Question

What is the pH at the point of neutralization when 25.00 mL of 0.1000 molar acetic acid (Ka = 1.8 x 10-5) reacts with 25.00 mL of 0.1000 molar sodium hydroxide?
a) 8.60
b) 8.72
c) 9.76
d) 10.33
e) 11.26

Answer 1

\[pH = pKA + Log[\frac{ [A^{-}] }{ [HA] }]\]

Well, the pH at the equivalence point is going to be greater than 7. when titrating with a strong base with a weak acid.

Answer 2

@Cuanchi @aaronq

Answer 3

so which one is it?

Answer 4

@aaronq what do you think?

Answer 5

no, in this case at the equivalence point you dont have the buffer anymore, you only have the salt sodium acetate, then it is a basic salt and you have to calculate the [OH-]

\[[OH-]=\sqrt{K _{b} \times C _{s}}\]

see example 3 in attached