50.0 mL of 0.100 M KOH (aq) is titrated with 0.100 M HNO3 (aq). What is the pH... a) initially? b) halfway to the equivalence point? c) at the equivalence point? d) 1 mL past the equivalence point?

Answer 1

KOH + HNO3 ==> KNO3 + H2O .. balanced equation

(a). Initial pH is the pH of 0.100 M KOH since no reaction has taken place yet.

pH + pOH = 14

pH = 14 - pOH

pOH = -log [OH-] = -log 0.100 = 1

pH = 14 - 1

pH = 13

(b). Half way to equivalence means 1/2 the moles of KOH have reacted.

Initial moles KOH = 50.0 ml x 1 L / 1000 ml x 0.100 mol / L = 0.005 mols KOH

Since the [HNO3] is the same as the KOH (0.100 M), to neutralize 1/2 of the KOH will require 25.00 mls

Total volume will be 50.00 mls + 25.00 mls = 75.00 mls = 0.07500 L

Moles KOH left will be 0.005 mols / 2 = 0.0025 mols

Final [KOH] = 0.0025 mols / 0.07500 L = 0.0333 M

pOH = -log 0.0333 = 1.48

pH = 14 - 1.48

pH = 12.5

(c). At equivalence all of the KOH has reacted to produce KNO3 and H2O. Since this is a reaction between a strong base (KOH) and a strong acid (HNO3), the pH will be neutral, or pH = 7

(d). At 1 ml past equivalence, there will be 1 ml excess of the HNO3, so the solution will now be acidic.

For equivalence, it takes 50.0 mls HNO3 so final volume @ equivalence = 50.0 ml + 50.0 ml = 100 mls

At 1 ml past equivalence, total volume = 101.00 mls = 0.101 L

Moles of excess HNO3 = 1 ml HNO3 x 1 L / 1000 mls x 0.100 mol / L = 1x10^-4 mols HNO3

Final [HNO3] = 1x10^-4 mols / 0.101 L = 9.90x10^-4 M

pH = -log [HNO3] = -log 9.90x10^-4

pH = 3.00