banner

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.

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.

5/5 - (16 votes)