A 50.00 mL sample of 0.0950 M acetic acid (Ka = 1.8 x 10-5) is being titrated with 0.106 M NaOH.
What is the pH at the half-way point of the titration? (22.41 mL of 0.106 M NaOH has been added)
A) 3.06
B) 5.04
C) 4.44
D) 3.18
E) 4.74
In the halfway titration, we always have a buffer
pH = pKa + log(conjguate/acid)
and since it is half way then
conjugate = acid
so
pH = pKa + log(1) = pKa
pH = pKa
pKa = -log(Ka) = -log(1.8*10^-5) = 4.75
neareswt answer is E pH = 4.74
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