配制50ml0.5mol/L的NaCl溶液的标准实验报告
可以取5.85g已灼烧至恒重的工作基准试剂氯化钠,溶解定容到200ml的容量瓶中.绛旓細浠ょ█閲婂悗鐨勬阿姘у寲閽犳憾娑茬殑娴撳害涓篶锛屾牴鎹█閲婂畾寰嬶紝绋閲婂墠鍚庢憾璐ㄧ殑鐗╄川鐨勯噺涓嶅彂鐢熶笉鍙橈紝鍒欙細50ml脳0.5mol/L=250ml脳c瑙ｅ緱c=0.1mol/L锛庢晠閫夛細A
绛旓細瑕閰嶅埗 50mL 0.5mol/L 鐨 NaCl 婧舵恫锛岄渶瑕佺殑鎽╁皵鏁颁负锛50 mL 脳 0.5 mol/L = 25 mmol NaCl 鐨勬懇灏旇川閲忎负 58.44 g/mol锛屽洜姝ら渶瑕佺殑璐ㄩ噺涓猴細25 mmol 脳 58.44 g/mol = 1.461 g 灏 1.461 g 鐨 NaCl 鍔犲叆鍒板閲忎负 50 mL 鐨勭儳鏉腑锛岀劧鍚庡姞鍏ヨ冻澶熺殑钂搁姘翠互浣挎讳綋绉揪鍒 50 mL銆...
绛旓細闇瑕侊細鎺ヨ繎50 ml姘
绛旓細鍒欓渶瑕丯a2SO4路10H2O鍏2.5*10^(-2)mol m=n*M=2.5*10^(-2)*322=8.05g
绛旓細銆1銆戣绠楋細鑽夐吀閾电殑绉伴噺璐ㄩ噺 m m = cvM = 50mlx0.5mol/ml梅1000ml/Lx142.11g/mol = 3.553 g 銆2銆閰嶅埗锛氬皢绉伴噺鐨勮崏閰搁摰鐢ㄥ皯閲忔按婧惰В锛屽畾閲忚浆绉诲埌50ml瀹归噺鐡朵腑銆傚湪杞Щ鍒拌瘯鍓傜摱涓紝璐村ソ鏍囩銆
绛旓細鈶閰嶅埗0.5mol/L NaOH婧舵恫50mL m(NaOH)=cV*M=0.5*50*10^-3*40=1(g)鈶￠厤鍒0.1mol/L CuSO4婧舵恫500mL n(CuSO4路5H2O)=n(CuSO4)=cV=0.1*500*10^-3=0.05(mol)m(CuSO4路5H2O)=n(CuSO4路5H2O)*M=0.05*250=12.5(g)鈶㈤厤鍒0.2mol/L H2SO4婧舵恫500mL m(H2SO4)=cV*M=0.2*...
绛旓細瑕侀厤缃0.5mol/L鐨凬aCL婧舵恫50ML,搴旂О鍙栧浐浣揘aCL澶氬皯鍏?鐢ㄥ寲瀛﹀叕寮忚绠 n(NaCl)=0.5Lx1mol.L-=0.5mol m(NaCl)=0.5molx58.5g.mol_=29.25g,鍥犱负鎵樼洏澶╁钩鐨勭簿搴0.1g,鎵浠ラ渶瑕佺О鍙29.3gNaCl.
绛旓細闇瑕佺～閰哥殑鐗╄川鐨勯噺n=cV=0.5*0.05=0.025mol 闇瑕佹祿纭吀鐨勮川閲弇=nM=0.025*98=2.45g 闇瑕98%娴撶～閰革紙瀵嗗害鏄1.84g/mL锛夌殑浣撶НV=m/蟻=2.45/1.84=1.33mL 灏嗘祿纭吀鍊掑叆灏戦噺姘翠腑鎼呮媽锛岀劧鍚庤浆绉绘礂娑ゅ埌50mL瀹归噺鐡朵腑锛屾渶鍚庣敤鑳跺ご婊寸瀹氬鍒50mL銆
绛旓細杩欎釜渚濇嵁璐ㄩ噺瀹氬緥锛岄鍏堝緱鍑虹～閰哥殑鎽╁皵閲05*0.5=0.025mol锛岀劧鍚庣～閰稿拰姘㈡哀鍖栭挔鏄寜鐓1:2閰嶆瘮锛屽氨闇瑕0.05mol姘㈡哀鍖栭挔銆傛渶鍚庝箻涓婃阿姘у寲閽犵墿璐ㄧ殑閲忋
绛旓細n(AlCl3)=0.05*0.5=0.025mol ; n[Al(OH)3]=1.56/78=0.02mol (1)NaOH涓嶈冻閲,Al3+娌℃湁娌夋穩瀹,寰楀埌0.005molAl3+涓0.02molAl(OH)3:c=0.02*3/0.05=1.2mol/L (2)NaOH瓒抽噺,鏈夐儴鍒咥l(OH)3娌夋穩婧惰В,寰楀埌0.005molAlO2-涓0.02molAl(OH)3:c=(0.02*3+0.005*4)/0....
