60.0 mL of 0.50 MHF(Ka=6.3 × 10^-4) is titrated with 0.50 M NaOH . Calculate the pH at the equiv

The titration reaction is going to be one mole of sodium hydroxide reacting with one mole of HF

Hi, so for this problem we are going to begin by getting the amount of OH - that we are going to

Question

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. \[ \begin{aligned} {\left[\mathrm{OH}^{-}\right] } & =x=\sqrt{\mathrm{K}_{\mathrm{b}}\left[\mathrm{~F}^{-}\right]} \quad \mathrm{K}_{\mathrm{b}}=1.6 \times 10^{-11} \quad[\mathrm{~F}-]=0.25 \mathrm{M} \\ \mathrm{pH} & +\mathrm{pOH}=14.00 \\ \mathrm{pH} & =14.00-\mathrm{pOH} \\ \mathrm{pH} & =14.00-\left(-\log \left[\mathrm{OH}^{-}\right]\right) \\ \mathrm{pH} & =14.00-\left(-\log \sqrt{\mathrm{K}_{\mathrm{b}}\left[\mathrm{~F}^{-}\right]}\right) \end{aligned} \] Predict the pH of the solution at the equivalence point. -Then, proceed with the video to see if your answer matches the correct answer. 8.30 5.70 1.90 7.00 12.10

          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.
\[
\begin{aligned}
{\left[\mathrm{OH}^{-}\right] } & =x=\sqrt{\mathrm{K}_{\mathrm{b}}\left[\mathrm{~F}^{-}\right]} \quad \mathrm{K}_{\mathrm{b}}=1.6 \times 10^{-11} \quad[\mathrm{~F}-]=0.25 \mathrm{M} \\
\mathrm{pH} & +\mathrm{pOH}=14.00 \\
\mathrm{pH} & =14.00-\mathrm{pOH} \\
\mathrm{pH} & =14.00-\left(-\log \left[\mathrm{OH}^{-}\right]\right) \\
\mathrm{pH} & =14.00-\left(-\log \sqrt{\mathrm{K}_{\mathrm{b}}\left[\mathrm{~F}^{-}\right]}\right)
\end{aligned}
\]

Predict the pH of the solution at the equivalence point. -Then, proceed with the video to see if your answer matches the correct answer.
8.30
5.70
1.90
7.00
12.10

60.0 mL of 0.50 MHF(Ka=6.3 × 10^-4) is titrated with 0.50 M NaOH . Calculate the pH at the equivalence point.

[OH^-] =x=√(Kb[ F^-]) Kb=1.6 × 10^-11 [ F-]=0.25 M
pH +pOH=14.00
pH =14.00-pOH
pH =14.00-(-log[OH^-])
pH =14.00-(-log√(Kb[ F^-]))

Predict the pH of the solution at the equivalence point. -Then, proceed with the video to see if your answer matches the correct answer.
8.30
5.70
1.90
7.00
12.10

Added by Gregg M.

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