘鎶戝埗姘寸數绂伙紝鎵浠ユ贩鍚堟憾娑蹭腑鐢辨按鐢电鍑虹殑c锛圚 + 锛夛紴0.1mol?L -1 NaOH婧舵恫涓敱姘寸數绂诲嚭鐨刢锛圚 + 锛夛紟鏁呯瓟妗堜负锛氾紴锛庯紙3锛夊皢閰稿拰纰辩瓑浣撶Н娣峰悎鍚庯紝婧舵恫鐨勪綋绉墿澶т竴鍊嶏紝娴撳害缂╁皬涓哄師鏉ョ殑涓鍗婏紝鎵浠ユ牴鎹墿鏂欏畧鎭掑緱锛 1 2 脳0.1mol/L =0.05mol/L锛庢晠绛旀涓猴細0...
绛旓細鎵浠ヨ繖鍑犵婧舵恫pH鐢卞皬鍒板ぇ鐨勯『搴忎负鈶犫憽鈶モ懁鈶ｂ憿.锛2锛塊HAc CH 3 COO - +H+, ,璇ユ俯搴︿笅CH 3 COONa鐨勬按瑙ｅ钩琛′负CH 3 COO - +H 2 O CH 3 COOH+OH - 銆傛按瑙ｅ钩琛″父鏁 鎵浠 銆傦紙3锛25鈩冩椂锛pH锛3鐨勯唻閰革紝c(H + )=10 -3 mol/L, pH锛11鐨勬阿姘у寲閽犳憾娑,c(H + )=10...
绛旓細锛2锛夋隘鍖栭摰涓紝閾垫牴绂诲瓙姘磋В鐢熸垚涓姘村悎姘ㄥ拰姘㈢瀛愶紝婧舵恫鏄剧ず閰告э紝鏁呯瓟妗堜负锛氶吀锛沶h4++h2o?nh3?h2o+h+锛涳紙3锛夋皑姘村睘浜庡急鐢佃В璐紝涓嶅畬鍏ㄧ數绂伙紝姘寲閾典腑锛岄摰鏍圭瀛愪細鍙戠敓姘磋В鍙嶅簲锛屾墍浠ヤ簩鑰呯殑娴撳害鍧囦細灏忎簬0.1mol/l锛鏁呴塨锛涳紙4锛夋牴鎹憾娑蹭腑鐨勭數鑽峰畧鎭掞細c锛坣h4+锛+c锛坔+锛=c锛坥h-锛+2c锛坰o...
绛旓細锛1锛夌洂閰稿拰閱嬮吀涓洪吀锛屾阿姘у寲閽′负纰憋紝纰抽吀閽犮佹隘鍖栭捑鍜屾隘鍖栭摰涓虹洂锛岀⒊閰搁挔涓哄己纰卞急閰哥洂锛屽叾婧舵恫鍛堢⒈鎬э紝姘寲閽句负寮洪吀寮虹⒈鐩愶紝涓轰腑鎬э紝姘寲閾典负寮洪吀寮辩⒈鐩愶紝鍏舵憾娑蹭负閰告э紝鐩稿悓娴撳害鐨勭洂閰稿拰閱嬮吀锛岄唻閰镐腑姘㈢瀛愭祿搴﹀皬浜庣洂閰革紝鎵浠H鐩愰吀灏忎簬閱嬮吀锛屽垯杩欏嚑绉嶆憾娑茬殑pH鐢卞皬鍒板ぇ鐨勯『搴忔槸鈶犫憽鈶モ懁鈶ｂ憿锛屾晠...
绛旓細锛1锛25鈩冩椂锛K sp [Mg锛圤H锛 2 ]=1.8脳10 -11 锛孠 sP [Cu锛圤H锛 2 ]=2.2脳10 -20 锛庢阿姘у寲閾滅殑婧跺害绉父鏁板皬浜庢阿姘у寲闀佺殑婧跺害绉父鏁帮紟鎵浠ュ湪25鈩冧笅锛屽悜娴撳害鍧囦负0.1 mol?L -1 鐨凪gCl 2 鍜孋uCl 2 娣峰悎婧舵恫涓愭淮鍔犲叆姘ㄦ按锛屽厛鐢熸垚姘㈡哀鍖栭摐娌夋穩锛庡弽搴旂殑鍖栧鏂圭▼寮忔槸锛欳uCl 2 +2...
绛旓細鍒欑敓鎴愭阿姘旂殑浣撶Н鈶⑩憽鈶犫懀锛屾晠绛旀涓猴細鈶⑩憽鈶犫懀锛涳紙4锛夌～閰告憾娑蹭腑锛岀～閰哥涓姝ョ數绂诲嚭姘㈢瀛愶紝绗簩姝ュ張鐢电鍑烘阿绂诲瓙锛岀涓姝ョ數绂诲嚭鐨勬阿绂诲瓙鎶戝埗浜嗙浜屾鐨勭數绂伙紝鑰岀～閰告阿閽犳憾娑蹭腑纭吀姘㈡牴绂诲瓙鍙湁涓姝ョ數绂伙紝鎵浠.1mol/L H2SO4涓殑HSO4-鐨勭數绂诲害灏忎簬0.1mol/L NaHSO4涓璈SO4-鐨勭數绂诲害锛...
绛旓細锛1锛夌█鐩愰吀涓庢皑姘村弽搴旂敓鎴愪簡姘寲閾碉紝姘寲閾垫槸寮洪吀寮辩⒈鐩愶紝婧舵恫鍛堥吀鎬э紝搴旈夋嫨閰告у彉鑹茶寖鍥村唴鐨勬寚绀哄墏锛屾晠閫夌敳鍩烘锛屾晠绛旀涓猴細鐢插熀姗欙紱锛2锛25鈩冩椂姝ゆ皑姘翠腑C锛圚+锛=1脳10-11 mol/L锛鍒機锛圤H-锛=1脳10?141脳10?11mol/L=10-3mol/L锛屽嵆宸茬數绂绘皑姘寸殑娴撳害涓10-3mol/L锛屾墍浠ユ皑姘寸殑鐢电搴ξ...
绛旓細A锛嶩F閰镐负寮遍吀锛屼笉鑳藉畬鍏ㄧ數绂伙紝鍒25鈩冩椂锛0.1 mol?L-1HF婧舵恫涓璸H锛1锛屾晠A閿欒锛汢锛嶬sp鍙笌娓╁害鏈夊叧锛屽垯Ksp锛圕aF2锛夐殢娓╁害鐨勫彉鍖栬屽彉鍖栵紝涓庢祿搴︽棤鍏筹紝鏁匓閿欒锛汣锛嶬a锛圚F锛=3.6脳10-4 mol?L-1锛屾憾搴︾Н甯告暟Ksp锛圕aF2锛=1.46脳10-10mol3?L-3锛孠a脳Ksp鈮1锛屾晠C閿欒锛汥锛庝袱婧舵恫...
