You have 775 mL of an 0.15 M acetic acid solution. What volume (V) of 1.80...
You have 775 mL of an 0.15 M acetic acid solution. What volume (V) of 1.80 M NaOH solution must you add in order to prepare an acetate buffer of pH = 4.65? (The pKa of acetic acid is 4.76.)
Homework Answers
pH = 4.65
moles of acetic acid = 775 x 0.15 / 1000 = 0.116
moles of NaOH = 1.80 x V
CH3COOH + NaOH -------------------> CH3COONa + H2O
0.11625 1.80V 0 0
0.11625 - 1.80V 0 1.80 V 1.80 V
pH = pKa + log [salt / acid]
4.65 = 4.76 + log [1.80V / 0.11625 - 1.80 V]
[1.80V / 0.11625 - 1.80 V] = 0.776
V = 0.0282 L
volume of NaOH = 28.2 mL
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