物理化学I-第11回(13).ppt

Similar documents
H22応用物理化学演習1_濃度.ppt

後期化学_04_酸塩基pH

Microsoft Word - 目次注意事項2.doc

ONS60409_gencyo.indd

Microsoft PowerPoint - D.酸塩基(2)

2 Zn Zn + MnO 2 () 2 O 2 2 H2 O + O 2 O 2 MnO 2 2 KClO 3 2 KCl + 3 O 2 O 3 or 3 O 2 2 O 3 N 2 () NH 4 NO 2 2 O + N 2 ( ) MnO HCl Mn O + CaCl(ClO

物理化学I-第12回(13).ppt

1320M/161320M

1

untitled

Microsoft Word - 酸塩基

H22環境地球化学4_化学平衡III_ ppt

近畿中国四国農業研究センター研究報告 第7号

土壌の観察・実験テキスト −土壌を調べよう!−

後期化学_01_濃度

δδ 1 2 δ δ δ δ μ H 2.1 C 2.5 N 3.0 O 3.5 Cl 3.0 S μ

<576F F202D F94BD899E8EAE82CC8DEC82E895FB5F31325F352E6C7770>

▲ 電離平衡

Microsoft Word - 座談会「小泉政治」.doc

文字化力強化合宿 No.1用

文字化力強化合宿

2

CuSO POINT S 2 Ni Sn Hg Cu Ag Zn 2 Cu Cu Cu OH 2 Cu NH CuSO 4 5H 2O Ag Ag 2O Ag 2CrO4 Zn ZnS ZnO 2+ Fe Fe OH 2 Fe 3+ Fe OH 3 2 Cu Cu OH 2 Ag Ag


(a) (b) X Ag + + X AgX F < Cl < Br < I Li + + X LiX F > Cl > Br > I (a) (b) (c)

<4D F736F F F696E74202D208D918E8E91CE8DF481698E5F89968AEE816A F38DFC97702E707074>

第4次酸性雨全国調査報告書(平成16年度) -(Ⅱ)付表編-

2011年度 化学1(物理学科)

基礎化学 ( 問題 ) 光速 c = m/s, プランク定数 h = J s, 電気素量 e = C 電子の質量 m e = kg, 真空中の誘電率 ε 0 = C 2 s 2 (kg


2004 年度センター化学 ⅠB p1 第 1 問問 1 a 水素結合 X HLY X,Y= F,O,N ( ) この形をもつ分子は 5 NH 3 である 1 5 b 昇華性の物質 ドライアイス CO 2, ヨウ素 I 2, ナフタレン 2 3 c 総電子数 = ( 原子番号 ) d CH 4 :6

H8.6 P


PDF

ドコモP-01Hカタログ

第2章 有機物汚濁指標の概要と問題点

Microsoft Word docx

無印良品のスキンケア

高 1 化学冬期課題試験 1 月 11 日 ( 水 ) 実施 [1] 以下の問題に答えよ 1)200g 溶液中に溶質が20g 溶けている この溶液の質量 % はいくらか ( 整数 ) 2)200g 溶媒中に溶質が20g 溶けている この溶液の質量 % はいくらか ( 有効数字 2 桁 ) 3) 同じ

Contents 1 1. ph ph ph 7 a) ph 9 b) 10 c) ph 11 a) 11 b) 12 c) ph 14 d) 15 ph a) 18 b) / 19 c)

スライド 1

( )

Ⅲ-2 酸 塩基の電離と水素イオン濃度 Ⅲ-2-1 弱酸 Ex. 酢酸 CH 3 COOH 希薄水溶液 (0.1mol/L 以下 ) 中では 一部が解離し 大部分は分子状で存在 CH 3 COOH CH 3 COO +H + 化学平衡の法則より [CH 3 COO ][H + ] = K [CH 3

クレイによる、主婦湿疹のケア

(3)(4) (3)(4)(2) (1) (2) 20 (3)

... 6

untitled

群馬県野球連盟


<82B582DC82CB8E7188E782C48A47967B41342E696E6464>

untitled



裁定審議会における裁定の概要 (平成23年度)


<91E F1938C966B95FA8ECB90FC88E397C38B5A8F708A778F7091E589EF8EC08D7388CF88F5837D836A B E696E6464>

Microsoft Word - 入居のしおり.doc

untitled

和県監査H15港湾.PDF

( )

