用0.10mol/LHAc溶液和0.10mol/LNaAc溶液配制PH=4.95的缓冲溶液100m 计算由0.080mol/LHAc溶液和0.20mol/LNa...
\u75280.10mol/LHAc\u548c0.20mol/LNaAc\u7b49\u4f53\u79ef\u6df7\u5408\u914d\u5236\u6210500 ml\u7684\u7f13\u51b2\u6eb6\u6db2\uff0c\u6c42\u6eb6\u6db2pH\u503c\u3002\u8bf7\u7ed9\u51fa\u8be6\u7ec6\u8fc7\u7a0b\u7f13\u51b2\u6eb6\u6db2pH=pKa+lg(\u583f/\u9178),\u81ea\u5df1\u5e26\u6570\u636e\u8fdb\u53bb\u7b97.
Ca=C\uff08HAc\uff09=0.080mol/L\uff1b Cb=C\uff08NaAc\uff09=0.20mol/L\uff1b PH=PKa-lg\uff08Ca/Cb\uff09=4.74-lg\uff080.08/0.2\uff09=\u7b54\u6848 \u5f88\u4e0d\u597d\u610f\u601d\u54e6\uff0c\u6211\u6ca1\u5e26\u8ba1\u7b97\u5668\uff0c\u6240\u4ee5\u4f60\u81ea\u5df1\u7b97\u4e0b\u54e6\uff01(*^__^*) \u563b\u563b\u2026\u2026
缓冲溶液中：pH=pKa + lg c（Ac-）/c（HAc）
lg c（Ac-）/c（HAc）=4.95-4.74=0.21
c（Ac-）/c（HAc）=1.32
需要醋酸的体积：
100/2.23=43.1ml
需要醋酸钠的体积：
100-43.1=56.9ml
绛旓細纰辩殑閲忔槸灏戜竴鐐圭殑锛屽洜姝锛瀊锛岄敊璇紱C銆丒鐐瑰搴旂殑鐐规槸鏄鹃吀鎬х殑锛岀粨鍚堢數鑽峰畧鎭掞紝鍏剁瀛愭祿搴︾敱澶у埌灏忕殑椤哄簭涓: c锛圕H 3 COO 锛 锛> c锛圢a 锛 锛> c锛圚 锛 锛> c锛圤H 锛 锛夛紝姝ｇ‘锛汥銆佸浘2瀹為獙閫夌敤鐢插熀姗欎笉鍚堥傦紝鍥犱负鍒氬ソ鍙嶅簲瀹屽叏鏃讹紝婧舵恫鏄剧⒈鎬э紝搴旈夌敤閰氶厼浣滄寚绀哄墏锛岄敊璇
绛旓細鏃犳按纰抽吀閽犲拰鐩愰吀鐨勫弽搴斿紡濡備笅锛歂a2CO3+HCI==2NaCI+CO2+H20 璁綱琛ㄧず鐢ㄥ幓鐩愰吀鐨勪綋绉紙ml)锛孋琛ㄧず鏃犳按纰抽吀閽犵殑娴撳害锛坈om/L锛塵琛ㄧず绉板彇鏃犳按纰抽吀閽犵殑璐ㄩ噺锛堝厠)锛屽垯鏈変笅鍒楀叧绯伙細C*V=m*1000/53.00 (53.00鏄1/2Na2CO3鐨勬懇灏旇川閲忥級璁＄畻鍚庢棤姘寸⒊閰搁挔鐨勭О鏍疯寖鍥存槸锛0.106鈥0.1325锛堝厠锛夊噺閲...
