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How many moles of \( \mathrm{NaF}(\mathrm{s}) \) should be added to 1.0 L . of 0.10 MI IIF (aq) to make a buffer at pH - \( 3.50^{\circ} \) : A ssume no volume change. \[ \begin{array}{l} \mathrm{pH}=\mathrm{pKa}+\log \frac{|\mathrm{F}|}{[\mathrm{HF}]} \frac{|\mathrm{F}|}{|\mathrm{HF}|}=2.0 \\ 3.50=3.20+\log \frac{\left[\mathrm{F}^{*}\right]}{[\mathrm{HF}]}|\mathrm{F}|=2.0[\mathrm{HF} \mid \\ \log \frac{[\mathrm{F}]}{[\mathrm{HF}]}=0.30|\mathrm{~F}|=2.0(0.10 \mathrm{M}) \\ \frac{[\mathrm{F}]}{[\mathrm{HF}]}=10^{\circ 0} 0 {\left[\mathrm{~F}^{-}\right]=0.20 \mathrm{M} } \\ 1.0 \mathrm{~L}\left(\frac{0.20 \mathrm{~mol}}{1 \mathrm{~L}}\right)=0.20 \mathrm{~mol} \mathrm{~F} \\ \end{array} \] \( 4: 18 \) \( 4: 37 \) earning re we done with this calculation? If not, what steps remain? Yes, we now have the quantity that the original question asked for. No, we still have to convert molarity to moles. No, we still have to convert molarity to moles, and convert moles of \( \mathrm{F}^{-} \)to moles of NaF . No, we still have to find the total volume and then convert molarity to moles.

          How many moles of \( \mathrm{NaF}(\mathrm{s}) \) should be added to 1.0 L . of 0.10 MI IIF (aq) to make a buffer at pH - \( 3.50^{\circ} \) : A ssume no volume change.
\[
\begin{array}{l} 
\mathrm{pH}=\mathrm{pKa}+\log \frac{|\mathrm{F}|}{[\mathrm{HF}]} \frac{|\mathrm{F}|}{|\mathrm{HF}|}=2.0 \\
3.50=3.20+\log \frac{\left[\mathrm{F}^{*}\right]}{[\mathrm{HF}]}|\mathrm{F}|=2.0[\mathrm{HF} \mid \\
\log \frac{[\mathrm{F}]}{[\mathrm{HF}]}=0.30|\mathrm{~F}|=2.0(0.10 \mathrm{M}) \\
\frac{[\mathrm{F}]}{[\mathrm{HF}]}=10^{\circ 0} 0 {\left[\mathrm{~F}^{-}\right]=0.20 \mathrm{M} } \\
1.0 \mathrm{~L}\left(\frac{0.20 \mathrm{~mol}}{1 \mathrm{~L}}\right)=0.20 \mathrm{~mol} \mathrm{~F} \\
\end{array}
\]
\( 4: 18 \)
\( 4: 37 \)
earning
re we done with this calculation? If not, what steps remain?
Yes, we now have the quantity that the original question asked for.
No, we still have to convert molarity to moles.
No, we still have to convert molarity to moles, and convert moles of \( \mathrm{F}^{-} \)to moles of NaF .
No, we still have to find the total volume and then convert molarity to moles.

How many moles of NaF(s) should be added to 1.0 L . of 0.10 MI IIF (aq) to make a buffer at pH - 3.50^∘ : A ssume no volume change.

pH=pKa+log(|F|)/([HF])(|F|)/(|HF|)=2.0

3.50=3.20+log([F^*])/([HF])|F|=2.0[HF|
log([F])/([HF])=0.30| F|=2.0(0.10 M)
([F])/([HF])=10^∘ 0 0 [ F^-]=0.20 M

1.0 L((0.20 mol)/(1 L))=0.20 mol F

4: 18
4: 37
earning
re we done with this calculation? If not, what steps remain?
Yes, we now have the quantity that the original question asked for.
No, we still have to convert molarity to moles.
No, we still have to convert molarity to moles, and convert moles of F^-to moles of NaF .
No, we still have to find the total volume and then convert molarity to moles.

Added by Aaron M.

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