ESPEC Technical Report 12

syogaku


-26-

Taro12-希少樹種.jtd

untitled

日本経大論集 第45巻 第1号


untitled

河川砂防技術基準・基本計画編.PDF

4 100g

09_organal2



攪拌シール攪拌棒 P50-1 P50-1 P , , , P50-6 P50-6 P , ,000 P50-2 P50-3 P L 500

303合成の定義事例集[読み取り専用]

橡マニュアルv4c.PDF

―目次―

木村の理論化学小ネタ 熱化学方程式と反応熱の分類発熱反応と吸熱反応化学反応は, 反応の前後の物質のエネルギーが異なるため, エネルギーの出入りを伴い, それが, 熱 光 電気などのエネルギーの形で現れる とくに, 化学変化と熱エネルギーの関

通 276

4 1 Ampère 4 2 Ampere 31

341-4改

untitled

untitled


untitled

SC Ł\†EŒÚ M-KL.ec6

Taro-化学3 酸塩基 最新版


life_6.pdf

DVIOUT-酸と塩

London -van der Waals Coulomb Fe, Mn

本組よこ/根間:文11-029_P377‐408

1 c H O. H C C O H. CH 3 COOH ( ). C 2 H 4 O 2. CH 2 O H [. CH2 CH 2 CH CH ]. CH 3 COOH 3 1 C H ( ) CH 3 CH 3 CH CH CH

i ( 23 ) ) SPP Science Partnership Project ( (1) (2) 2010 SSH

CH 2 CH CH 2 CH CH 2 CH CH 2 CH 2 COONa CH 2 N CH 2 COONa O Co 2+ O CO CH 2 CH N 2 CH 2 CO 9 Change in Ionic Form of IDA resin with h ph CH 2 NH + COO

温泉の化学 1

untitled

= = = = = = 1, 000, 000, 000, = = 2 2 = = = = a,

P P P P P P

Transcription:

I- 11-1 10 10.5 H 3 O + (aq) + OH (aq) + Na + (aq) + Cl (aq) Na + (aq) + Cl (aq) + 2H 2 O(l) - or - [A] V A = [B] V B 22 mol dm 3 equiv dm 3 - H 2 SO 4 (aq) + 2NaOH(aq) 2Na + (aq) + SO 4 2 (aq) + 2H 2 O(l) 2[H 2 SO 4 ] V H2 SO 4 = [NaOH] V NaOH ph = 7 ph 0.1 mol dm 3 10 cm 3 11-2 1) V NaOH = 9 cm 3 n HCl = 1 10 3 9 10 4 = 1 10 4 mol, V = 0.019 dm 3 [H 3 O + ] = 1 10 4 mol 0.019 dm 3 = 5.26 10 3 mol dm 3, ph = 2.28 2) V NaOH = 9.9 cm 3 n HCl = 1 10 3 9.9 10 4 = 1 10 5 mol, V = 0.0199 dm 3 [H 3 O + ] = 1 10 5 mol 0.0199 dm 3 = 5.03 10 4 mol dm 3, ph = 3.30 3) V NaOH = 10.1 cm 3 n NaOH = 1.01 10 3 1 10 3 = 1 10 5 mol, V = 0.0201 dm 3 [OH ] = 1 10 5 mol 0.0201 dm 3 = 4.98 10 4 mol dm 3 ph = 14 poh = 14 3.30 = 10.70 4) V NaOH = 11 cm 3 n NaOH = 1.1 10 3 1 10 3 = 1 10 4 mol, V = 0.021 dm 3 [OH ] = 1 10 4 mol 0.021 dm 3 = 4.76 10 3 mol dm 3 ph = 14 poh = 14 2.32 = 11.68 ph 1

