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    The example they gave in my textbook for a buffer solution is really confusing.

    'In the CH3COOH/CH3COONa buffer system:

    the weak acid dissociates partially:
    CH3COOH(aq) <==> H+(aq)

    the salt dissociates completely generating the conjugate base
    CH3COO-Na+ --> CH3COO-(aq) + Na+(aq)

    The equilibrium mixture formed containes a high concentration of the undissociated weak acid and its conjugate base. The high concentration of the conjugate base pushes the equilibrium to the left, so the concentration of H+ ions is very small'

    What is the purpose of CH3COO-Na+? I know it shifts the equilibrium to the left, but what is the purpose of this? What does shifting the equilibrium even do?
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    Is the purpose of it to make a buffer solution that is basic? Is that it?
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    (Original post by celina10)
    The example they gave in my textbook for a buffer solution is really confusing.

    'In the CH3COOH/CH3COONa buffer system:

    the weak acid dissociates partially:
    CH3COOH(aq) <==> H+(aq)

    the salt dissociates completely generating the conjugate base
    CH3COO-Na+ --> CH3COO-(aq) + Na+(aq)

    The equilibrium mixture formed containes a high concentration of the undissociated weak acid and its conjugate base. The high concentration of the conjugate base pushes the equilibrium to the left, so the concentration of H+ ions is very small'

    What is the purpose of CH3COO-Na+? I know it shifts the equilibrium to the left, but what is the purpose of this? What does shifting the equilibrium even do?
    I'm not sure if I understand your questions but here's an explanation about buffers:

    Buffers can be formed either by a weak acid and its conjugate base or a weak base and its conjugate acid. The example that your textbook is giving is of a weak acid CH3COOH and its conjugate base CH3COO-

    CH3COOH is a weak acid so it will just partially dissociate into CH3COO- and H+, this is the reaction (if reacting with NaOH):
    CH3COOH + NaOH --> CH3COONa + H2O.

    In other words, what happens to the CH3COOH is this:
    CH3COOH --> CH3COO- + H+ (It dissociates).


    Your textbook says that "The high concentration of the conjugate base pushes the equilibrium to the left, so the concentration of H+ions is very small". At equilibrium both CH3COOH and CH3COO- will be the same, but if the concentration of CH3COO- is higher, then the equilibrium shifts to the left to counteract this. You can think about it as the reaction trying to maintain both CH3COOH and CH3COO- equal. So if there is more CH3COO- then you need to convert it back to CH3COOH in order to restore the equilibrium. Because of this the equilibrium is said to be pushed to the left (the backward reaction is favoured), as more CH3COOH is produced in order to counteract the amount of CH3COO- present to try to achieve a reaction where CH3COOH = CH3COO-. This also affects the amount of H+ because in order to produce CH3COOH the CH3COO- needs to "combine" with the H+ which means that less H+will be available. Because of this the solution will be slightly basic (pH over 7.0) as the concentration of H+ will be less (if you think about pH, it is defined as the -log[H+], so a higher concentration of hydrogen ions will mean a lower pH and a lower concentration of hydrogen ions a higher pH, or a basic solution).

    I hope this helps
 
 
 
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