Question

Calculate the ratio of naf to hf required to create a buffer with ph = 4.05.

Answer 1

Answer:

Ratio of NaF to HF required to create a buffer with pH=4.15 is 8.13

Explanation:

HF is an weak acid and NaF is a strong electrolyte containing $F^{-}$ (Conjugate base of HF)

Hence mixture of NaF and HF creates a buffer systems.

Applying Henderson-Hasselbalch equation to this buffer system-

$pH=pK_{a}\left ( HF \right )+log(\frac{[F^{-}]}{[HF]})$

Here, species inside third bracket represents their concentrations.

It is given that pH of buffer is 4.05 and literature value suggests $pK_{a}$ of HF is 3.14

Plug-in these two values in the above equation-

$4.05=3.14+log\frac{[F^{-}]}{[HF]}$

$\Rightarrow \frac{[F^{-}]}{[HF]}=8.13$

As 1 molecule of NaF contains 1 molecule of $F^{-}$ therefore $\frac{[NaF]}{[HF]}=8.13$

Answer 2

A buffer system with a pH of 4.05 has a ratio of NaF to HF of 7.58:1.

A buffer system is formed by a weak acid (HF) and its conjugate base (F⁻ from NaF). Given the pKa of HF is 3.17, we can calculate the ratio of NaF to HF required to make a solution with a pH of 4.05 using the Henderson-Hasselbach equation.

$pH = pKa + log \frac{[NaF]}{[HF]} \\\\4.05 = 3.17 + log \frac{[NaF]}{[HF]} \\\\\frac{[NaF]}{[HF]} = 7.58$

A buffer system with a pH of 4.05 has a ratio of NaF to HF of 7.58:1.

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