Equilibrium Question 750

Question: When 10 ml of 0.1 M acetic acid ( pKa . = 5.0) is titrated against 10ml of 0.l Mammonia solution ( pKb =5.0), the equivalence point occurs at pH

Options:

A) 5.0

B) 6.0

C) 7.0

D) 9.0

Show Answer

Answer:

Correct Answer: C

Solution:

pKa=−log⁡KapKa,pKb=−log⁡Kb

pH=−12[logKa+logKw−logKb]

=−12[−5+log(1×10−14)−(−5)]

=−12[−5−14+5]=−12(−14)=7

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