Bài giải hóa phân tích.pdf - CHƯƠNG CHUẨN ĐỘ...

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CHƯƠNG: CHUẨN ĐỘ AXIT- BAZ III.4: Chất chỉ thị được dùng? Metyl da cam (pH= 3,3 – 4,4); metyl đỏ(4,4-6,2); p.p(8-10). a) Chuẩn độ HCl 0,1M bằng NaOH 0,1M HCl + NaOH NaCl + H 2 O C o V o CV => pH = 7 pH 1 = 9 , 99 100 ) 9 , 99 100 ( 1 , 0 lg + - - 2 2 10 . 2 10 lg - - = 2 10 lg 4 - - = = 4,3 pH 2 = + - - - 1 , 100 100 ) 100 1 , 100 ( 1 , 0 lg 14 = 9,7 => Bước nhảy pH = 4,3 9,7 Do đó: cct= metyl da cam,metyl đỏ, p.p
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b) Chuẩn độ HCOOH 0,1M bằng NaOH 0,1M, pK a (HCOOH)= 3,75 HCOOH + NaOH HCOONa + H 2 O pH = ½(pK n + pK a +lgC m )=½(14+3,75+lg0,05) pH = 8,25 pH 1 = 9 , 99 . 1 , 0 ) 9 , 99 100 ( 1 , 0 lg 75 , 3 - - = CV CV V C pK a - - 0 0 1 lg =6,75 pH 2 = + - - - V V V C CV 0 0 0 lg 14 + - - - = 1 , 100 100 ) 100 1 , 100 ( 1 , 0 lg 14 =9,7 => Bước nhảy pH= 6,75 9,7 => Chỉ dùng p.p
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NH 4 OH + HCl NH 4 Cl + H 2 O pH =½(pK n -pK b -lgC m )=½(14-4,75-lg0,05) =5,275 pH 1 = - - - = 9 , 99 . 1 , 0 ) 9 , 99 100 ( 1 , 0 lg 75 , 4 14 =6,25 pH 2 = 1 , 100 100 ) 100 1 , 100 ( 1 , 0 lg + - - = = 4,3 => Bước nhảy pH = 6,25 4,3 => Cct = metyl da cam, metyl đỏ c)Chuẩn độ NH 3 0,1M(pK b =4,75) bằng HCl0,1M . - - - CV CV V C pK b 0 0 lg 14 V V V C CV + - - 0 0 0 lg
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III.5:a) Chuẩn độ 25ml HCl bằng NaOH 0,05M. Tính nồng độ HCl nếu V NaOH =17,5ml b) Kết thúc chuẩn độ ở pT=4 => S%=? c) Bước nhảy chuẩn độ nếu S%= ± 0,2% Giải a) HCl + NaOH NaCl + H 2 O C o V o = CV => C o =CV/V o =0,05.17,5/25 =0,035N b) pH =7 => pH c =pT=4< pH :S(-);dd(HCl) S% = - 0,485% 2 0 0 10 . . ) ( 10 % C C C C S pT + - = - 2 4 10 . 035 , 0 . 05 , 0 ) 035 , 0 05 , 0 ( 10 + - = -
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c) 2 , 0 10 035 , 0 . 05 , 0 ) 035 , 0 05 , 0 ( 10 % 2 - = + - = - pT S => pT=4,38 2 , 0 10 035 , 0 . 05 , 0 ) 035 , 0 05 , 0 ( 10 % 2 14 + = + + = - pT S => pT=9,62 => Bước nhảy pH = 4,38 9,62 III.6:a) Chuẩn độ 50ml CH 3 COOH hết 24,25ml NaOH 0,025M. Tính nồng độ CH 3 COOH. b) Tính S% nếu pT = 10. c) Tính pH nếu V NaOH = 24,5ml Giải a) CH 3 COOH + NaOH CH 3 COONa + H 2 O C o V o = CV => C o = CV/V o =0,025.24,25/50 = 0,012125M
