105 mL x 0.4M = 42.0 millimoles HOCl
105 mL x 0.36M = 37.8 mmoles NaOCl.
13.4 mL x 1.5M NaOH = 20.1 mmoles NaOH.
.........HOCl + OH^- ==> OCl^- + H2O
Initial..42.0...0........37.8.....
add............20.1.................
change..-20.1..-20.1.....+20.1
equil.....?.....0.........?
Substitute the equil line into the Henderson-Hasselbalch equation and solve for pH.
For 105ml of a buffer that is .40 M in HOCL and .36M in NaOCL what is the pH after 13.4 ml of 1.5M NaOH is added?
Ka for HOCL is 3.5x10^-8
1 answer