怎样配置浓度为0.1M，ph=4.75的醋酸钠和醋酸的缓冲液，具体算法 0.1M PH=4的醋酸钠缓冲液 怎么配置？醋酸钠为三水醋酸...
\u600e\u4e48\u914d\u52360.1mol/L pH\u4e3a4.8\u7684\u918b\u9178-\u918b\u9178\u94a0\u7f13\u51b2\u6db2\uff1f\u5e94\u8be5\u5148\u914d\u52360.1mol/L\u918b\u9178\u94a0\uff0c
\uff08\u6839\u636e\u4e09\u6c34\u5408\u918b\u9178\u94a0\u7684\u6469\u5c14\u8d28\u91cf\uff0c\u9664\u6c34\u7684\u8d28\u91cf\uff0cm/M=n\uff09
\u518d\u5b9a\u5bb9\uff08\u6839\u636e\u7f13\u51b2\u6db2\u7684\u7528\u91cf\u6765\u786e\u5b9a\u662f\u8981250ml\u3001500ml\u30011000ml\uff09
\u7136\u540e\u7528PH\u8ba1\u6d4b\u5b9a\u9178\u5ea6\uff0c
\u518d\u7528\u51b0\u918b\u9178\u8c03\u8282PH\u503c\u81f34.8\uff08\u8fb9\u7528\u80f6\u5934\u6ef4\u7ba1\u6ef4\u52a0\u51b0\u918b\u9178\uff0c\u8fb9\u7528PH\u8ba1\u6d4b\u5b9a\uff0c\u9632\u6b62\u8fc7\u9178\uff0c\u8fc7\u9178\u53ef\u7528\u7a00\u6c22\u6c27\u5316\u94a0\u6eb6\u6db2\u6765\u8c03\u8282\uff09
\u4f7f\u7528\u918b\u9178-\u918b\u9178\u94a0\u914d\u5236\u8be5\u7f13\u51b2\u6eb6\u6db2\u3002\u8be5\u7f13\u51b2\u6eb6\u6db2\u7684pH\u503c\u7684\u8ba1\u7b97\u516c\u5f0fPH=PKa+lg[C(NaAc)/C(HAc)]\uff08\u5728\u6b64\uff0cKa\u662f\u918b\u9178\u7684\u89e3\u79bb\u5e38\u6570\uff0cPKa=-lgKa\uff0c\u918b\u9178\u7684Ka=1.8*10-5(1.8\u4e5810\u7684-5\u6b21\u65b9)\uff0cC(NaAc)\u6307\u918b\u9178\u94a0\u7684\u6d53\u5ea6\u53730.1mol/L\uff0cC(HAc)\u6307\u918b\u9178\u7684\u6d53\u5ea6\u672a\u77e5\u5f85\u6c42\uff09\u3002\u5c06PH=4\u4ee3\u5165\uff0cC(HAc)=0.56mol/L\u3002
