医学部専門予備校 YMS 医学部進学予備校メビオ

解答速報

日本医科大学（前期） 化学

2020年 2月 2日実施

[I]A࢔㸸 C = 0.20 mol/L ࡜ࡍࡿࠋ[CH3 COOH] ; C ࡜㏆ఝࡋ࡚ࡼ࠸ࡢ࡛㸪pH ࡣḟࡢࡼ࠺࡟࡞ࡿࠋpp[H+ ] ; CKa = 0.20 mol/L × 2.0 × 105 mol/L = 2.0 × 103 mol/L (¿ C)! pH = 3  log10 2 = 2.70··㸦· ⟅㸧࠙ὀࠚpH ࡣᑐᩘ್࡞ࡢ࡛ᑠᩘ㒊ศࡀ᭷ຠᩘᏐ࡟ᑐᛂࡍࡿࠋ௨ୗࡶྠᵝ࡛࠶ࡿࠋ࢖㸸 pH = 4.70 ࡢ࡜ࡁ㸪[H+ ] = 104.70 mol/L = 100.305 mol/L = 2.0 × 105 mol/L ࡛࠶ࡿࠋᕼ㔘ᚋࡢࣔࣝ⃰ᗘࢆC 0 ࡜ࡍࡿࠋᕼ㔘࡟ࡼࡾゎ㞳ᗘࡀቑຍࡍࡿࡢ࡛๓ၥࡢࡼ࠺࡟㏆ఝ࡛ࡁ࡞࠸࡜⪃࠼ࡽࢀࡿࠋ[CH3 COO ] = [H+ ] ࡼࡾ C 0 ࡀồࡵࡽࢀࡿࠋ[CH3 COO ][H+ ][H+ ]2= 0[CH3 COOH]
C  [H+ ]¶¶µµ+[H ]
2.0 × 105 mol/L0+× 2.0 × 105 mol/L = 4.0 × 105 mol/L
C = 1+[H ] = 1 +Ka
2.0 × 105 mol/L

Ka =ࡼࡗ࡚㸪ồࡵࡿᕼ㔘ࡢ๭ྜࡣḟࡢࡼ࠺࡟࡞ࡿࠋ

C
0.20 mol/L== 5.0 × 103
C0
4.0 × 105 mol/L࢘㸸 ゎ㞳ᗘࢆ
0=0··㸦· ⟅㸧࡜ࡍࡿ࡜ [H+ ] = C 0 0 ࡼࡾ㸪ḟࡢࡼ࠺࡟ồࡲࡿࠋ

[H+ ]2.0 × 105 mol/L== 0.500C
4.0 × 105 mol/L··㸦
· ⟅㸧B࢚㸸 K ࡀ CH3 COO ࡢຍỈศゎᐃᩘ㸦ሷᇶ㟁㞳ᐃᩘ㸧ࡢ㏫ᩘ࡛࠶ࡿࡇ࡜࡟Ẽ௜ࡃ࡜ࡸࡾࡸࡍ࠸ࠋ[CH3 COO ][H+ ][CH3 COO ]2.0 × 105 mol/L
Ka[CH3 COOH]
K=== 2.0 × 109 (mol/L)1==
+
[CH3 COOH][OH ][H ][OH ]Kw
1.0 × 1014 (mol/L)2㸦⟅㸧2.0 × 109࢜㸸 [CH3 COOH] = [OH ] ࡛࠶ࡾ㸪ຍỈศゎ㸦୰࿴ࡢ㏫཯ᛂ㸧ࡣࢃࡎ࠿ࡋ࠿㉳ࡁ࡞࠸ࡢ࡛ [CH3 COO ] ; C ࡜㏆ఝ࡛ࡁࡿ࡜ࡋ࡚ࡼ࠸ࠋsr
C0.20 mol/L[OH ] == 1.0 × 105 mol/L (¿ 0.20 mol/L)=K
2.0 × 109 (mol/L)1ࡼࡗ࡚㸪ồࡵࡿᮍ཯ᛂ⋡ࡣḟࡢࡼ࠺࡟࡞ࡿࠋ

