欲使滴定时消耗0.10mol/lhcl溶液2025ml，问应称取基准试剂naco3多少克 理工学科 -> 化学
\u7406\u5de5\u5b66\u79d1->\u5316\u5b66\u89e3\uff1a
\u5728\u6e29\u5ea6\u538b\u5f3a\u76f8\u540c\u7684\u60c5\u51b5\u4e0b\uff0c\u540c\u6837\u4f53\u79ef\u7684\u6c2e\u6c14\u4e0e\u6c27\u6c14\u8d28\u91cf\u6bd4\u4e3a\u5176\u76f8\u5bf9\u5206\u5b50\u8d28\u91cf\u4e4b\u6bd4\uff0c\u6c2e\u6c14\u7684\u76f8\u5bf9\u5206\u5b50\u8d28\u91cf\u662f 14.0067\u00d72 = 28.0134 \u3002\u6c27\u6c14\u7684\u76f8\u5bf9\u5206\u5b50\u8d28\u91cf\u662f 15.9994\u00d72 = 29.9988 \u3002\u4e5f\u5c31\u662f\u8bf4\uff0c\u540c\u6837\u4f53\u79ef\u6c2e\u6c14\u4e0e\u6c27\u6c14\u8d28\u91cf\u6bd4\u4e3a 28.0134 : 29.9988 \u3002
\u5728\u7a7a\u6c14\u4e2d\uff0c\u6c2e\u6c14\u4e0e\u6c27\u6c14\u7684\u8d28\u91cf\u6bd4\u4e3a 75.47 : 23.20
\u8bbe\u6c27\u6c14\u4f53\u79ef\u4e3a V \uff0c\u6c2e\u6c14\u4f53\u79ef\u662f\u6c27\u6c14\u4f53\u79ef\u7684 x \u500d\u3002\u90a3\u4e48\u6c2e\u6c14\u4f53\u79ef\u662f Vx
(V\u00b7\u6c2e\u6c14\u5bc6\u5ea6) : (V\u00b7\u6c27\u6c14\u5bc6\u5ea6) = 28.0134 : 29.9988
(Vx\u00b7\u6c2e\u6c14\u5bc6\u5ea6) : (V\u00b7\u6c27\u6c14\u5bc6\u5ea6) = 75.47% : 23.20%
\u4e8c\u5f0f\u76f8\u9664\uff0c\u5f97\u5230
1/x = (28.0134 \u00f7 29.9988) \u00f7 (75.47% \u00f7 23.20%)
\u7b97\u5f97 x = 3.483(6)
\u6709\u6548\u6570\u5b57\u5e94\u8be5\u662f\u56db\u4f4d\uff0c\u591a\u7559\u4e00\u4f4d\u4ee5\u4fbf\u4e4b\u540e\u8ba1\u7b97\u3002
\u5982\u679c\u4f7f\u7528\u6574\u6570\u7684\u76f8\u5bf9\u539f\u5b50\u8d28\u91cf\uff0cx = 3.48538\u2026\u2026
\u8bbe\u5176\u5b83\u6c14\u4f53\u4f53\u79ef\u5206\u6570 a% \uff0c\u5bc6\u5ea6\u4e3a \u03c1 \u3002
