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

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

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