Calculate the mass of NaF that must be added to 300ml if a 0.25m HF solution to form a buffer solution with a ph of 3.5

6.574 g NaF into 300ml (0.25M HF) => Bfr with pH ~3.5

Explanation:

For buffer solution to have a pH-value of 3.5 the hydronium ion concentration [H⁺] must be 3.16 x 10⁻⁴M ( => [H⁺] = 10^-pH = 10⁻³°⁵ =3.16 x 10⁻⁴M).

Addition of NaF to 300ml of 0.25M HF gives a buffer solution. To determine mass of NaF needed use common ion analysis for HF/NaF and calculate molarity of NaF, then moles in 300ml the x formula wt => mass needed for 3.5 pH.

HF ⇄ H⁺ + F⁻; Ka = 6.6 x 10⁻⁴

Ka = [H⁺][F⁻]/[HF] = 6.6 x 10⁻⁴ = (3.16 x 10⁻⁴)[F⁻]/0.25 => [F⁻] = (6.6 x 10⁻⁴)(0.25)/(3.16x10⁻⁴) = 5.218M in F⁻ needed ( = NaF needed).

For the 300ml buffer solution, moles of NaF needed = Molarity x Volume(L)

= (5.218M)(0.300L) = 0.157 mole NaF needed x 42 g/mole = 6.574 g NaF needed.

Check using the Henderson - Hasselbalch Equation...

pH = pKa + log ([Base]/[Acid]); pKa (HF) = 3.18

Molarity of NaF = (6.572g/42g/mole)/(0.300 L soln) = 0.572M in NaF = 0.572M in F⁻.

pH = 3.18 + log ([0.572]/[0.25]) ≅ 3.5.

One can also back calculate through the Henderson -Hasselbalch Equation to determine base concentration, moles NaF then grams NaF.

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explanation:

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