1) A titration is performed by adding 0.727 M KOH to 80 mL of 0.156 M HNO 3 . a)
ID: 776005 • Letter: 1
Question
1) A titration is performed by adding 0.727 M KOH to 80 mL of 0.156 M HNO3.
a) Calculate the pH before addition of any KOH.
b) Calculate the pH after the addition of 3.43, 8.59 and 16.17 mL of the base.(Show your work in detail for one of the volumes.)
c) Calculate the volume of base needed to reach the equivalence point.
d) Calculate the pH after adding 5.00 mL of KOH past the endpoint.
1a) pH = _____________ (before adding any KOH)
1b) pH =_____________ (after adding 8.59 mL)
1c) Volume = ____________ mL (volume at the equivalence point)
1d) pH = _______________ (after adding 5.00 mL of KOH past the end point)
Explanation / Answer
Both HNO3 and KOH are strong acid and strong base hence they dissocite completely to give H+ and OH- ions respectively.
[HNO3] = 0.156 M
[KOH] = 0.727 M
(a).
initial [H+] = 0.156 M
pH = - log [H+]
= - log ( 0.156)
= 0.80
(b). 3.43 mL of base is added.
moles of H+ = 0.156*0.080
= 0.01248
moles of OH- = 0.00343*0.727
= 0.00249
excess moles of H+ = 0.01248 - 0.00249
= 0.00999
total volume = 80 + 3.43
= 83.43 mL = 0.08343 L
[H+] = 0.00999 / 0.08343
= 0.119 M
pH = - log[H+]
= - log (0.119)
= 0.92
(c).
at equivalence point moles of HNO3 = moles of KOH
80*0.156 = V*0.727
V = 17.16 mL
(d).
after adding 5.00 mL of KOH past the endpoint
excess moles of OH- = 0.727*0.005
= 0.003635 moles
total volume = 80 + 17.16 + 5
= 102.16 mL = 0.10216 L
[OH-] = 0.003635 / 0.10216
= 0.0355 M
pOH = - log[OH-]
= 1.45
pH = 14 - pOH
= 14 - 1.45
= 12.55
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