[CH3 COOH]
1.0 × 105 mol/L=× 100 % = 5.0 × 103 %C0.20 mol/L㸦⟅㸧5.0 × 103࢝㸸 ୚ᘧࡣḟࡢࡼ࠺࡟ኚᙧ࡛ࡁࡿࠋ

K=[CH3 COO ][NH4 + ][CH3 COO ][H+ ][NH4 + ][OH ]1K a Kb=×× +=
[CH3 COOH][NH3 ][CH3 COOH][NH3 ][H ][OH ]
Kw··㸦
· ⟅㸧

医学部専門予備校 YMS࣭࢟ࢡ㸸ࠉࠉᙜ㔞Ⅼ Q ࡛ࡣ CH3 COONH4 Ỉ⁐ᾮ࡟࡞ࡗࡓ࡜ぢ࡞ࡏࡿࡢ࡛ [CH3 COOH] = [NH3 ] ࡛࠶ࡿࠋࡇࡢ࡜ࡁ㸪CཎᏊ࠾ࡼࡧ N ཎᏊࡢ≀㉁཰ᨭࢆ⪃࠼ࡿ࡜ [CH3 COO ] = [NH4 + ] ࡢ㛵ಀࡶᡂࡾ❧ࡘࡇ࡜ࡀศ࠿ࡿࠋ
¶2 µ¶2µ[CH3 COO ][NH4 + ]K==[CH3 COOH][NH3 ]
࢝ࡢ⤖ᯝ࡜㐃❧ࡍࡿ࡜ḟࡢࡼ࠺࡟ィ⟬࡛ࡁࡿࠋsr$[CH3 COO ]Ka Kb
2.0 × 105 mol/L × 2.0 × 105 mol/L[NH4 + ]=== K =
[CH3 COOH][NH3 ]
Kw
1.0 × 1014 (mol/L)2= 2.0 × 102··㸦· ⟅㸧㸦࣭࢟ࢡ࡜ࡶ࡟㸧ࢣ㸸 ᙜ㔞Ⅼ Q ࡢ pH ࡣḟࡢࡼ࠺࡟ồࡵࡽࢀࡿࠋ

[H+ ]
Ka
[CH3 COO ][H+ ][NH3 ][H+ ]2==×=+Kb[CH3 COOH][OH ]Kw[NH4 ][OH ]sr1425Kw Ka
1.0 × 10(mol/L) × 2.0 × 10 mol/L[H+ ] ==
= 1.0 × 107 mol/LKb
2.0 × 105 mol/L! pH = 7.00··㸦· ⟅㸧Cࢥ㸸 ୚ᘧࡼࡾ㸪᮲௳ᘧࡣḟࡢࡼ࠺࡟࡞ࡿࠋ1[XH+ ]Kb
Kb [H+ ]
5 10
5==

10[X]
[OH ]
Kw
Kw
10Kw
5 [H+ ] 510Kb
Kb
10Kw
Kw
 log105  log10 [H+ ] 5  log10Kb
10Kb! 14 + log10 Kb  1 5 pH 5 14 + log10 Kb + 1
㸦⟅㸧14 + log10 Kb ± 1ࢧ㸸 ⁲ᐃ⋡ 0.99 ࡛ࡣᙅ㓟ࡀ㐣๫࡞ࡢ࡛㸪㓑㓟ࡢ Ka ࡢᘧ࠿ࡽồࡵࡿࠋ

[CH3 COOH]
1  0.99= 2.0 × 105 mol/L ×= 2.0 × 107 mol/L[CH3 COO ]0.99· ⟅㸧pH = 7  log10 2 = 6.70 · · 㸦[H+ ] = Ka ×ࢩ㸸 ⁲ᐃ⋡ 1.01 ࡛ࡣᙉሷᇶࡀ㐣๫࡛࠶ࡾ㸪㓑㓟ࢼࢺ࣒ࣜ࢘ࡢຍỈศゎࡣ↓ど࡛ࡁࡿࡢ࡛ḟࡢࡼ࠺࡟ồࡵࡽࢀࡿࠋ

