chem126--Acids IV Buffers

chem126--Acids IV Buffers - COMPARISON OF ACIDIC SOLUTION...

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A buffer consisting of 0.58 M sodium acetate and 0.33 M acetic acid has a pH of 5.00. Diluting 50 ml of the buffer has no effect on the pH 1.0x10 -5 M HCl has a pH of 5. Diluting 50 ml of the acid to 100 ml Æ pH = 5.3 Diluting 50 ml of the acid to 500 ml Æ pH = 6.0 Adding 10 ml 0.10 M HCl to 50 ml of the acid Æ pH = 1.78 Adding 10 ml 0.10 M NaOH to 50 ml of the acid Æ pH = 12.2 Adding 10 ml 0.10 M HCl to 50 ml of the buffer Æ pH = 4.96 Adding 10 ml 0.10 M NaOH to 50 ml of the buffer Æ pH = 5.05 COMPARISON OF ACIDIC SOLUTION TO BUFFER SOLUTION
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A BUFFER—A SOLUTION THAT CONTAINS BOTH A WEAK ACID AND ITS CONJUGATE BASE. IT RESISTS CHANGES IN pH BY NEUTRALIZING EITHER AN ADDED ACID OR AN ADDED BASE. ( ) ( )( ) ( ) 33 2 CH COOH aq OH aq CH COO aq H O l −− +→ + () ( ) ( ) 3 2 HO aq CHCOO aq CHCOOH aq HOl +− + AT EQUILIBRIUM K b = 5.75x10 -10 M K a = 1.74x10 -5 M
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FINDING Ph OF BUFFER SOLUTION USING METHOD OF SUCCESSIVE APPROXIMATIONS K a = 1.74x10 -5 M EXAMPLE 16 : Find pH of 0.33 M CH 3 COOH and 0.58 M CH 3 COO - . Init. Change Eqm. 0.33 M 0 0.58 M -x +x +x 0.33-x x 0.58+x ( ) 5 0.58 1.74 10 0.33 a xx x Kx + == Solve quadratic equation or use method of successive approximations
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( ) 5 1.74 10 0.33 0.58 x x x x = + 0 ENTER ( ) ( ) E- 1.74 5 .33 ANS / 0.58 ANS −+ ENTER ENTER 0 9.90 E -6 9.90 E -6 pH = -log [H 3 O + ] = -log(9.90 E - 6) = 5.00 [H 3 O + ] = 9.90 x 10 -6 M TI-83 E6
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DERIVATION OF EASY METHOD OF FINDING pH OF BUFFER SOLUTION—HENDERSON-HASSELBALCH EQUATION () ( ) ( )( ) 23 a HB aq H O l H O aq B aq K +− ++ ± Let HB be a weak acid and B - it conjugate base. Let [HB] 0 and [B - ] 0 be their respective initial concentrations. Find the pH. Init. Change Eqm. [HB] 0 0 [B - ] 0 -x +x +x [HB] 0 -x x [B - ] 0 +x
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DERIVATION OF EASY METHOD OF FINDING pH OF BUFFER SOLUTION—HENDERSON-HASSELBALCH EQUATION Init. Change Eqm. [] ( ) 0 3 0 a xB x K HB HO B H x B + ⎡⎤ ⎣⎦ = ⎤ + = [HB] 0 0 [B - ] 0 -x +x +x [HB] 0 -x x [B - ] 0 +x Let HB be a weak acid and B - it conjugate base. Let [HB] 0 and [B - ] 0 be their respective initial concentrations. Find the pH. () ( ) ( )( ) 23 a HB aq H O l H O aq B aq K +− ++ ±
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DERIVATION OF EASY METHOD OF FINDING pH OF BUFFER SOLUTION—HENDERSON-HASSELBALCH EQUATION Init. Change Eqm. [] ( ) 0 3 0 a xB x K HB HO B H x B + ⎡⎤ ⎣⎦ = ⎤ + = [ ] () [ ] 00 0 0 aa H Bx H B xK K B =≈ + [HB] 0 0 [B - ] 0 -x +x +x [HB] 0 -x x [B - ] 0 +x Let HB be a weak acid and B - it conjugate base. Let [HB] 0 and [B - ] 0 be their respective initial concentrations. Find the pH. ( ) ( )( ) 23 a HB aq H O l H O aq B aq K +− ++ ±
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[ ] 0 3 0 a HB HO K B + ⎡⎤ = ⎣⎦
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[] 0 3 0 log log log a HB HO K B + ⎛⎞ ⎜⎟ ⎡⎤ =+ ⎣⎦ ⎝⎠ [ ] 0 3 0 a HB K B + =
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[] 0 3 0 log log log a HB HO K B + ⎛⎞ ⎜⎟ ⎡⎤ =+ ⎣⎦ ⎝⎠ 0 3 0 log log log a HB pH H O K B + =− [ ] 0 3 0 a HB K B + =
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[] 0 3 0 log log log a HB HO K B + ⎛⎞ ⎜⎟ ⎡⎤ =+ ⎣⎦ ⎝⎠ [ ] 0 3 0 a HB K B + = 0 3 0 log log log a HB pH H O K B + =− 0 0 log a B pH pK HB CHECK METHOD : Find pH of 0.33 M CH 3 COOH and 0.58 M CH 3 COO - .
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This note was uploaded on 04/07/2008 for the course CHEM 126 taught by Professor Biolsi during the Spring '08 term at NJIT.

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chem126--Acids IV Buffers - COMPARISON OF ACIDIC SOLUTION...

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