\u6c27\u6c14\u4f53\u79ef\u5206\u6570\u4e3a (1 - a%)/(x + 1)\uff0c\u5219\u6c2e\u6c14\u4f53\u79ef\u5206\u6570\u4e3a x(1 - a%)/(x + 1) \u3002\u8fd9\u91cc\u8fb9 x \u662f\u5df2\u7ecf\u6c42\u51fa\u6765\u7684\uff0c\u662f\u7cfb\u6570\uff0ca \u624d\u662f\u672a\u77e5\u6570\u3002
\u6240\u6709\u6c14\u4f53\u7684\u4f53\u79ef\u5206\u6570\u5206\u522b\u4e58\u4ee5\u5176\u5bc6\u5ea6\u4e4b\u548c\uff0c\u7b49\u4e8e\u7a7a\u6c14\u7684\u5bc6\u5ea6\u3002\u6839\u636e\u6761\u4ef6\uff0c\u53ef\u4ee5\u5217\u51fa\u65b9\u7a0b\u7ec4
\u03c1a% + (1 - a%)/(x + 1)\u03c1\u6c27\u6c14 + x(1 - a%)/(x + 1)\u03c1\u6c2e\u6c14 = \u03c1\u7a7a\u6c14
\u03c1a% / \u03c1\u7a7a\u6c14 = 1.33%
\u7531\u4e0b\u8fb9\u8fd9\u4e2a\u65b9\u7a0b\u5f97\u5230 \u03c1 = 0.0133\u03c1\u7a7a\u6c14 / a%
\u4e8e\u662f\u4e0a\u8fb9\u90a3\u4e2a\u65b9\u7a0b\u53d8\u6210\u5982\u4e0b\u5f62\u5f0f
1.33\u03c1\u7a7a\u6c14 + (1 - a%)/(x + 1)\u03c1\u6c27\u6c14 + x(1 - a%)/(x + 1)\u03c1\u6c2e\u6c14 = \u03c1\u7a7a\u6c14
\u5176\u4e2d\u03c1\u7a7a\u6c14\u3001\u03c1\u6c27\u6c14\u3001\u03c1\u6c2e\u6c14 \u8fd8\u6709 x \u90fd\u662f\u5df2\u77e5\u7684\u4e86\u3002\u5e26\u5165\u6240\u6709\u5df2\u77e5\u91cf\uff0c\u5f97\u5230 a% = 1.154% \uff1b
\u5982\u679c\u4f7f\u7528\u6574\u6570\u7684\u76f8\u5bf9\u539f\u5b50\u8d28\u91cf\uff0ca% = 0.01152984\u2026\u2026 \u2248 1.15%
\u6c2e\u6c14\u4f53\u79ef\u5206\u6570 x(1 - a%)/(x + 1) = 76.80%
\u6c27\u6c14\u4f53\u79ef\u5206\u6570 (1 - a%)/(x + 1) = 22.05%
\u5982\u679c\u4f7f\u7528\u6574\u6570\u76f8\u5bf9\u539f\u5b50\u8d28\u91cf\uff0c\u6c2e\u6c14\u4f53\u79ef\u5206\u6570\u4e3a76.81%\uff0c\u6c27\u6c14\u4f53\u79ef\u5206\u6570\u4e3a22.04%
(1)\u8bbe\u539f\u6eb6\u6db2\u4e2d\u786b\u9178\u8d28\u91cf\u4e3aX.
H2SO4+Ba(OH)2==BaSO4+2H2O
X=98*20*17.1%/171=1.96
\u6240\u4ee5\uff0c\u539f\u6eb6\u6db2\u4e2d\u786b\u9178\u8d28\u91cf\u5206\u6570\u4e3a1.96/20*100%=9.8%