[OH ] = 0.20 mol/L × (1.01  1) = 2.0 × 103 mol/LpOH = 3  log10 2! pH = 14  pOH = 11 + log10 2 = 11.30 · · 㸦· ⟅㸧[II]࢔㸸 ≧ែ᪉⛬ᘧࡼࡾ a ࡣḟࡢࡼ࠺࡟⾲ࡏࡿࠋa=8.30 × 103 Pa · L/(K · mol) × 273 K× nHA = 22.4 L/mol × nHA
1.01 × 105 Pa࢘㸸 ࣮࣊ࣥࣜࡢἲ๎ࡼࡾḟࡢࡼ࠺࡟࡞ࡿࠋnApA=HnA
1.01 × 105 Pa࢚㸸 ࢔࡜࢘ࡢᘧࡼࡾ nHA ࢆᾘཤࡍࡿࠋnA = nHA ×pAapA=×= 4.42 × 107 mol/(L · Pa) × apA51.01 × 10 Pa
22.4 L/mol
1.01 × 105 Pa࢜㸸 ᕼⷧ⁐ᾮ࡞ࡢ࡛ n + nA ; n ࡜㏆ఝ࡛ࡁࡿࠋ
nAnA
xA =;
n + nAn

医学部専門予備校 YMS࢝㸸 ⣧Ỉ 1 L ࡢ≀㉁㔞 n ࡣḟࡢࡼ࠺࡟࡞ࡿࠋ

n=

1 L × 1.00 × 103 g/L
500=mol = 55.6 mol
18.0 g/mol
9࢟㸸 ࢘࡜࢜ࡢᘧ㸪࠾ࡼࡧ࣮࣊ࣥࣜࡢἲ๎ࢆ㐃❧ࡍࡿࠋnHA = nA ×

1.01 × 105 Pa
1.01 × 105 Pa
500
1.01 × 105 Pa
5.61 × 106 mol · Pa= nxA ×=mol ×=pA
KxA
9KKࢡ㸸 ࢔ࡢᘧࡼࡾ nHA ࢆᾘཤࡍࡿࠋ

K=

5.61 × 106 mol · Pa
5.61 × 106 mol · Pa
1.26 × 108 Pa · L==aHnAa22.4 L/molၥ 1 㸦⟅㸧࢔㸸22.4

࢘㸸1.01 × 105

ၥ 2 㸦⟅㸧࢚㸸4.42 × 107 apA

࢝㸸55.6࢟㸸5.61 × 106

ࢡ㸸1.26 × 108࢜㸸nၥ 3 㸦⟅㸧࣮࣊ࣥࣜၥ 4 ≧ែ᪉⛬ᘧࡼࡾ n0O2 ࡣḟࡢࡼ࠺࡟⾲ࡏࡿࠋ

n0O2 =p [Pa] × 60.0 × 103 L= 2.4 × 108 p [mol]8.30 × 103 Pa · L/(K · mol) × 300 K··㸦· ⟅㸧

ၥ 5 27 H C ࡢ O2 ࡟ࡘ࠸࡚ࡣ a = 0.0283 L ࡛࠶ࡿࠋA ࢆ O2 ࡟ࡋ࡚࢚ࡢᘧ࡟௦ධࡍࡿࠋ⁐፹ࡢయ✚ࡢ㐪࠸࡟ὀពࡍ
ࡿࡇ࡜ࠋ
0.5400 L
nA = 4.42 × 107 mol/(L · Pa) × 0.0283 L ×( )× pO2 = 6.75 × 109 mol/Pa × pO21L953= 6.75 × 10 mol/Pa × 1.50 × 10 Pa = 1.0 × 10 mol · · 㸦· ⟅㸧
ၥ 6 ᖹ⾮≧ែ࡟࠾࠸࡚㸪⁐ゎࡋ࡚࠸ࡿ O2 ࡢ≀㉁㔞 nO2 ࡣ ( ) ᘧ࡛୚࠼ࡽࢀࡿࠋẼ┦࡟Ꮡᅾࡋ࡚࠸ࡿ O2 ࡢ≀㉁㔞ࢆnO2 (g) ࡜ࡍࡿ࡜㸪ၥ 5 ࡢ⤖ᯝࡼࡾḟࡢࡼ࠺࡟⾲ࡏࡿࠋnO2 (g) = 1.01 × 103 mol  6.75 × 109 mol/Pa × pO2୍᪉㸪≧ែ᪉⛬ᘧࡼࡾḟᘧࡀᡂࡾ❧ࡘࠋ

