Caluclate the pH at the equivalence point for the titration of 40mL 0.1M HNO2 wi
ID: 504671 • Letter: C
Question
Caluclate the pH at the equivalence point for the titration of 40mL 0.1M HNO2 with 0.2M KOH.I found the moles of KOH required to be 0.004mol and the volume to be 20mL. The answer I'm getting is pH=11.74, but my textbook says the answer is 18.08? I didn't think that was possible. Please show work on how to calculate pH at equivalence point! Caluclate the pH at the equivalence point for the titration of 40mL 0.1M HNO2 with 0.2M KOH.
I found the moles of KOH required to be 0.004mol and the volume to be 20mL. The answer I'm getting is pH=11.74, but my textbook says the answer is 18.08? I didn't think that was possible. Please show work on how to calculate pH at equivalence point!
I found the moles of KOH required to be 0.004mol and the volume to be 20mL. The answer I'm getting is pH=11.74, but my textbook says the answer is 18.08? I didn't think that was possible. Please show work on how to calculate pH at equivalence point!
Explanation / Answer
at equivalence point millimoles of acid = millimoles of base
C1 V1 = C2 V2
40 x 0.1 = 0.2 X V2
V2 = 20 mL
volume of base needed = 20 mL
HNO2 + KOH --------------------->KNO2 + H2O
at equivalence point only salt remains . total acid and base consumed
salt concentration = millimoles / total volume
= 40 x 0.1 / (40 +20)
= 0.067 M
HNO2 , pKa value = 3.25
the salt is from strong base and weak acid so the pH is more than 7 .
formula : pH = 7 + 1/2 [pKa + logC]
pH =7 + 1/2 [3.25 + log 0.067]
pH = 8.04 --------------------> answer
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