绛旓細缁堢偣璇樊
绛旓細缂撳啿婧舵恫涓細pH=pKa + lg c锛圓c-锛/c锛圚Ac锛塴g c锛圓c-锛/c锛圚Ac锛=4.95-4.74=0.21 c锛圓c-锛/c锛圚Ac锛=1.32 闇瑕侀唻閰哥殑浣撶Н锛100/2.23=43.1ml 闇瑕侀唻閰搁挔鐨勪綋绉細100-43.1=56.9ml
绛旓細瑙ｏ細闇瑕丯aoH鐨勬懇灏旈噺涓0.5L*0.1mol/L=0.05mol 锛屾晠NaoH鐨勮川閲忎负0.05mol*NaoH鐨勭浉瀵硅川閲=0.05mol*40g/mol=2g,鏁呯敤澶╁钩绉板彇2鍏嬶紝
绛旓細褰撴憾娑蹭腑NaH2PO4鍜孨a2HPO4鐗╄川鐨勯噺鐩哥瓑鏃讹紝婧舵恫鐨凱H=7.2锛屽彲鏍规嵁涓嬪紡璁＄畻锛2H3PO4+3NaOH=NaH2PO4+Na2HPO4+3H2O 2---3---1---1 0.02---0.03---0.02---0.02 闇瑕0.10mol/L姘㈡哀鍖栭挔婧舵恫鐨勪綋绉负1000*0.03/0.1=300ml.
绛旓細璁＄畻鐢∟aOH娑(0.10mol/l)婊村畾HCOOH(0.10mol/l)鍒板寲瀛﹁閲忕偣鏃舵憾娑茬殑PH,璋㈠暒 甯府蹇欏晩 璇﹁В 鍢垮樋 鎴戞潵绛 1涓洖绛 #鐑# 楣ゅ矖鐖嗙伀鑳屽悗鐨勫師鍥犳槸浠涔?绾卞涓69 2011-10-28 路 TA鑾峰緱瓒呰繃137涓禐 鐭ラ亾绛斾富 鍥炵瓟閲:115 閲囩撼鐜:0% 甯姪鐨勪汉:35.6涓 鎴戜篃鍘荤瓟棰樿闂釜浜洪〉 鍏虫敞 灞曞紑鍏ㄩ儴 ...
绛旓細銆愮瓟妗堛戯細涓や釜銆傜涓涓獊璺冧负HCl琚淮瀹氥丠3PO4琚淮瀹氬埌H2PO4锛涚浜屼釜绐佽穬涓篐3PO4-琚淮瀹氬埌HPO42-锛汬PO42-涓烘瀬寮遍吀婊村畾鏇茬嚎鏃犳槑鏄剧獊璺冦備袱涓傜涓涓獊璺冧负HCl琚淮瀹氥丠3PO4琚淮瀹氬埌H2PO4锛涚浜屼釜绐佽穬涓篐3PO4-琚淮瀹氬埌HPO42-锛汬PO42-涓烘瀬寮遍吀,婊村畾鏇茬嚎鏃犳槑鏄剧獊璺冦
绛旓細涔抽吀閽犲垎瀛愰噺涓112锛屽嵆鎽╁皵璐ㄩ噺涓112g/mol銆傞偅涔0.10mol/L涔抽吀閽犳祿搴︿篃灏辨槸112*0.10=11.2g/L锛屽嵆0.0112g/ml銆250ml涔抽吀閽犳憾娑叉墍鍚憾璐ㄤ负0.0112*250锛坓锛夋墍浠ラ渶鐢112g/L涔抽吀閽犳憾娑茬殑浣撶Н涓0.0112*250/112=0.025锛圠锛0.025L涔熷氨鏄25ml銆
绛旓細瑕侀厤鍒100mL娴撳害绾︿负0.10mol/L鐨 K2Cr2O7鏍囧噯婧舵恫锛岄渶瑕佹寜鐓т笅闈㈢殑姝ラ鎿嶄綔锛1. 璁＄畻鎵闇鐨 K2Cr2O7 鐨勮川閲忥細鎽╁皵璐ㄩ噺(M) = 294.18 g/mol 鎵闇鐗╄川鐨勬懇灏旀暟(n) = 0.10 mol/L 脳 0.100 L = 0.010 mol 鎵闇鐗╄川鐨勮川閲(m) = M 脳 n = 29.418 g 2. 灏嗚绠楀嚭鐨 K2Cr2O7 璐ㄩ噺绉...
