Posted by June on Thursday, July 19, 2012 at 8:43pm.
mol NH3 = M x L = 0.19*1 = 0.190
mol NH4^+ = 0.25*1 = 0.250
-----------------
........NH4^+ + OH^- ==> NH3 + H2O
I.....0.250.....0......0.190
add..............x............
C.......-x......-x.......+x
E.....0.250-x.....0......0.190+x
Substitute into the Henderson-Hasselbalch equation and solve for x = mol NaOH added to 1L buffer to make pH = 9.25
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