1 Function of the Conjugate Base The function of the acetate ion C 2 H 3 O 2  is to neutralize added H 3 O +. The acetic acid produced by the neutralization.

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Presentation transcript:

1 Function of the Conjugate Base The function of the acetate ion C 2 H 3 O 2  is to neutralize added H 3 O +. The acetic acid produced by the neutralization contributes to the available weak acid. C 2 H 3 O 2  (aq) + H 3 O + (aq) HC 2 H 3 O 2 (aq) + H 2 O(l) acetate ion acid acetic acid water

2 Summary of Buffer Action Buffer action occurs because the weak acid in a buffer neutralizes base the conjugate base in the buffer neutralizes acid the pH of the solution is maintained

3 pH of a Buffer The [H 3 O + ] in the K a expression is used to determine the pH of a buffer. Weak acid + H 2 O H 3 O + + conjugate base K a = [H 3 O + ][conjugate base] [weak acid] [H 3 O + ] = K a x [weak acid] [conjugate base] pH =  log [H 3 O + ]

Guide to Calculating pH of a Buffer 4

5 Example of Calculating Buffer pH The weak acid H 2 PO 4  in a blood buffer H 2 PO 4  /HPO 4 2  has a K a = 6.2 x 10  8. What is the pH of the buffer if [H 2 PO 4  ] = 0.20 M and [HPO 4 2  ] = 0.20 M? STEP 1 Write the K a expression for: H 2 PO 4  (aq) + H 2 O(l) HPO 4 2  (aq) + H 3 O + (aq) K a = [HPO 4 2  ][H 3 O + ] [H 2 PO 4  ] STEP 2 Rearrange the K a for [H 3 O + ]: [H 3 O + ] = K a x [H 2 PO 4  ] [HPO 4 2  ]

6 Example of Calculating Buffer pH (continued) STEP 3 Substitute [HA] and [A  ]: [H 3 O + ] = 6.2 x 10  8 x [0.20 M] = 6.2 x 10  8 [0.20 M] STEP 4 Use [H 3 O + ] to calculate pH: pH =  log [6.2 x 10  8 ] = 7.21

7 Learning Check What is the pH of a H 2 CO 3 buffer that is 0.20 M H 2 CO 3 and 0.10 M HCO 3  ? K a (H 2 CO 3 ) = 4.3 x 10  7 1) ) ) 6.07

8 Solution What is the pH of a H 2 CO 3 buffer that is 0.20 M H 2 CO 3 and 0.10 M HCO 3  ? K a (H 2 CO 3 ) = 4.3 x 10  7 STEP 1 Write the K a expression for: H 2 CO 3 (aq) + H 2 O(l) HCO 3  (aq) + H 3 O + (aq) K a = [HCO 3  ][H 3 O + ] [H 2 CO 3 ] STEP 2 Rearrange the K a for [H 3 O + ]: [H 3 O + ] = K a x [H 2 CO 3 ] [HCO 3  ]

9 Solution (continued) STEP 3 Substitute [HA] and [A  ]: [H 3 O + ] = 4.3 x 10  7 x [0.20 M] = 8.6 x 10  7 M [0.10 M] STEP 4 Use [H 3 O + ] to calculate pH: pH =  log [8.6 x 10  7 ] = 6.07