\u6839\u636e\u6240\u9700\u7684\u7f13\u51b2\u6eb6\u6db2\u7684\u4f53\u79ef\u7b97\u51fa\u53d6\u7528\u7684\u918b\u9178\u53ca\u918b\u9178\u94a0\u7684\u91cf\u3002\u5047\u8bbe\u6240\u914d\u7f13\u51b2\u6eb6\u6db2\u7684\u4f53\u79ef\u4e3aV\u5347\uff0c\uff081\uff09\u79f0\u91cf\u4e09\u6c34\u918b\u9178\u94a0\uff1a\u6240\u9700\u79f0\u91cf\u7684\u4e09\u6c34\u918b\u9178\u94a0\u7684\u8d28\u91cf\u4e3am=C(NaAc)*V*M(NaAc.3H2O)(M\u6307\u4e09\u6c34\u918b\u9178\u94a0\u7684\u6469\u5c14\u8d28\u91cf\uff0c\u5373136g/mol\uff09\uff0c\u5373m=13.6*V\uff08\u514b\uff09\uff0cV\u7684\u5355\u4f4d\u662f\u5347\u3002
\uff082\uff09\u914d\u52360.56mol/L\u918b\u9178\uff1a\u51b0\u918b\u9178\u662f\u6db2\u4f53\uff0c\u51b0\u918b\u9178\u7684\u6469\u5c14\u6d53\u5ea6\u7ea6\u4e3a17.5mol/L\uff0c\u6839\u636e\u521a\u624d\u7b97\u51fa\u7684C(HAc)=0.56mol/L\uff0c\u914d\u5236V\u5347\u6eb6\u6db2\u65f6\uff0c\u91cf\u53d6\u51b0\u918b\u9178\u7684\u91cfV'=C(HAc)*V*1000/17.5(mL)=32*V(\u6beb\u5347)\u3002
\uff083\uff09\u5c06\u79f0\u597d\u7684\u918b\u9178\u94a0\u548c\u91cf\u597d\u7684\u51b0\u918b\u9178\u52a0\u5165V\u5347\u6c34\u4e2d\u6eb6\u89e3\u3001\u6405\u62cc\u5747\u5300\u5373\u53ef\u3002
醋酸的Ka=1.75x10^-5，PKa=4.76
4.75=4.76+lgc盐/c酸
c盐/c酸=1.023
那样就是0.2molL-的HAc与0.2046molL-的NaAc等体积混合
粗略说就是02molL-的醋酸和醋酸钠等体积混合
绛旓細PH=PKa+lgc鐩/c閰 閱嬮吀鐨凨a=1.75x10^-5锛孭Ka=4.76 4.75=4.76+lgc鐩/c閰 c鐩/c閰=1.023 閭ｆ牱灏鏄0.2molL-鐨凥Ac涓0.2046molL-鐨凬aAc绛変綋绉贩鍚 绮楃暐璇村氨鏄02molL-鐨勯唻閰稿拰閱嬮吀閽犵瓑浣撶Н娣峰悎
绛旓細浣跨敤閱嬮吀-閱嬮吀閽犻厤鍒惰缂撳啿婧舵恫銆傝缂撳啿婧舵恫鐨刾H鍊肩殑璁＄畻鍏紡PH=PKa+lg[C(NaAc)/C(HAc)]锛堝湪姝わ紝Ka鏄唻閰哥殑瑙ｇ甯告暟锛孭Ka=-lgKa锛岄唻閰哥殑Ka=1.8*10-5(1.8涔10鐨-5娆℃柟)锛孋(NaAc)鎸囬唻閰搁挔鐨勬祿搴﹀嵆0.1mol/L锛孋(HAc)鎸囬唻閰哥殑娴撳害鏈煡寰呮眰锛夈傚皢PH=4浠ｅ叆锛孋(HAc)=0.56mol/L銆傛牴鎹墍...
绛旓細鍏堣绠0.1M鐨勭７閰搁挔鐩愰渶瑕佸灏戠函鐨勭７閰搁挔鐩愶紝纾烽吀閽犵洂婧舵恫搴旇鏄⒈鎬х殑锛屽強PH鍊煎ぇ浜7锛鎵浠ヨ鍔犱竴鐐圭７閰革紝浣垮緱PH鍊=7锛岃绠楃敤澶氬皯浠涔堟祿搴︾殑纾烽吀锛岀劧鍚庣敤瀹归噺鐡堕櫔灏卞彲浠ヤ簡銆傘佸鏋滀翰璁や负瀵癸紝璇蜂翰閲囩撼锛岃阿璋
绛旓細灏嗕袱鑰呭姞鍒1000姣崌钂搁姘翠腑,鎼呮媽鍧囧寑鍗冲彲.
