This question aims to find the ratio of **Sodium Fluoride** (NaF) to **Hydrogen Fluoride** (HF) that is used to create a buffer having pH 4.20 .

The **pH of a solution** determines whether a solution is **basic or acidic.** pH is measured by a pH scale which ranges from 0-14. A solution giving a pH reading of 7 is considered neutral whereas a solution giving a pH greater than 7 is considered a basic solution. Similarly, a solution having a pH less than 7 is considered an acidic solution. **Water** has a pH of 7.

A **buffer solution** is a solution that **resists** the pH from changing. If a small concentration of an acid or base is added to the solution, it helps to maintain the pH of the solution. Buffer solution consists of a **weak acid** and its **conjugate base** or a weak base or its conjugate acid.

## Expert Answer

To derive the expression for the given data:

\[ pH = pK_a + log \frac {[F]} {[HF]} \]

\[ pH = pK_a + log \frac {[NaF]}{[HF]}\]

\[ pH – pK_a = log \frac{[NaF]}{[HF]}\]

Taking **anti-log** on both sides of the expression:

\[ 10 ^ {pH} – pK_a = \frac {[NaF]}{[HF]} \]

This ratio of $ NaF $ to $ HF $ can be found by further simplification of the above mentioned expression:

\[ \frac {[NaF]}{[HF]} = 10 ^ {pH} – pK_a \]

\[ = 10 ^{{pH} – ( – log K_a )} \]

\[ = 10^{{pH} + log K_a } \]

## Numerical Solution

By putting values of $ pH $ and $ K_a $ for $ HF $ is $3.5 \times 10 ^{-4}$ :

\[ = 10 ^{{4.20} + log(3.5 \times 10 ^{-4})}\]

\[ \frac{[NaF]}{[HF]} = 5.5 \]

**The ratio of $ NaF $ to $ HF $ is $ 3.5 $ when a buffer solution having $ pH $ of $ 4.0 $ is used.**

## Example

Consider the $pH$ of the **buffer solution is $4.0$.** Calculate the ratio of $NaF$ to $HF$ required to make this buffer solution.

\[ pH = pK_a + log \frac { [F] } { [HF] } \]

\[pH = pK_a + log \frac{ [NaF] } { [HF] } \]

\[pH – pK_a = log \frac{ [NaF] } { [HF] } \]

\[10 ^ {pH} – pK_a = \frac{ [NaF] } { [HF] } \]

This ratio of $NaF$ to $HF$ can be found by:

\[\frac { [NaF] } { [HF] } = 10 ^ {pH} – pK_a \]

\[= 10 ^ {{pH} – (- log K_a ) } \]

\[= 10 ^ {{pH} + log K_a } \]

By putting values:

\[ =10 ^ {{4.20} + log (3.5 \times 10 ^{-4)}}\]

\[ \frac{[NaF]}{[HF]} = 3.5 \]

**The ratio of $NaF$ to $HF$ is $3.5$ when a buffer solution having $pH$ of $4.0$ is used.**

*Image/Mathematical drawings are created in Geogebra.*