If the ratio of weak base to conjugate acid is #10:1#, to what extent does the #"pH"# change and does it increase or decrease?

Question

Peyton Adams · Accepted Answer

Suppose the buffer started with a #"pH"# of, well, "#"pH"#", with a #"pKa"# of "#"pKa"#", with general concentrations of acid #"HA"# and conjugate base #"A"^(-)# (or if you prefer, the salt #"NaA"#). Then the Henderson-Hasselbalch equation shows: #"pH" = "pKa" + log\frac(["A"^(-)])(["HA"])# Now, if we have #10# times the #"A"^(-)# (noting that no identity of anything changes, so the #"pKa"# of the acid stays the same), we simply get a new #"pH"# of: #color(blue)("pH"') = "pKa" + log\frac(10["A"^(-)])(["HA"])# And if we recall... #log(ab) = log a + log b#. Thus... #=> "pKa" + log\frac(["A"^(-)])(["HA"]) + log 10# #= "pH" + log 10# #= color(blue)("pH" + 1)# So, the #"pH"# increases by #1#, and the solution becomes #10# times more basic. That SHOULD make sense... we did, after all, make it so we had #10# times the conjugate base.

Olivia Anton · Answer

undefined If the ratio of weak base to conjugate acid is 10:1, the pH will increase. This is because the weak base will react with water to produce more hydroxide ions, which will shift the equilibrium towards higher pH, making the solution more basic.