The Henderson-Hasselbalch equation is pH = pKa + log [(base)/(acid)]. Technically, those are concentrations for base and acied; however, I will take a shortcut and use mols since the volume cancels. Your prof may not like the shortcut so beware. Also, I will let HB stand for benzoic acid and B^- is benzoate.
mols HCl added = M x L = 0.06
...................B^- + H^+ ==> HB
I................0.5......0.............0.4
add.....................0.06..............
C...........-0.06..-0.06...........+0.06
E............0.44.......0.............0.46
Plug the E line into the HH equation and solve for the new pH.
Post your work if you get stuck.
A buffer was prepared by dissolving 0.40 mol of benzoic acid (Ka=6.3*10^-5) and 0.50 mol of sodium benzoate in sufficient pure water to form a 1.00 L solution. To this solution was added 30.00 mL of 2.00 M HCl solution. What was the pH of the new solution?
C6H5COOH (aq) + H20 (l) <--------> H3O+ (aq) + C6H5COO- (aq)
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