Question

How many grams of NaF must be added to \(70.00 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{HNO}_{3}\) to obtain a buffer with a pH of \(4.68 ?\)

Short answer

Answer: 0.844 grams of NaF must be added.

Step-by-step solution

Find the Ka of the weak acid (HNO3)

First, we need to determine the Ka of the weak acid HNO3. Using a reference table, we find that the Ka of HNO3 is \(3.4 \times 10^{-4}\).

Calculate the amount of HA (HNO3) in the solution

Calculate the moles of HNO3 in the 70.00 mL of 0.150 M HNO3 solution. $$ \mathrm{moles \ of \ HNO}_{3}=\mathrm{volume \ of \ HNO}_{3} \times \mathrm{concentration \ of \ HNO}_{3} $$ $$ \mathrm{moles \ of \ HNO}_{3} = 70.00 \, \mathrm{mL} \times \frac{0.150 \, \mathrm{mol}}{1000 \, \mathrm{mL}} $$ $$ \mathrm{moles \ of \ HNO}_{3} = 0.0105 \, \mathrm{mol} $$

Use the Henderson-Hasselbalch equation to find the ratio of A- to HA

Plugging in the given pH and the Ka we found earlier, we can solve for the ratio of A-/HA: $$ 4.68 = -\log(3.4 \times 10^{-4}) + \log \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]} $$ Solve for the ratio of A-/HA: $$ \mathrm{ratio}=\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}=10^{(4.68+\log(3.4 \times 10^{-4}))} = 1.912 $$

Calculate the moles of A- (F-) needed

Now that we have the ratio, we can determine the moles of F- needed: $$ \mathrm{moles \ of \ F}^{-} = \mathrm{moles \ of \ HNO}_{3} \times \mathrm{ratio} $$ $$ \mathrm{moles \ of \ F}^{-} = 0.0105 \, \mathrm{mol} \times 1.912 = 0.0201 \, \mathrm{mol} $$

Convert the moles of F- to grams of NaF

To find the grams of NaF, we first need to relate F- to NaF (1:1 stoichiometry). Then, use the molar mass of NaF to convert moles to grams: $$ \mathrm{grams \ of \ NaF} = \mathrm{moles \ of \ F}^{-} \times \mathrm{molar \ mass \ of \ NaF} $$ $$ \mathrm{grams \ of \ NaF} = 0.0201 \, \mathrm{mol} \times \frac{41.99 \, \mathrm{g}}{\mathrm{mol}} = 0.844 \, \mathrm{g} $$ To obtain a buffer with a pH of 4.68, 0.844 grams of NaF must be added to 70.00 mL of 0.150 M HNO3 solution.