Question

Determine the pH increase of a solution after 0.10 mol of NaOH is added to 1.00 L of a solution containing 0.15 M HC2H3O2 and 0.20 M NaC2H3O2. If this same amount of NaOH was added to 1.00 L of pure water then what would be the pH increase of the resulting solution? Compare these two values.

pH increase in buffer: ?

pH increase in pure water: ?

Answer 1

in the buffer

pH = pKa + log(conjguate/acid) = 4.75 + log(0.2/0.15)= 4.8749

after 0.1 mol addition

0.15 + 0.15 = 0.30 acetate

0.15- 0.10 = 0.05 acid left

pH = pKa + log(conjguate/acid) = 4.75 + log(0.3/0.05)= 5.53

pH change = 5.53-4.8749 = 0.656

then

i nwater

pH initial = 7

pH After = [OH] = 0.1/1 = 0.1

pOH = -log(0.1) =1

pH = 14-1 = 13

13 vs 0.656 for the same amount!