You have 275 mL of an 0.13 M acetic acid solution. What volume (V) of 2.30 M NaOH solution must you add in order to prepare an acetate buffer of pH = 4.84? (The pKa of acetic acid is 4.76.)
Here we use Henderson hasselbalch equation to solve this problem.
We are given volume and molarity of acetic acid and molarity of NaOH.
Concentration = mol solute / volume of solution in L .
This buffere is formed from weak acid CH3COOH and NaOH
Conjujgate base is CH3COO- which is formed from CH3COONa
Since CH3COONa dissociates complete and so conjugate base has same concentration as that of CH3COONa.
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