can gap giup minh voi

Question

bai 1: tinh nong do ion OH- trong cac dd sau:
a/ dd hon hop NH3 0,01M va NaOH 0,0001M. cho KbNH3 = 1,8x10^{-5}
b/ Tron 200ml dd CH3COOH 0,1M voi 300ml dd KOH 0,01M duoc dd A. cho Ka= 1,75x10^{-5}

bai2: tinh nong do H+ trong
a/ dd CH3COOH 10^{-2}M va H2SO4 10^{-4}M. Cho Ka = 1,75x 10^{-5}

Lê Quang Trung · Answer 1 · 7/11/2013 3:16:41 PM

Bình chọn tăng 1 Bình chọn giảm

Bài $2$ :
$H_2SO_4 \rightarrow 2H^+ + SO4^{2-}$
$10^{-4}            2.10^{-4}$
         $CH_3COOH \rightleftharpoons CH3COO^- + H^+$
bd         $10^{-2}                                       2.10^{-4}$
pư            $x                        x         x$
CB        $10^{-2}-x               x        2.10^{-4}+x$
Ta có $K_a = \frac{[CH_3COO^-].[H^+]}{[CH_3COOH]} = x.(2.10^{-4} + x)/(10^{-2} - x) = 1,75.10^{-5}$
=> $x = 3,235.10^{-4} => [H^+] = 3,235.10^{-4} + 2.10^{-4} = 5,235.10^{-4}$

Trả lời 11-07-13 03:16 PM

Lê Quang Trung
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Lê Quang Trung · Answer 2 · 7/11/2013 3:07:58 PM

$NaOH \rightarrow Na^+ + OH^-$
$0,0001                            0,0001$
       $NH_3 + H_2O \rightleftharpoons NH4^+ + OH^-$
bd    $0,01                                   0,0001$
pư   $x                             x       x$
Cb   $0,01-x                x     0,0001+x$
Ta có $K_b = \frac{[NH_4^+].[OH^-]}{[NH_3]} = \frac{x.(0,0001+x)}{0,01-x} = 1,8.10^{-5}$
=> $x = 3,693.10^{-4} => [OH^-] = 0,0001 + 3,693.10^{-4} = 4,693.10^{-4}$
b, $n_{CH_3COOH} = 0,02$ mol ; $n_{KOH} = 0,003$ mol , $V_{dd mới} = 0,5$ lít
      $CH_3COOH + KOH \rightarrow CH_3COOK + H_2O$
Bđ       $0,02             0,003$
CB      $0,017            0                       0,003$
Dung dịch sau gồm $CH_3COOK$ và $CH_3COOH $
$C_MCH_3COOK = 0,003/0,5 = 0,006$ M; $C_MCH_3COOH = 0,017/0,5 = 0,034$M
           $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$
bd         $0,006                                    0,034$
pư         $y                                             y                  y$
CB       $0,006-y                             0,034 + y    y$
$K_b = \frac{[CH_3COOH].[OH^-]}{[CH_3COO^-]} = \frac{(0,034+y).y}{0,006-y} = 10^{-14}/K_a = 5,71.10^{-10}$
=> $y = 1,0076.10^{-10} = [OH^-]$