Asked by hershi

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)?

11 years ago

Answered by DrBob222

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^-

11 years ago

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