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Tuyển tập các đề thi học sinh giỏi Olympic môn hóa học & sinh học lớp 10 (kèm đáp án chi tiết) (by Dameva)

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<strong>Tuyển</strong> <strong>tập</strong> <strong>các</strong> <strong>đề</strong> <strong>thi</strong> <strong>Olympic</strong> <strong>hóa</strong> <strong>học</strong> <strong>lớp</strong> <strong>10</strong> ( có <strong>đáp</strong> <strong>án</strong> <strong>chi</strong> <strong>tiết</strong>)<br />

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Nơi bồi dưỡng kiến thức To<strong>án</strong> - Lý - Hóa cho <strong>học</strong> <strong>sinh</strong> cấp 2+3 /<br />

Diễn Đàn To<strong>án</strong> - Lý - Hóa <strong>10</strong>00B Trần Hưng Đạo Tp.Quy Nhơn Tỉnh Bình Định<br />

HCOO + H O⇀<br />

- -<br />

2 ↽ HCOOH + OH<br />

CH COO + H O⇀<br />

- -<br />

3 2 ↽ CH3COOH + OH<br />

Töông töï nhö dung dòch hoãn hôïp 2 ñôn axit, ôû ñaây coi nhö hoãn hôïp 2 ñôn bazô, ta coù:<br />

-<br />

2<br />

-<br />

⎡OH<br />

⎤ = ⎡HCOO<br />

⎤<br />

⎣ ⎦ ⎣ ⎦ bñ -<br />

. K - + ⎡CH HCOO 3COO<br />

⎤<br />

⎣ ⎦ bñ . K<br />

-<br />

CH3COO<br />

- -14 - -14<br />

⎡HCOO ⎤ bd.<strong>10</strong> ⎡CH3COO ⎤ bd.<strong>10</strong><br />

=<br />

⎣ ⎦<br />

+<br />

⎣ ⎦<br />

K<br />

K<br />

HCOOH<br />

CH COOH<br />

2<br />

-14 -14<br />

- 0,06.<strong>10</strong> 0,06.<strong>10</strong><br />

-<strong>10</strong><br />

⎤<br />

-4 -5<br />

⎡<br />

⎣<br />

OH<br />

⎦<br />

= + = 0,367.<strong>10</strong><br />

1,78.<strong>10</strong> 1,80.<strong>10</strong><br />

5<br />

⎡OH − ⎤<br />

⎣ ⎦<br />

= 0,6058.<strong>10</strong><br />

−<br />

-14<br />

<strong>10</strong><br />

⎡H ⎤<br />

⎣ ⎦<br />

= =1,6507.<strong>10</strong><br />

-5<br />

0,6058.<strong>10</strong><br />

Suy ra: pH = - lg 1,6507.<strong>10</strong> -9 = 8,794<br />

Khoái löôïng muoái:<br />

0,3V.68 + 0,3V.82 = 4,5<br />

V = 0,1(l) = <strong>10</strong>0(ml)<br />

Caâu 4<br />

+ -9<br />

1. Ñaây laø muoái axit, laø chaát ñieän li löôõng tính :<br />

MHA ↽ ⇀ M + + HA –<br />

HA –<br />

HA – +<br />

↽ ⇀ H + + A 2– K 2 (1)<br />

H + ↽ ⇀<br />

–1<br />

H 2 A K 1<br />

3<br />

(2)<br />

pH phuï thuoäc hai quaù trình (1) vaø (2). Neáu K 2 caøng lôùn vaø K 1 caøng lôùn thì dung dòch coù pH<br />

caøng beù vì quaù trình nhöôøng proton (1) xaûy ra maïnh, quaù trình thu proton (2) xaûy ra yeáu. So saùnh ôû<br />

treân ta thaáy:<br />

pH (NaHC 2 O 4 ) < pH (NaHSO 3 ) < pH (NaHCO 3 ) < pH (NaHS).<br />

Neáu aùp duïng coâng thöùc gaàn ñuùng ñeå tính pH cuûa caùc muoái ñiaxit cho caùc heä treân<br />

pK1 + pK 2<br />

pH =<br />

2<br />

Ta thaáy pH(NaHC 2 O 4 ) = (2 + 5 ) / 2 = 3,5<br />

pH (NaHSO 3 ) = (2 +6) / 2 = 4,0<br />

pH (NaHCO 3 ) = (7 +11) / 2 = 9,0<br />

pH (NaHS) = (7 +13) / 2 = <strong>10</strong>,0<br />

Keát quaû naøy phuø hôïp vôùi caùch saép ôû treân.<br />

2. Caùc quaù trình xaûy ra :<br />

HCl → H + + Cl –<br />

DIỄN ĐÀN TOÁN - LÍ - HÓA <strong>10</strong>00B TRẦN HƯNG ĐẠO TP.QUY NHƠN<br />

Pb(NO 3 ) 2 → Pb 2+ + 2NO 3<br />

–<br />

Pb 2+ + 2Cl – ⇀ ↽ PbCl 2 ↓ ( K S ) –1 = ( <strong>10</strong> –4,8 ) –1<br />

C <strong>10</strong> –3 C ?<br />

∆C – ( <strong>10</strong> –3 - <strong>10</strong> –5 ) –1,98 . <strong>10</strong> –3<br />

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