A buffer contains the conjugate acid and base in similar amounts S OURCE OF

# A buffer contains the conjugate acid and base in

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A buffer contains the conjugate acid and base in similar amounts. S OURCE OF PROTONS (HA) to neutralize incoming base S INK OF PROTONS (A - ) to neutralize incoming acids
Compare two 1L solutions with the same pH (4.74), when we add 0.01mol NaOH: 1) The 0.1M acetic + 0.1M acetate solution from above 2) 1.8 x 10 -5 M HCl pH pH H 2 O( l ) + CH 3 CO 2 - ( aq ) CH 3 CO 2 H( aq ) + OH - ( aq )
Conjugate base: A - Weak acid: HA H 3 O + ( aq ) + Base( aq ) Acid( aq ) + H 2 O( l ) [ Acid ] [ Base ] [H 3 O + ] = K a K a = [H 3 O + ][ Base ] [ Acid ] H 3 O + ( aq ) + CH 3 CO 2 - ( aq ) CH 3 CO 2 H( aq ) + H 2 O( l ) -log([H 3 O + ]) = -log( K a ) - log [ Acid ] [ Base ] pH= p K a + log [ Base ] [ Acid ] H ENDERSON H ASSELBALCH EQUATION : pH of a buffer solution has a value close to the pK a . The Henderson-Hasselbalch equation provides a short-cut for calculating pH of buffers of known concentrations or determining how to make a buffer at a desired pH.
Buffers can’t tolerate addition of infinite amounts of strong acid or base! B UFFER C APACITY : is the molar amount of acid or base which the buffer can handle without significant changes in pH. After enough acid or base has been added to react with all the conjugate base or acid in the system, the buffer is destroyed. In CHM135, we assume that our buffer should maintain pH 1.00 around pK a pH= p K a + log [A - ] [HA] pH- p K a = log [A - ] [HA] That means that the relative concentrations can change from 1:10 to 10:1 (100x) and the pH of the buffer will only vary from pK a -1 to pK a +1 Large changes in acid or base concentration results in small changes in pH! 1 = log 10 -1 = log 1/10
M AKING A BUFFER : we generally want to maintain a solution at a particular pH. How do we know which buffer to use? 1. Select a weak acid with a p K a similar to desired pH. 2. Mix an equal amount of the acid with the conjugate base (e.g. for CH 3 COO - , use a salt with an inert counter-ion, like NaCH 3 COO) equal amounts H 3 O + ( aq ) + CH 3 CO 2 - ( aq ) CH 3 CO 2 H( aq ) + H 2 O( l ) equal amounts H 2 O( l ) + CH 3 CO 2 H ( aq ) + NaOH( aq ) Na CH 3 CO 2 ( aq ) OR Start with a weak acid and neutralize half of it with a strong base.

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