# Calculation for buffer of pH 5 in sodium fluoride solution

From a journal:

$\pu{500 mL}$ of $\pu{50 ppm}$ fluoride at $\mathrm{pH}~5$ was prepared by mixing $\pu{0.0552 g}$ of sodium fluoride with $\pu{4.1057 g}$ of sodium acetate and $\pu{1.65 mL}$ of acetic acid.

May I know how to get the solution of calculation for amounts of sodium acetate and acetic acid which are $\pu{4.1057 g}$ and $\pu{1.65 mL}$, respectively?

For sodium fluorid calculation is

$$\frac{\pu{50 mg}}{\pu{1 L}} \times \frac{\pu{1 g}}{\pu{1000 mg}} \times \frac{\pu{41.98 g mol-1}}{\pu{18.988 g mol-1}} \times \pu{0.5 L} = \pu{0.0552 g}$$

• I want to know the formula to get 4.1057 g of sodium acetate and 1.65 mL of acetic acid – jess Dec 4 '17 at 10:52
• I mean how to know the amount of sodium acetate must be 4.1057g and acetic acid is 1.65ml? I don't know where 4.1057g and 1.65ml came from – jess Dec 4 '17 at 11:14

The journal entry is sloppy.
It only specifies the final "concentration" of sodium fluoride. Note that the parts-per notation is highly discouraged! They neither specified the buffer capacity, nor the concentration of the other components.

Without knowing what kind of chemicals were used, you have to guess. So in this case I would guess they used anhydrous sodium acetate, which is a solid, $M=\pu{83 g/mol}$. With $m(\ce{NaOAc}) = \pu{4.1057 g}$ as given in the journal, we calculate that they used about $n(\ce{NaOAc})=\pu{0.05 mol}$. We can use the Henderson-Hasselbalch equation (ignoring sodium fluoride) to determine the amount of substance of acetic acid we need, $\mathrm{p}K_\mathrm{a}=4.75$, $\mathrm{pH}=5$. \begin{align} \mathrm{pH} &= \mathrm{p}K_\mathrm{a} + \log_{10} \left(\frac{c(\ce{NaOAc})}{c(\ce{HOAc})}\right)& \Longleftrightarrow&& \mathrm{pH} - \mathrm{p}K_\mathrm{a} &= \log_{10} \left(\frac{n(\ce{NaOAc})}{n(\ce{HOAc})}\right)\\ \end{align}

Therefore we need about $n(\ce{HOAc}) = \pu{0.03 mol}$.

Now we have to guess again that they have used glacial acetic acid. This is quite reasonable because it's a small volume and the purest form of acetic acid. According to Sigma Aldrich it has a concentration of about $c(\ce{HOAc})=\pu{17.4 mol/L}$. Therefore we need about $V(\ce{HOAc}\text{, glac.}) = \pu{1.7 mL}$.

I hope you learned from this that you should not leave incomplete notes in your lab journal.

• Thanks so much! But can I know that how to know n(NaOAc)=0.05 mol? and amount for sodium acetate =83g/mol x 0.05mol = 4.1015? – jess Dec 6 '17 at 13:24
• @jess Like I said, I guessed. I simply took the values from the entry and reverse-engineered it. – Martin - マーチン Dec 6 '17 at 13:30

Acetic acid is a liquid. It has a $\mathrm{p}K_\mathrm{a} = 4.75$. A mixture of acetic acid and sodium acetate form a buffer solution which means that the solution resists $\mathrm{pH}$ changes. If the acetic acid and sodium acetate are present in equal molar quantities, then the $\mathrm{pH}$ of the buffer solution will be $4.76$. Since $\mathrm{pH}$ of buffer was $5$, there are more moles of acetate ($\ce{OAc^-}$) than acetic acid ($\ce{HOAc}$).

$$\mathrm{p}K_\mathrm{a} = 4.76$$ $$K_\mathrm{a} = 1.74 \times 10^{-5} = \frac{\ce{[H^+][OAc^-]}}{\ce{[HOAc]}}$$ $$\mathrm{pH} = - \log{\ce{[H^+]}}$$