绛旓細鐢ㄩ唻閰稿拰閱嬮吀閽犻厤鍒讹紝缂撳啿婧舵恫鐨PH=PKa+lg[C(NaAc)/C(HAc)]锛堝湪姝わ紝C(HAc)鎸囬唻閰哥殑娴撳害锛C(NaAc)鎸囬唻閰搁挔鐨娴撳害,Ka鏄閱嬮吀鐨勮В绂诲父鏁帮紝PKa=-lgKa锛岄唻閰哥殑Ka=1.8*10-5(1.8涔10鐨-5娆℃柟)锛屽皢PH=4浠ｅ叆锛屽彲寰桟(NaAc)/C(HAc)鐨勫硷紝鍙圕(NaAc)=0.1M锛堝嵆0.1mol/L锛,鍙緱鍒伴唻閰哥殑娴撳害...
绛旓細0.1M PBS 婧舵恫,浠1L閲忎负渚: NaCl 80g Na2HPO4路12H2O 32.3g NaH2PO4路2H2O 4.5g (鍗佷簩姘村悎纾烽吀姘簩閽犲拰浜屾按鍚堢７閰镐簩姘㈤挔,鏈甯歌,涓鎬庝箞鍚告疆, 濡傛灉浣犳湁鏃犳按鐨勮繖涓ょ纾烽吀鐩愪篃琛,鎹㈢畻涓涓嬪嵆鍙,鏃犳按鐨勫鏄撳惛娼) 鍔犲弻钂告按 900mL 宸﹀彸,鐢℉Cl/NaOH 璋pH鑷 7.2,鏈鍚庡畾瀹逛负 1000mL銆 鏈洖绛旂敱绉戝鏁欒偛鍒...
绛旓細16.25鍏嬩笁姘寲閾侊紝0.833姣崌娴撶洂閰革紝鍔犳按鑷1鍗
绛旓細鍑嗙‘绉板彇绾0.16g浜800鈩冪伡鐑ц嚦鎭掗噺鐨勫熀鍑哯nO,缃簬灏忕儳鏉腑,鍔犲叆0.4mL鐩愰吀,婧惰В鍚庣Щ鍏200mL瀹归噺鐡,鍔犳按绋閲婅嚦鍒诲害,娣峰寑.3锛 EDTA婧舵恫娴撳害鐨勬爣瀹 鍚稿彇30.00~35.00mL閿屾爣鍑嗘憾娑蹭簬250mL閿ュ舰鐡朵腑,鍔犲叆70mL姘,鐢40%姘ㄦ按涓拰鑷硃H涓7~8,鍐嶅姞10mL姘ㄦ按-姘寲閾电紦鍐叉恫锛pH=10锛,鍔犲叆灏戣閾粦T鎸囩ず鍓,...
绛旓細0.1M PBS 婧舵恫锛屼互1L閲忎负渚嬶細NaCl 80g Na2HPO4路12H2O 32.3g NaH2PO4路2H2O 4.5g 锛堝崄浜屾按鍚堢７閰告阿浜岄挔鍜屼簩姘村悎纾烽吀浜屾阿閽狅紝鏈甯歌锛屼笉鎬庝箞鍚告疆锛屽鏋滀綘鏈夋棤姘寸殑杩欎袱绉嶇７閰哥洂涔熻锛屾崲绠椾竴涓嬪嵆鍙紝鏃犳按鐨勫鏄撳惛娼傦級鍔犲弻钂告按 900mL 宸﹀彸锛岀敤HCl/NaOH 璋pH鑷 7.2锛屾渶鍚庡畾瀹逛负 1000mL銆
绛旓細PH=PKa^2-lg[H2PO4^-]/[HPO4^2-]7.8=7.2-lg[H2PO4^-]/[HPO4^2-]lg[H2PO4^-]/[HPO4^2-]=0.6 [H2PO4^-]/[HPO4^2-]=4/1 [H2PO4^-]=0.1脳4/5=0.08mol/L [HPO4^2-]=0.1脳1/5=0.02mol'L 鍋囪閰嶅埗杩欐牱婧舵恫1鍗.閭ｄ箞闇瑕佹棤姘碞a2HPO4 120脳0.1=12鍏 ...
