Calculate the ratio of NaF to HF required to create a buffer with pH=4.15.

The main objective of this question is to calculate the ratio $NaF$ to $HF$ required to create a buffer with a given $pH$.

A buffer is an aqueous solution that sustains noticeable variation in $pH$ levels when a small amount of acid or alkali is added, which is made up of a weak acid and its conjugate base, or vice versa. When the solutions are mixed with a strong acid or base, a rapid change in $pH$ can be observed. A buffer solution then facilitates neutralizing some of the added acid or base, allowing the $pH$ to change more progressively.

Each buffer has a fixed capacity, which is defined as the amount of strong acid or base required to change the $pH$ of $1$ liter of the solution by $1$ $pH$ unit. Alternatively, buffer capacity is the amount of acid or base that can be added before the $pH$ significantly changes.

Buffer solutions can neutralize up to a certain limit. Once the buffer has reached its capacity, the solution will behave as if there is no buffer present, and the $pH$ will begin to fluctuate substantially again. The Henderson-Hasselbalch equation is used to estimate the $pH$ of a buffer.

Expert Answer

Now, using the Henderson-Hasselbalch equation:

$pH=p{K}_a+\log {\displaystyle \frac{[F]}{[HF]}}$

$pH=p{K}_a+\log {\displaystyle \frac{[NaF]}{[HF]}}$

$pH-p{K}_a=\log {\displaystyle \frac{[NaF]}{[HF]}}$

$\log ({10}^{(pH-p{K}_a)})=\log {\displaystyle \frac{[NaF]}{[HF]}}$

Applying anti-log on both sides, we get:

${10}^{(pH-p{K}_a)}={\displaystyle \frac{[NaF]}{[HF]}}$

Since $p{K}_a=-\log K_a$, so:

${\displaystyle \frac{[NaF]}{[HF]}}={10}^{pH-(-\log K_a)}$

${\displaystyle \frac{[NaF]}{[HF]}}={10}^{pH+\log K_a}$

${\displaystyle \frac{[NaF]}{[HF]}}={10}^{4.00+\log (3.5\times {10}^{-4})}$

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

Example 1

Suppose that there is a solution of $3M$ $HCN$. Find the concentration of $NaCN$ needed in order for $pH$ to be $8.3$, provided that the $K_a$ for $HCN$ is $4.5\times {10}^{-9}$.

Solution

Using the Henderson-Hasselbalch equation, we get:

$pH=p{K}_a+\log {\displaystyle \frac{[C{N}^{-}]}{[HCN]}}$

$8.3=p{K}_a+\log {\displaystyle \frac{[C{N}^{-}]}{[HCN]}}$

Since, $K_a$ of $HCN$ is $4.5\times {10}^{-9}$, so $p{K}_a$ of $HCN$ will be

$p{K}_a=-\log (4.5\times {10}^{-9})=8.3$

So, we will have the above equation as:

$8.3=8.3+\log {\displaystyle \frac{[C{N}^{-}]}{[HCN]}}$

or  $\log {\displaystyle \frac{[C{N}^{-}]}{[HCN]}}=0$

It is given that $HCN=3M$, therefore:

$\log {\displaystyle \frac{[C{N}^{-}]}{[3]}}=0$

${\displaystyle \frac{[C{N}^{-}]}{[3]}}=1$

$[C{N}^{-}]=3M$

Consequently, a concentration of $3M$ $NaCN$ allows the $pH$ of the solution to be $8.3$.

Example 2

Find the ratio of conjugate base to acid, if the solution of acetic acid has a $pH$ of $7.65$ and $p{K}_a=4.65$.

Solution

Since, $pH=p{K}_a+\log {\displaystyle \frac{[A^{-}]}{[HA]}}$

Substituting the given data:

$7.65=4.65+\log {\displaystyle \frac{[A^{-}]}{[HA]}}$

$\log {\displaystyle \frac{[A^{-}]}{[HA]}}=3$

${\displaystyle \frac{[A^{-}]}{[HA]}}={10}^3=1000$

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