Question

Show Calculations used in mkaing 250mL of 0.5M sodium acetate ph 4.7.

You are given:
1M acetic acid (weak acid)
1M sodium hydroxide (strong base)

Sodium acetate buffer: ph 4.7
acetic acid pKa: 4.7

So far this is what I've gotten:
pH = pKa + log (A-/HA-)
4.7=4.7 + log (A-/HA-)
0 = log (A-/HA-)
(A-/HA-) = 1
A- = HA -

0.5M x 0.25L = 0.125 moles of each to 250 mL of water.

Not sure if correct.
Thank you!

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