Let's call benzoic acid HB and sodium benzoate is NaB. NaB is the base; HB is the acid.
.......HB + NaOH ==> NaB + H2O
I......0.1....0.......0.1......
add.........0.01..............
C....-0.01..-0.01......+0.01
E.....0.09.....0.......0.11
Substitute the E line into the Henderson-Hasselbalch equation and solve for pH.
A 500 mL buffer solution is 0.1 M benzoic acid and 0.10 M in sodium benzoate and has an initial pH of 4.19. What is the pH of the buffer upon addition of 0.010 mol of NaOH?
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