What would the [OH-] be after addition of 17.47 mL of 0.1000 M NaOH to 25.00 mL of 0.1000 HA (a weak acid, Ka = 3.16e-4)?
millimols NaOH = mL x m = 17.47 x 0.1 = 1.747
millimols HA = 25 x 0.1 = 2.5
.........HA + NaOH ==> NaA + H2O
I........2.5....0.......0......0
added.........1.747............
C......-1.747.-1.747...1.747.....1.747
E......0.753....0......1.747
Substitute into the Henderson-Hasselbalch equation and solve for pH then convert to pOH and OH^-