(2) ph=7\u65f6\uff0c\u6eb6\u8d28\u4e3a\u6c2f\u5316\u94a1\uff0c\u8bbe\u4e3ay
2HCl+Ba(OH)2==BaCl2+H2O
y=208*40*17.1%/171=8.32
绛旓細鎵闇Na2CO3鐨勬懇灏旇川閲忥細500 ml * 0.1000 mol/l = 0.0500mol Na2CO3鐨勬懇灏旇川閲忥細23 * 2 + 12 + 16 *3锛106 g/mol 鎵浠ュ彇0.0500 mol * 106g/mol = 5.3g
绛旓細鍗0.106鍏嬪埌0.1325鍏嬩箣闂
绛旓細浣犵О鍙栫⒊閰搁挔0.106g-101325g鑼冨洿鍐呯殑浠绘剰涓涓笺傛妸纰抽吀閽犻厤鎴愭憾娑茬敤0.1鎽╁皵姣忓崌鐨勭洂閰婊村畾锛屾牴鎹娑堣鐨勭洂閰搁噺灏卞彲浠ヨ绠楃О鍙栫殑纰抽吀閽犵殑閲嶉噺銆傝繖鏍峰氨璁＄畻鍑虹О閲忚宸簡銆
绛旓細鎽╁皵鐩愬寲瀛﹀紡(NH4)2Fe(SO4)2.6H2O锛屽弽搴斿氨鎸塊MnO4+5Fe锛圛I锛=Mn锛圛I锛+5Fe锛圛II锛夌畻灏卞ソ浜
绛旓細銆愯В鏋愩慛a2CO3 + 2HCl ===2NaCl + H2O + CO2(姘斾綋) 璁惧弽搴娑堣HCl Xmol 1 2 0.16/106 X ===>X=0.0030 鏃佹秷鑰桯Cl 0.003mol锛屾墍浠ラ渶瑕佺洂閰哥殑浣撶Н涓0.003mol梅0.1mol/L=0.03L=30ml 闇瑕30ml 0.1mol/L鐨勭洂閰搞傛晠閫塁 鍥炵瓟婊℃剰璇烽噰绾硚
绛旓細10mL 0.05mol?L-1鐨凬aOH婧舵恫涓姘㈡哀鍖栭挔鐨勭墿璐ㄧ殑閲=0.01L脳0.05mol/L=5脳10-4mol锛岀敱NaOH+HCl=NaCl+H2O鍙煡锛娑堣HCl鐨勭墿璐ㄧ殑閲=5脳10-4mol锛屾晠闇瑕佺洂閰哥殑浣撶Н=5脳10?4mol0.1mol/L=0.005l=5mL锛岃嫢閰稿紡婊村畾绠℃湭娑︽礂锛岀洂閰哥殑娴撳害鍋忎綆锛屾秷鑰楃洂閰镐綋绉亸澶э紝鍗冲浜5mL锛屾晠閫塁锛
绛旓細鏁呯瓟妗堜负锛0.115mol/L锛涳紙3锛夎鍥惧彲鐭ワ紝缁堢偣璇绘暟涓22.28mL锛屾晠绛旀涓猴細22.28mL锛涳紙4锛堿锛庨吀寮婊村畾绠′娇鐢ㄥ墠锛屾按娲楀悗鏈敤寰呮祴鐩愰吀娑︽礂锛屽緟娴嬫恫鐨勭墿璐ㄧ殑閲忓亸灏忥紝閫犳垚V锛堟爣鍑嗭級鍋忓皬锛屾牴鎹甤锛堝緟娴嬶級鍏紡鍒嗘瀽锛屽彲鐭锛堟爣鍑嗭級鍋忓皬锛屾晠A閿欒锛汢锛庨敟褰㈢摱娲楁钉鍚庝笉骞茬嚗锛屽缁撴灉鏃犲奖鍝嶏紝鏁匓閿欒锛...
绛旓細鎵浠锛堥吀锛=c(纰)脳V(纰)V(閰)=0.10mol/L脳20.00mL20.00mL=0.10 mol/L锛屾晠绛旀涓猴細0.10 mol/L锛涳紙3锛夌敱浜庢皵娉＄殑浣撶Н璁″叆浜嗘爣鍑嗘恫娑堣鐨勪綋绉紝浣垮緱鏍囧噯娑蹭綋绉瘮瀹為檯浣撶Н澧炲ぇ锛屾祴瀹氱粨鏋滃亸楂橈紝鏁呯瓟妗堜负锛氬亸楂橈紱锛4锛夐珮閿伴吀閽惧叿鏈夊己姘у寲鎬э紝搴旂洓鏀惧湪閰稿紡婊村畾绠＄敳涓紱楂橀敯閰搁捑婧舵恫姘у寲浜氶搧...
绛旓細]2锛涘噺灏忥紱锛3锛夆憼鍙嶅簲娑堣鐨勯珮閿伴吀閽剧殑鐗╄川鐨勯噺涓猴細0.10mol?L-1脳0.01025L=0.001025mol锛岃鍙傚姞鍙嶅簲鐨勭绂诲瓙鐨勭墿璐ㄧ殑閲忎负nmol锛 2MnO4-+10I-+16H+=2Mn2++5I2+8H2O 2 10 0.001025mol n鍒2 0.001025mol =10n锛岃В寰楋細n=0.005125mol锛屾墍浠锛圛-锛=nV=0.005125mol0...
绛旓細NaOH婧舵恫锛=19.98+20.02+20.003mL=20.00mL锛屾墍浠锛堥吀锛=c(纰)脳V(纰)V(閰)=0.10mol/L脳20.00mL20.00mL=0.10 mol/L锛涙晠绛旀涓猴細0.10 mol/L锛涳紙5锛夐敟褰㈢摱姘存礂鍚庢湭骞茬嚗锛屽緟娴嬫恫鐨勭墿璐ㄧ殑閲忎笉鍙橈紝V锛堟爣锛変笉鍙橈紝鏍规嵁c锛堝緟娴嬶級=V(鏍)脳c(鏍)V(寰呮祴)鍒嗘瀽鍙煡c锛堝緟娴嬶級...
