{"id":3598,"date":"2022-03-21T15:16:51","date_gmt":"2022-03-21T07:16:51","guid":{"rendered":"https:\/\/www.hnyt99.com\/index.php\/2022\/03\/21\/%e6%b0%b4%e8%b4%a8%e7%9b%91%e6%b5%8b%e4%b8%ad%e6%bb%b4%e5%ae%9a%e5%88%86%e6%9e%90%e8%bf%87%e7%a8%8b\/"},"modified":"2022-03-26T10:10:51","modified_gmt":"2022-03-26T02:10:51","slug":"980","status":"publish","type":"post","link":"https:\/\/www.hnyt99.com\/baike\/shuizijc\/980.html","title":{"rendered":"\u6c34\u8d28\u76d1\u6d4b\u4e2d\u6ef4\u5b9a\u5206\u6790\u8fc7\u7a0b"},"content":{"rendered":"\n
\u3000\u3000\u2460\u6ef4\u5b9a\u524d HCl\u662f\u5f3a\u7535\u89e3\u8d28\uff0c\u5168\u90e8\u7535\u79bb\uff0c\u56e0\u6b64\u6eb6\u6db2\u4e2dH+\u7684\u6d53\u5ea6\u5c31\u7b49\u4e8eHCl\u7684\u6d53\u5ea6\uff1a<\/p>\n\n\n\n
\u3000\u3000[H+]=[HCl]=0.1mol\/L pH=1<\/p>\n\n\n\n
\u3000\u3000\u2461\u8ba1\u91cf\u70b9\u524d \u5f00\u59cb\u6ef4\u5b9a\u540e\uff0c\u6eb6\u6db2\u4e2d\u52a0\u5165\u4e86\u4e00\u90e8\u5206NaOH\u6eb6\u6db2\uff0c\u76f8\u5e94\u4e2d\u548c\u4e86\u4e00\u90e8\u5206HCl,\u76f4\u5230\u8ba1\u91cf\u70b9\u524d\uff0c\u6eb6\u6db2\u4e2d\u8fd8\u6709\u4e00\u90e8\u5206\u672a\u88ab\u4e2d\u548c\u7684HCl\uff0c\u6b64\u65f6\u6eb6\u6db2\u4e2dH+\u6d53\u5ea6\u53d6\u51b3\u4e8e\u5269\u4f59HCl\u7684\u6d53\u5ea6\uff0c\u5373\uff1a<\/p>\n\n\n\n