请问如何配制250.0mL浓度为1.50%的氯化钠溶液 理工学科->化学
\u7406\u5de5\u5b66\u79d1 -> \u5316\u5b66(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
\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%
我居然都忘了浓度是怎么计算了，好像是质量吧，不对不对，好像是涉及到什么摩尔质量啥的，1000毫升溶液的质量x里1.5%x是盐，咦，盐溶解进水里对体积有没有影响来着，忘了离子状态的怎么算不算体积了，好吧，我忘了学傻了我复习复习去
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绛旓細0.5000V=0.05000脳0.2500L锛孷=0.02500L =25.00mL锛屽彇娴撴憾娑25.00mL
绛旓細nH3PO4=300/1000X0.5mol/L=1.5mol nNaOH=250/1000X0.3mol/L=0.75mol H3PO4+NaOH=NaH2PO4+H2O 涓よ呭彂鐢熷弽搴旂敓鎴恘NaH2PO4=0.75mol,鍓╀綑nH3PO4=0.75mol锛宯NaH2PO4=nH3PO4 鐢变簬鍦ㄥ悓涓婧舵恫浣撶郴涓袱鑰呯殑鐗╄川鐨勯噺娴撳害鐩稿悓 鐢辩紦鍐叉憾娑睴H鍏紡寰楋紝PH=Pka-lg(cH3PO4/cNaH2PO4)=2.148-lg1...