pO2 × 60.0 × 103 L = nO2 (g) × 8.30 × 103 Pa · L/(K · mol) × 300 Kࡇࢀࡽࡢᘧࡼࡾ nO2 (g) ࢆᾘཤࡋ࡚ pO2 ࡀồࡵࡽࢀࡿࠋ

pO2 = 3.3 × 104 Pa··㸦
· ⟅㸧

[III]
ၥ 1 ࢜ࣝࢺ㸪࣓ࢱ㸪ࣃࣛࡣ㸪ࡑࢀࡒࢀ 2㸪2㸪1 ࠿ᡤ࠶ࡿࠋ㸦⟅㸧x = 2, y = 2, z = 1ၥ 2 㸦⟅㸧࢔㸸㸦࠶㸧 ࢖㸸㸦ࡃ㸧
ၥ 3 ࢽࢺࣟᇶࡣ㟁Ꮚ྾ᘬᇶ࡛⨨᥮཯ᛂࡢ㏿ᗘࢆῶᑡࡉࡏࡿࡢ࡛㸪㧗㸦⟅㸧2,4,6-ࢺࣜࢽࢺࣟࢺ࢚ࣝࣥ࡟ࡋ࡞࠸࡜」ᩘᅇࡢ⨨᥮཯ᛂࡀ㉳ࡁ࡞࠸ࠋၥ 4 OH ࡶ CH3 ࡶ㟁Ꮚ౪୚ᇶ࡛࠶ࡿࡀ㸪㟁Ꮚ౪୚ᛶࡣ๓⪅ࡢ᪉ࡀᙉ࠸ࡢ࡛㸪OH ࡢ఩⨨࡟ᑐࡋ࡚࢜ࣝࢺࡢ఩⨨࡟⨨᥮ࡀ㉳ࡇࡿ࡜⪃࠼ࡽࢀࡿࠋ㸦⟅㸧ୖᘧࡢ ‫ ڧ‬ෆࢆཧ↷ࡢࡇ࡜ࠋၥ 5 ࢡ࣓࡛ࣥࡣ⨨᥮ᇶࡢ࢖ࢯࣉࣟࣆࣝᇶࡀ㸪ࢺ࢚ࣝࣥࡢ⨨᥮ᇶ࡛࠶ࡿ࣓ࢳࣝᇶࡼࡾࡶ࠿ࡉ㧗࠸ࡢ࡛㸪❧య㞀ᐖ࡟ࡼࡾ࢜ࣝࢺ఩࡟⨨᥮ࡀ㉳ࡁ࡟ࡃࡃ࡞ࡿࠋ㸦⟅㸧㸦࠺㸧
ၥ 6 㸦⟅㸧ղ㸦ゎ⟅౛㸧ⰾ㤶᪘࢔࣑ࣥࡣ㓟໬ࡉࢀࡸࡍ࠸ࡢ࡛㸪KMnO4 ࢆ⏝࠸ࡿ࡜࢔࣑ࣀᇶࡀ཯ᛂࡍࡿྍ⬟ᛶࡀ࠶ࡿࠋ

