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 . Calculate the pH at the equivalence point.
\[
\left[\mathrm{OH}^{-}\right]=x=\sqrt{K_{b}\left[\mathrm{~F}^{-}\right]} \quad K_{b}=1.6 \times 10^{-11}
\]
\[
\left[\mathrm{F}^{-}\right]=\frac{\text { moles }}{\mathrm{li}^{\mathrm{n}} \mathrm{~s}}
\]
How do you think the moles of \( \mathrm{F}^{-} \)at the -equivalence point are related to the initial moles of HF?
moles \( \mathrm{F}^{-}= \)initial moles HF - moles C
moles \( \mathrm{F}^{-}= \)initial moles \( \mathrm{HF}^{-} \)
moles \( \mathrm{F}^{-}=\frac{\text { initial moles HF }}{2} \)
moles \( \mathrm{F}^{-}=2 \) (initial moles HF)