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A 1 L buffer solution of 0.25 m HF and 0.25 m NaF. The addition of 0.1 mol of NaOH can make the pH of solution 3.82 .
The buffer is made up of weak hydrofluoric acid (HF), sodium fluoride (NaF), and the fluoride anion (F), which is its conjugate base.
The weak acid and the strong base will neutralize each other when the sodium hydroxide solution is added, supposing there is no change in the buffer's overall volume.
More conjugate bases will be produced by this reaction.
The balanced chemical equation for the reaction between sodium hydroxide and hydrofluoric acid thus looks like this.
HF(aq) + NaOH(aq) ---> NaF(aq) + H₂O(l)
In the reation,1 mole of hydrofluoric acid will react with 1 mole of sodium hydroxide and produce 1 mole of sodium fluoride.
Determine how many moles you have in solution by multiplying the hydrofluoric acid's molarity by the volume of the buffer.
c = [tex]\frac{ n }V}[/tex]
n = c . V
n(HF) = 0.25 M × 1 L
n(HF) = 0.25 moles
n(F⁻) = 0.25 M × 1 L
n(F⁻) = 0.25 moles
1 mole of sodium hydroxide are now being added to the buffer. The sodium hydroxide will be completely consumed since there are fewer moles of strong base than weak acid.
The concentration of hydrofluoric acid will be changed from moles to
n(HF) = 0.25 - 0.1
n(HF) = 0.15 moles
The concentration of fluoride anions will be changed from moles to
n(F⁻) = 0.25 + 0.1
n(HF) = 0.35 moles
Calculate the new molarities of the weak acid and its conjugate base using the volume of the buffer.
[HF] = 0.15 moles / 1 L = 0.15 M
[F⁻] = 0.35 moles / 1 L = 0.35 M
To determine the pH of the buffer using the Henderson-Hasselbalch equation.
pH = pKₐ + log ([tex]\frac{ conjugate base }{ weak acid }[/tex])
Use the acid dissociation constant, Kₐ, to get the value of pKₐ
pKₐ = - log (Kₐ)
The pH of the solution will be,
pH = - log (Kₐ) + log ( [tex]\frac{[F]}{[HF]}[/tex] )
pH = - log ( 3.5 × 10⁻⁴ ) + log ( [tex]\frac{0.35}{0.15}[/tex] )
pH = 3.46 + 0.36
pH = 3.82
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