医学部専門予備校 YMS[IV]ၥ 1 ࣉࣟ࣌ࣥ࡬ࡢ Br2 ࠾ࡼࡧ HBr ࡢ௜ຍ࡛࠶ࡿࠋ㸦⟅㸧

ၥ 2 C4 H8 ࠾ࡼࡧ C5 H10 ࡢ␗ᛶయࡣ௨ୗࡢ㏻ࡾ࡛࠶ࡿࠋ

㸦⟅㸧࢘㸸4ࢡ㸸6ၥ 3 ࣐ࣝࢥࣇࢽࢥࣇ๎ࢆ⪃៖ࡋ࡚࢔ࣝࢣࣥࢆ≉ᐃ࡛ࡁࡿࠋ㸦⟅㸧࠙໬Ꮫ㸦ㅮホ㸧ࠚ᭷ᶵ໬Ꮫࡣྲྀࡾ⤌ࡳࡸࡍ࠸ၥ㢟࡛࠶ࡗࡓࠋ౛ᖺ㏻ࡾ㸪タၥ࡟ࡼࡗ࡚㞴᫆ᗘ࡟኱ࡁ࡞ᕪࡀ࠶ࡗࡓࠋ᫬㛫ⓗ࡟᏶⟅ࡣཝࡋ࠸ࡢ࡛㸪ၥ㢟ࡢྲྀᤞ㑅ᢥࡀ㔜せ࡛࠶ࡿࠋ[I] ࡣ㟁㞳ᖹ⾮࡟㛵ࡍࡿ⥲ྜၥ㢟࡛࠶ࡿࠋ඾ᆺⓗ࡛ࡣ࠶ࡿࡀ㸪㏆ఝᘧࢆ౑࠼࡞࠸ሙྜࡢィ⟬ࡶ࠶ࡿࡢ࡛㸪⩦⇍ᗘ࡛ᕪࡀ௜ࡃၥ㢟࡛࠶ࡿࠋ[II] ࡣẼయࡢ⁐ゎᗘ࡟㛵ࡍࡿၥ㢟࡛㸪๓༙ࡣㄏᑟ࡟ᚑࡗ࡚ᘧኚᙧࢆࡋ࡚࠸ࡃᙧᘧ࡛࠶ࡿࠋ㐣ཤၥ࡟ࡶఝࡓᙧᘧࡀ࠶ࡗࡓࠋࡓࡔ㸪ㄏᑟࡢ୰࡛ࡶᩘ್ィ⟬ࢆྵࡳ㸪ᘧࡢ┦஫㛵ಀࢆᤊ࠼ࡿࡢ࡟ከᑡⱞປࡍࡿࠋᚋ༙ࡣᐦ㛢⣔࡛࣮࣊ࣥࣜࡢἲ๎ࢆ㐺⏝ࡍࡿၥ㢟࡛㸪㢮㢟ࢆゎ࠸ࡓࡇ࡜ࡣ࠶ࡿ࡜ᛮ࠺ࠋ❅⣲࡟㛵ࡍࡿࢹ࣮ࢱࡣ୚࠼ࡽࢀ࡚࠸ࡿࡀ㸪ゎ⟅࡟ᐇࡣᚲせ࡞࠸ࠋ[III] ࡢෆᐜࡣⓎᒎⓗ࡛࠶ࡿࡀ㸪Ꮫ⩦ࡋࡓࡇ࡜ࡀ࠶ࡿேࡀ࡯࡜ࢇ࡝ࡔ࡜ᛮ࠺ࡢ࡛㸪ゎ⟅ࡣᐜ࡛᫆࠶ࡿ࡜⪃࠼ࡽࢀࡿࠋ[IV] ࡶ࣐ࣝࢥࣇࢽࢥࣇ๎ࡣⓎᒎⓗ࡛࠶ࡿࡀ㸪඲య࡜ࡋ࡚ᖹ࡛᫆࠶ࡾ㸪᏶⟅ࡣᚲ㡲࡛࠶ࡿࠋ᭷ᶵ໬Ꮫࡢ [III] ࡜ [IV] ࡣ᏶⟅ࡋ㸪[I] ࡛ᑡ࡞ࡃ࡜ࡶ༙ศ㸪[II] ࡣ㒊ศⓗ࡟࡛ࡶ࡛ࡁࡿ࡜ࡼ࠸ࠋ඲య࡛ 7 ๭ࡣᚓⅬࡋࡓ࠸ࠋ

医学部専門予備校 YMS 医学部進学予備校メビオ

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