(NaOH) = 0.130 x (500/1125) = ?= approx 0.06
(HA) = 0.250 x (625/1125) = ? = approx 0.14
.......... HA(aq)+OH−(aq)⟶H2O(l)+A−(aq)
I..........0.14.......0..................0..........0
add,,,,,,,,,,,,,,,,..0.06..............................
C.....-0.06.....-0.06...........0.06........0.06
E.........?...........0...............0.06.......0.06
Note: all of these numbers are estimates.
Plug the E line into the HH equation and solve or pH.
Post your work if you get stuck.
500.0 mL of 0.130 M NaOH is added to 625 mL of 0.250 M weak acid (Ka=5.17×10−5). What is the pH of the resulting buffer?
HA(aq)+OH−(aq)⟶H2O(l)+A−(aq)
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