Chemistry: A BUFFER IS PREPARED BY MIXING 15g OF HF AND 25g OF NAF IN A 125mL water. what is the pH of this buffer solution? show hydrolysis?

Molar mass of HF = (1.0 + 19.0) g/mol = 20.0 g/mol
Molar mass of NaF = (23.0 + 19.0) g/mol = 42.0 g/mol

Before the dissociation of HF :
[HF] = [(15 g) / (20.0 g/mol)] / (125/1000 L) = 6 M
[F⁻] = [NaF] = [(25 g) / (42.0 g/mol)] / (125/1000 L) = 4.8 M

Refer to: https://depts.washington.edu/eooptic/links/acidstrength.html
Ka for HF = 6.6 × 10⁻⁴

Consider the dissociation of HF :
____________ HF(aq) ___ ⇌ ___ F⁻(aq) ___ + ___ H⁺(aq) ___ Ka = 6.6 × 10⁻⁴
Initial: _______ 6 M ___________ 4.8 M _________ 0 M
Change: ______ -y M __________ +y M ________ +y M
At eqm: ____ (6 - y) M ______ (4.8 + y) M ________ y M

As Ka is very small and due to the common ion effect in the presence of F⁻ ions, the dissociation of HF would be to a very small extent.
It is assumed that 6 > 4.76 ≫ y, i.e.
[HF] at equilibrium = (6 - y) M ≈ 6 M
[F⁻] at equilibrium = (4.8 + y) M ≈ 4.8 M

Ka = [F⁻] [H⁺] / [HF]
6.6 × 10⁻⁴ = 4.8 × [H⁺] / 6
[H⁺] = (6.6 × 10⁻⁴) × (6/4.8) = 8.3 × 10⁻⁴
pH = -log[H⁺] = -log(8.3 × 10⁻⁴) = 3.1


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Alternative method :

pH = pKa + log([F⁻]/[HF]) = -log(6.6 × 10⁻⁴) + log(4.8/6) = 3.1