# Calculate the ratio of NaF to HF required to create a buffer with pH =4.20.

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.

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.

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