500.0 mL of 0.120 M NaOH is added to 595 mL of 0.250 M weak acid (Ka = 2.95 × 10-5). What is the pH of the resulting buffer?
The answer that i got using HH equation is 4.14 but apparently it's wrong. Am I approaching this problem from a different angle ?
4 answers
I think it can be done that way. Show your work and I'll try to find the error. I think your answer is close.
.5*.12=.06 mols NaOH
.595*.25=.14875 mols Acid
(.14875)/(1.095)=.1358M Acid
(.06/1.095)=.05479M NaOH
pH=4.53+log(base/acid)=4.14
.595*.25=.14875 mols Acid
(.14875)/(1.095)=.1358M Acid
(.06/1.095)=.05479M NaOH
pH=4.53+log(base/acid)=4.14
The acid and base react. Why didn't you let them do that.
...........NaOH + HA ==> NaA + H2O
I.........0.055..0.135....0.....0
C........-0.055..-0.055..+0.055
E...........0.....?........?
Then plug those numbers into except I would plug in the "real" numbers and not my estimates. I get 4.35 for the pH.
The main problem with the way you worked it is that the (HA) is decreased by the amount of NaOH added and you didn't count that.
...........NaOH + HA ==> NaA + H2O
I.........0.055..0.135....0.....0
C........-0.055..-0.055..+0.055
E...........0.....?........?
Then plug those numbers into except I would plug in the "real" numbers and not my estimates. I get 4.35 for the pH.
The main problem with the way you worked it is that the (HA) is decreased by the amount of NaOH added and you didn't count that.
OHH Great !!! It makes much more sense now.Thanks !