chemistry

If you add 5.0 mL of 0.50 M NaOH solution to 20.0 mL to Buffer C, what is the change in pH of
the buffer?
(where buffer C is 8.203 g sodium acetate with 100.0 mL of 1.0 M acetic acid)

I have calculated the pH of buffer C to be 4.74.
Now what? =\

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  1. couple someone please still explain? and explain how the 4.74 pH was found? thanks

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  2. could someone please still explain? and explain how the 4.74 pH was found? thanks

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  3. pH = pKa + log ([A-]/[HA])
    We ned to determine the concentration of the sodium acetate:
    Molar mass CH3COONa = 82.0340 g/mol
    Therefore you are adding 1mol CH3COOH per litre = 1.00M
    pKa CH3COOH = 4.75
    pH = 4.75 + log (1/1)
    pH = 4.75 as you got - OK.

    Question 2: If you add 5 ml of 0.50M NaOH solution to 20ml of the buffer, what is the change in pH?

    The answer should be that the pH does not change very much - that is what a buffer solution is all about. I am sure that there is a simple way to make this calculation - but I do not know it. So let us try and do it from first principles:

    You have a buffer solution that is 1.00M in CH3COOH and 1.00M in CH3COONa.
    If you add some NaOH you react with the CH3COOH , reducing its concentration and increasing the concentration of CH3COONa

    Mol CH3COOH in 20 mL of 1.00M = 20/1000*1 = 0.02mol CH3COOH
    Mol CH3COONa in 20mL of 1.00M = 0.02mol CH2COONa
    Mol NaOH in 5.0mL of 0.50M = 5/1000*0.5 = 0.0025mol NaOH
    The NaOH reacts with the CH3COOH to form 0.0025 mol CH3COONa and 0.0025 mol CH3COOH is removed
    You end up with
    CH3COOH = 0.02mol - 0.0025 mol = 0.0175mol CH3COOH in 25mL total solution:
    Molarity of CH3COOH = 0.0175/25*1000 = 0.70M CH3COOH
    CH3COONa = 0.02mol + 0.0025mol = 0.0225 mol CH3COONa in 25mL total solution
    Molarity of CH3COONa = 0.0225/25*1000 = 0.9M CH3COONa

    pH = pKa + log([A-]/[HA])
    pH = 4.75 + log( 0.9/0.7)
    pH = 4.75 + log 1.286
    pH = 4.75 + 0.11
    pH = 4.86

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