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Chemistry Senior High

(2)から(4)まで解き方がわからないです。 画像中の解説では理解が出来なかったので分かりやすく教えて頂けると幸いです。お願いします!!

APINGAT STIMULO, HUG 例題 ⑤ 電離度と濃度 5 0.10mol/Lの酢酸水溶液がある。電離度を0.010として次の(1)~(4) の濃度を求めよ。(金) (1) 水素イオン濃度[H+] (2) 酢酸イオン濃度[CH3COO-] (4) 水酸化物イオン濃度[OH-] (3) 酢酸分子濃度[CH3COOH] |解説| cum CH3COOH - →H+ + CH3COOO (2) HOBK&Hしたと 酸c 0 OHN (6) 1つは 電離前 電離後 (1-α) (1) 「イオンはカモ電」を使う。 酢酸は1価の弱酸より, [H+] = 1 × 0.10 × 0.010 = 1.0 × 10-3mol/L (エ) よるの森 Hen(d) ca 1.0 × 10-14 [H+] 電離前 (オ)0年) (2) 酢酸の電離式は, CH3COOH CH3COO+H+だから, [H+]=[CH3COO-] となり [CH3COO-]=1×0.10×0.010=1.0×10-3mol/L/に近い (3) 電離せずに残った [CH3COOH] は, 初めの濃度から電離した分を引けばよい。 = 水 OeHold Dom (02.(s) D. CHICO C (I) < (RC) < (C) < (F) () <() () [H+][mol/L] = (価数) × (モル濃度) × (電離度)より,倍に濃くする。 0 えて薄め [CH3COOH]=0.10-0.10×0.010=0.10×(1-0.010)=9.9 × 102mol/L Ricmomo [H] (4) [OH-]は, 水のイオン積[H+][OH-]=1.0 × 10-14 (mol/L)2より、 toner [OH-]= CH3COOH 2+1 FORD 1.0 × 10-1400 = 1.0×10-11mol/L 1.0 × 103 ので 0.10 HOC 電離後 0.10(1-0.010 ) 解答 (1) [H+]=1.0×10-3mol/L XOHq ( na α = 0.010 COULD OOK 157~8-30+ OBA の味中 個円 (3) [CH3COOH] = 9.9 × 10-2mol/L LITOL LITOL, SIMOLEST (0) XH&m H+ CH3COOH 10:30 10倍濃くなり), pH=5-3=2 0 THE & 0 HUN 1 x 0.10 × 0.010 1 × 0.10 × 0.010 HO. (2) [CH3COO-] = 1.0 × 10-3mol/L Luty (SPC (4) [OH-]=1.0×10-11mol/L TEEKLINIKAH MARO

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English Senior High

2の答えがなぜ4になるのかが分かりません🙇‍♂️

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English Senior High

分かる所のみでも良いので、教えてください。お願いします。

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