60.0 mL of \( 0.50 \mathrm{M} \mathrm{HF}\left(K_{\mathrm{a}}=6.3 \times 10^{-4}\right) \) is titrated with 0.50 M NaOH . Calc \( \quad \) ate the pH at the equivalence point.
\( \left[\mathrm{OH}^{-}\right]=x=\sqrt{K} \)
\[
K_{b}=1.6 \times 10^{-11}
\]
\[
\left[\mathrm{F}^{-}\right]=\frac{1}{1}
\]
moles
fial moles \( \mathrm{HF}= \)
\[
\left[\frac{0.50 \mathrm{~mol}}{1 \mathrm{~L}}\right)=0.030 \mathrm{~mol}
\]
What do you think is the total -volume of the solution at the equivalence point?
120.0 mL
30.0 mL
60.0 mL .
180.0 mL