11-3 10.6 ph CH 3 COOH(aq) + NaOH(aq) CH 3 COO (aq) + Na + (aq) + H 2 O(l) CH 3 COO (aq) + H 2 O(l) CH 3 COOH(aq) + OH (aq) K h = [CH 3 COOH][OH ] [CH 3 COO ] = [CH 3 COOH][OH ][H 3 O + ] [CH 3 COO ][H 3 O + = ] K a = 5.71 10 10 mol dm 3 K h K b 2H 2 O(l) H 3 O + (aq) + OH (aq) 0.1 mol dm 3 10 cm 3 20 cm 3 [CH 3 COO ] = 0.05 mol dm 3, [CH 3 COOH] = [OH ] [OH ] 2 = K h [CH 3 COO ] = (5.71 10 10 mol dm 3 ) (0.05 mol dm 3 ) [OH ] = 5.34 10 6 mol dm 3, ph = 14 5.27 = 8.73 11-4 ph CH 3 COOH(aq) + NaOH(aq) CH 3 COO (aq) + Na + (aq) + H 2 O(l) (0.1 mol dm 3, 10 cm 3 ) (0.1 mol dm 3 ) 1) V NaOH = 9 cm 3 K a = [H 3O + ][CH 3 COO ] = [H 3 O + ] 9 [CH 3 COOH] [H 3 O + ] = K a / 9 = 1.94 10 6 mol dm 3 ph = 5.71 2) V NaOH = 10 cm 3, ph = 8.73 3) V NaOH = 11 cm 3 [OH ] = 1 10 4 mol 0.021 dm 3 = 4.76 10 3 mol dm 3 ph = 14 poh = 14 2.32 = 11.68 ph (a) 2

11-5 NH 3 HCl ph NH 3 (aq) + HCl(aq) NH 4 + (aq) + Cl (aq) NH 4 + (aq) + H 2 O(l) H 3 O + (aq) + NH 3 (aq) K h = [H 3 O+ ][NH 3 ] [NH 4 + ] = 5.6 10 10 mol dm 3 = [H 3 O+ ][OH ][NH 3 ] [NH + 4 ][OH = ] K b 0.1 mol dm 3 10 cm 3 20 cm 3 K h K a [NH 4 + ] = 0.05 mol dm 3, [NH 3 ] = [H 3 O + ] [H 3 O + ] 2 = K h [NH 4 + ] = (5.6 10 10 mol dm 3 ) (0.05 mol dm 3 ) [H 3 O + ] = 5.3 10 6 mol dm 3, ph = 5.3 ph (b) 11-6 CH 3 COONahydrolysis h K h CH 3 COONa CH 3 COO (aq) + Na + (aq) [c 0 (mol dm 3 )] CH 3 COO (aq) + H 2 O(l) CH 3 COOH(aq) + OH (aq) c 0 (1 h) c 0 h c 0 h K h = [CH 3 COOH][OH ] [CH 3 COO ] = [CH 3 COOH][OH ][H 3 O + ] [CH 3 COO ][H 3 O + = ] K a K a = [CH 3COO ][H + ], = [H 3 O + ][OH ] [CH 3 COOH] K h = (c 0 h)2 c 0 (1 h) = c 0 h2 1 h c 0 h2 (h << 1) ( ) h K h K h K b K h K b h = K h c 0 = K a c 0 (K h = / K a ) h [OH ] = c 0 h = K h c 0 = c 0 K a 3 [OH ] [H 3 O + ] = [OH ] = K a c 0 ph = log[h 3 O + ] = (1 / 2)(p + pk a + logc 0 ) at 25 o C, ph = 7 + (1 / 2)(pK a + logc 0 ) [H 3 O + ] ph 3

hydrolysis h K h NH 4 + 11-7 NH 4 Cl NH 4 + (aq) + Cl (aq) [c 0 (mol dm 3 )] NH 4 + (aq) + H 2 O(l) H 3 O + (aq) + NH 3 (aq) c 0 (1 h) c 0 h c 0 h K h = [H 3 O+ ][NH 3 ] [NH 4 + ] = [H 3 O+ ][OH ][NH 3 ] [NH + 4 ][OH = ] K b K b = [NH 4 + ][OH ], = [H 3 O + ][OH ] [NH 3 ] K h = (c 0 h)2 c 0 (1 h) = c 0 h2 1 h c 0 h2 (h << 1) ( ) h K h K h K a K h K a h = K h c 0 = K b c 0 (K h = / K b ) h [H 3 O + ] = c 0 h = K h c 0 = c 0 K b 5 [H 3 O + ] ph = log[h 3 O + ] = (1 / 2)[p (pk b + logc 0 )] at 25 o C, ph = 7 (1 / 2)(pK b + log c 0 ) ph 11-8 10.7 ph ph 10.4 ph 4