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b) pH =½(pK n +pK a +lgC m ) pT = 10 > pH => S(+): dd thừa NaOH 2 14 10 10 025 , 0 . 012125 , 0 ) 025 , 0 012125 , 0 ( 10 + + = - 2 0 0 14 10 . ) ( 10 % C C C C S pT + + = - = + 1,225% ) lg 75 , 4 14 ( 0 0 0 2 1 V V CV V C pH + = + + = ) 25 , 24 50 025 , 0 . 25 , 24 lg 75 , 4 14 ( 2 1 + + + = pH = 8,33
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c) V c = 24,5ml > V =24,25ml = 9,92 III.7:a) Chuẩn độ 25ml NH 3 0,05M bằng HCl 0,1M. pH ? pT = 4 => V HCl =? NH 4 OH + HCl NH 4 Cl + H 2 O C o V o = CV => V =C o V o /C=0,05.25/0,1= 12,5ml = 5,296 ) ) lg ( 14 0 0 0 2 V V V C CV pH + - - - = ) 5 , 24 50 ) 50 . 012 , 0 5 , 24 . 025 , 0 lg ( 14 + - - - = ) lg ( 0 0 0 2 1 V V V C pK pK pH b n + - - = ) 5 , 12 25 25 . 05 , 0 lg 75 , 4 14 ( 2 1 + - - = pH
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*pT = 4 < pH => F>1: dd thừa HCl 2 10 . . ) ( 10 % C Co C Co S pT + = - 2 5 10 . 1 , 0 . 05 , 0 ) 1 , 0 05 , 0 ( 10 + = - = + 0,03% b) pH khi thêm 12,3ml HCl :V c <V => dd NH 3 = 9,99 c) pT=5 <pH => S(+):dd HCl 4 lg 0 0 0 = + - - = = V V V C CV pT pH 4 0 0 0 10 - = + - V V V C CV 4 0 0 0 10 ) ( - + = - V V V C CV ) 10 ( ) 10 ( 4 0 0 4 - - + = - C V C V 4 4 0 0 10 ) 10 ( - - - + = C C V V 4 4 10 1 , 0 ) 10 05 , 0 ( 25 - - - + = = 12,5249ml ) lg ( 14 0 0 0 2 1 V V CV V C pK pH b + - - - = ) 3 , 12 25 3 , 12 . 1 , 0 25 . 05 , 0 lg 75 , 4 ( 14 2 1 + - - - =
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III.8: HCl 0,1M HA0,1M(pK a =6) + NaOH 0,2M a) pH khi F = 0 :HCl chuẩn độ trước HCl + NaOH NaCl + H 2 O pH o = -lgC o (HCl) = -lg0,1= 1 b) pH khi chuẩn độ 99,9% HCl Xem như HCl đã chuẩn độ hết(dd chỉ còn HA) pH 1 = ½[pK a -lgC o (HA)] C 01 .V 0 = C.V 1 => V 1 =C 01 .V 0 /C =0,1.50/0,2=25ml = 3,59 50ml ) lg 6 ( 1 0 0 01 2 1 V V V C + - = ) 25 50 50 . 1 , 0 lg 6 ( 2 1 + - =
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c) pH khi 2 axit đã trung hòa hết HA + NaOH NaA + H 2 O C 02 .V o =CV 2 => V 2 = C 02 .V o /C =0,1.50/0,2=25ml pH 2 = ½[pK n +pK a +lgC NaA ] = 0,05M pH 2 = ½(14+6+lg0,05) = 9,35 III.9: 50ml HA 0,05M(pK a1 =3,75 HB 0,1M(pK a2 =7,5 +NaOH 0,1M a) pH tđ1 :pK a2 -pK a1 = 7,5-3,75=3,75=> ch.độ riêng từng axit(xem HA và HB như 1 axit yếu 2 chức: H 2 X) 2 1 0 0 02 .
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