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

PH is the test of acidity or basicity of a solution. it follows the formula: 
pH = pKa + log [salt] / [acid] where NaF is the salt and HF is the acid in this case. 

By literature, Ka of HF is 3.5*10^-4 
pKa= -log(Ka)= 3.46 

pH = pKa + log [NaF / [HF] 

4.05 = 3.46 + log [NaF / [HF] 

log [NaF / [HF] = 0.59

[NaF / [HF] = 3.89