Question

The $p{K_a}$ of a weak acid is 4.8. What is the ratio of [salt] to [acid], if a buffer of pH=5.8 is to be prepared?

Answer 1

Hint: The letters in pH mean power of Hydrogen. pH of a solution indicates the measure of the molarity of hydrogen ions in a solution and thus we can say it is a direct measure of the alkalinity or acidity of a solution. The numerical value of pH is the negative of power of 10 of the molarity of H+ ions.

Complete answer:
The Henderson–Hasselbalch equation is generally used to determine the pH of a buffer solution and finding out the equilibrium pH in the acid-base reaction. According to the Henderson-Hasselbach equation, pH of buffer solution is calculated using the following formula:
$pH = p{K_a} + \log \dfrac{{\left[ {salt} \right]}}{{[acid]}}$
In the question, we are provided with the following information:
pH of buffer = 5.8 (Given)
pKa = 4.8 (Given)
$\dfrac{{[salt]}}{{[acid]}}$= ?
Now substituting the given values in the aforementioned formula:
$
\Rightarrow 5.8 = 4.8 + \log \dfrac{{\left[ {salt} \right]}}{{[acid]}} \\
\Rightarrow \log \dfrac{{\left[ {salt} \right]}}{{[acid]}} = 1.0 \\
\Rightarrow \dfrac{{\left[ {salt} \right]}}{{[acid]}} = \dfrac{{10}}{1} = 10:1 \\
$

Hence, the ratio of [salt] to [acid], if a buffer of pH=5.8 is to be prepared, is 10:1.

Note:
If you are provided with the molar concentrations of base and salt then you can use the following formula to calculate pH. And always remember that in case, the salt and base are having equal molarity then pH is directly equal to pKa.
$pH = p{K_a} - \log (\dfrac{{{b^ + }}}{{BOH}})$
or
$pH = (14 - p{K_b}) - \log (\dfrac{{{b^ + }}}{{BOH}})$
Here, $BOH$= molar concentration of weak base present in the solution, \[{b^ + }\]= molar concentration of base's anion in solution which is equal to the salt’s concentration.