phindicator 11-9 InH(aq) + H 2 O(l) H 3 O + (aq) + In (aq) K a = [H 3O + ][In ] [InH] at [InH] = [In ] 0.1 < [In ] [InH] = K a [H 3 O + ] < 10 pk a = 3.46 (K a = 3.47 10 4 mol dm 3 ) [H 3 O + ] = K a = 10 3.46 = 10 4 10 0.54 = 3.47( 3.5) 10 4 mol dm 3 [H 3 O + ] > 3.5 10 4 mol dm 3 (= K a ) [InH] > [In ] [H 3 O + ] < 3.5 10 4 mol dm 3 (= K a ) [InH] < [In ] ph = 3.4 6 [H 3 O + ] 10K a > [H 3 O + ] > 0.1K a 2.46 < ph < 4.46 (pk a ± 1) 11-10 HCl NaOH ph 10.2 pk a ± 1 5

11-11 10.8 ph ph 1 dm 3 1x10 4 mol (= 0.00365 g) ph = 4 1x10 4 mol (= 0.0040 g) ph = 10 ph = 7pH ±3 1 CH 3 COOH + CH 3 COONa 2 2 NH 3 + NH 4 Cl 2 ph (a) ph [NH 3 ] = [NH 4+ ] = 0.1 mol dm 3, ph = 9.25 (b) 1 dm 3 1x10 4 mol ph = 9.25 (c) 1 dm 3 1x10 4 mol ph = 9.25 12-14 11-12 2 NH 3 + NH 4 Cl NH 3 NH 4 Cl- NH 4 Cl NH 4 + (aq) + Cl (aq) NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH (aq) K b = [NH 4 + ][OH ], [NH 3 ([NH 3 ] = c base, [NH + 4 ] = c salt ) ] [H 3 O + ], ph NH 4 + [OH ] = K b [NH 3] [NH 4 + ] = K b c base c salt [H 3 O + ] = [OH ] = [NH 4 + ] K b [NH 3 ] = K a [NH 4 + ] [NH 3 ] = K a c salt c base K a = [NH 3][H 3 O + ] [NH 4 + ] [H 3 O + ] = K a [NH 4 + ] [NH 3 ] ( NH + 4 (aq) + H 2 O(l) NH 3 (aq) + H 3 O + (aq)) ph = (p pk b ) log c salt c base = pk a log c salt c base 6

2 ph (a) [H 3 O + ], ph [NH 3 ] = c base = 0.1 mol dm 3, [NH 4 + ] = c salt = 0.1 mol dm 3 NH 4 + (aq) + H 2 O(l) NH 3 (aq) + H 3 O + (aq) K a = = [NH 3 ][H 3 O+ ] K b [NH + = 5.6 10 10 mol dm 3 4 ] [H 3 O + ] = K a [NH 4 + ] [NH 3 ] = K a = 5.6 10 10 mol dm 3, ph = 9.25 (b) 1 dm 3 1x10 4 mol NH 3 (aq) + HCl(aq) NH 4 + (aq) + Cl (aq) [NH 4 + ] = 0.1 + 0.0001 = 0.1001 mol dm 3 [NH 3 ] = 0.1 0.0001 = 0.0999 mol dm 3 11-13 [H 3 O + ] = (5.6 10 10 ) 0.1001 0.0999 = 5.61 10 10 mol dm 3, ph = 9.25 (c) 1 dm 3 1x10 4 mol NH 4 + (aq) + OH (aq) NH 3 (aq) + H 2 O(l) [NH 4 + ] = 0.1 0.0001 = 0.0999 mol dm 3 [NH 3 ] = 0.1 + 0.0001 = 0.1001 mol dm 3 11-14 [H 3 O + ] = (5.6 10 10 ) 0.0999 0.1001 = 5.59 10 10 mol dm 3, ph = 9.25 1 CH 3 COOH + CH 3 COONa 2 CH 3 COONa CH 3 COO (aq) + Na + (aq) CH 3 COOH(aq) + H 2 O(l) CH 3 COO (aq) + H 3 O + (aq) K a = [CH 3COO ][H 3 O + ], [CH 3 COOH] = c acid, [CH 3 COO ] = c salt [CH 3 COOH] [H 3 O + ] = K a [CH 3 COOH] [CH 3 COO ] = K a c acid c salt ph = pk a + log c salt c acid ( ) 12 7