Question

The ${\text{p}}{{\text{K}}_{\text{a}}}$ of certain weak acid is $4.0$. What should be the salt to acid ratio if we have to prepare a buffer with ${\text{pH = 5}}$ using the acid and one of the salts?
A. $4:5$
B. $5:4$
C. $10:1$
D. $1:10$

Answer 1

Acid dissociation constant, ${{\text{K}}_{\text{a}}}$ is a quantitative measurement of the strength of an acid in solution. It is related to ${\text{pH}}$ and the salt to acid ratio with the help of an equation called Henderson-Hesselbach equation.

Complete step by step solution:
Consider the dissociation of a compound ${\text{HA}}$.
${\text{HA}} \rightleftharpoons {{\text{H}}^ + } + {{\text{A}}^ - }$
The acid dissociation constant can be expressed as:
${{\text{K}}_{\text{a}}} = \dfrac{{\left[ {{{\text{A}}^ - }} \right]\left[ {{{\text{H}}^ + }} \right]}}{{\left[ {{\text{HA}}} \right]}}$
${\text{p}}{{\text{K}}_{\text{a}}}$ is used because using ${{\text{K}}_{\text{a}}}$ will be inconvenient practically. ${\text{p}}{{\text{K}}_{\text{a}}}$ is the negative logarithm of ${{\text{K}}_{\text{a}}}$.
If the value of ${\text{p}}{{\text{K}}_{\text{a}}}$ value is small, then the acid will be stronger.
When the equation is rearranged, we get
$\left[ {{{\text{H}}^ + }} \right] = \dfrac{{{{\text{K}}_{\text{a}}}\left[ {{\text{HA}}} \right]}}{{\left[ {{{\text{A}}^ - }} \right]}}$
Multiplying both sides with $- \log$, we get
$- \log \left[ {{{\text{H}}^ + }} \right] = - \log {{\text{K}}_{\text{a}}} - \log \dfrac{{\left[ {{\text{HA}}} \right]}}{{\left[ {{{\text{A}}^ - }} \right]}}$
i.e. ${\text{pH}} = {\text{p}}{{\text{K}}_{\text{a}}} - \log \dfrac{{\left[ {{\text{HA}}} \right]}}{{\left[ {{{\text{A}}^ - }} \right]}} \Leftrightarrow {\text{pH}} = {\text{p}}{{\text{K}}_{\text{a}}} + \log \dfrac{{\left[ {{{\text{A}}^ - }} \right]}}{{\left[ {{\text{HA}}} \right]}}$
The salt is also known as the conjugate base in the case of acid. For example, acetate is the conjugate base of acetic acid.
While ${\text{pH}}$ of the basic buffer solution can be expressed as:
${\text{pOH}} = {\text{p}}{{\text{K}}_{\text{b}}} + \log \dfrac{{\left[ {\text{s}} \right]}}{{\left[ {\text{b}} \right]}}$
The salt is known as the conjugate acid in the case of base. For example, ${\text{HC}}{{\text{O}}_3}^ -$ is a base and its conjugate acid is ${{\text{H}}_2}{\text{C}}{{\text{O}}_3}$.
Thus ${\text{pH}}$ of the acidic buffer solution can be expressed as:
${\text{pH}} = {\text{p}}{{\text{K}}_{\text{a}}} + \log \dfrac{{\left[ {\text{s}} \right]}}{{\left[ {\text{a}} \right]}}$
It is given that ${\text{p}}{{\text{K}}_{\text{a}}} = 4.0$ and ${\text{pH = 5}}$
Substituting the values, we get
$5 = 4 + \log \dfrac{{\left[ {\text{s}} \right]}}{{\left[ {\text{a}} \right]}}$
Simplifying, we get
$1 = \log \dfrac{{\left[ {\text{s}} \right]}}{{\left[ {\text{a}} \right]}} \Leftrightarrow \dfrac{{\left[ {\text{s}} \right]}}{{\left[ {\text{a}} \right]}} = \dfrac{{10}}{1}$
Thus the ratio will be $10:1$

Hence, the correct option is C.

Additional information:
In this equation, we assume that, at equilibrium, the concentration of the acid and its conjugate base will be equal. Also, this does not consider the dissociation of acid, dissociation of water and the hydrolysis of base.

Note: Buffer solution is a solution which maintains its ${\text{pH}}$ even small amounts of acids or bases are added. They are of two types-acidic buffer and basic buffer. Henderson-Hasselbalch equation explains about the method for estimating ${\text{pH}}$ of a buffer solution.