Question

How will you write the Henderson-Hasselbalch equation for a solution of propanoic acid $(C{{H}_{3}}C{{H}_{2}}COOH)\,pKa=4.874$using $H{{A}^{-}},\,{{A}^{-}}$and the given $pKa$value in the expression.

Answer 1

Hint: Propanoic acid is a weak acid. As per the Henderson-Hasselbalch equation $pH$of an acidic buffer consisting of a weak acid and its salt depends on the $pKa$of acid and concentration of salt and acid in the concentration. The general form of the equation is:
 $pH=\,pKa+\log \dfrac{\left[ salt \right]}{\left[ acid \right]}$ ….. $(i)$

Complete step-by-step answer:
As per the information given in the question, propanoic acid $(C{{H}_{3}}C{{H}_{2}}COOH)$is represented by $HA$, $C{{H}_{3}}CO{{O}^{-}}$is represented by ${{A}^{-}}$
When propanoic acid $(HA)$is dissolved in water, it will undergo only partial dissociation as it is a weak acid.
The equation of its dissociation is represented as;
$HA(aq)\,\overset{{}}{leftrightarrows}\,{{H}^{+}}(aq)\,+\,{{A}^{-}}(aq)$
The equilibrium constant for the above reaction is represented by;
$Ka=\,\dfrac{\left[ {{H}^{+}} \right]\,\left[ {{A}^{-}} \right]}{\left[ HA \right]}$
On cross multiplying the above equation we get;
$\left[ {{H}^{+}} \right]=\,\dfrac{Ka\,\left[ HA \right]}{\left[ {{A}^{-}} \right]}$
Taking the negative logarithm on both the sides we get,
$-\log \left[ {{H}^{+}} \right]\,=\,\,-\log Ka\,-\,\log \dfrac{\left[ HA \right]}{\left[ {{A}^{-}} \right]}$
Now we know that $pH\,=\,-\log \left( {{H}^{+}} \right)$and $pKa=\,-\log \left( Ka \right)$
$
 \Rightarrow \,\,\,\,\,pH=\,pKa\,-\,\log \,\dfrac{\left[ HA \right]}{\left[ {{A}^{-}} \right]} \\
 \Rightarrow \,\,\,\,pH\,=\,pKa\,+\,\log \,\,\dfrac{\left[ {{A}^{-}} \right]}{\left[ HA \right]} \\
$
Now in the question the value of $pKa=\,4.874$
Substituting the value of $pKa=4.874$in the above equation we get;
$pH\,=\,4.874\,+\,\log \,\dfrac{\left[ {{A}^{-}} \right]}{\left[ HA \right]}$ ….. $\left( ii \right)$
Hence this is the required expression.

Note: It may be noted that the concentration of acetate ion $\left[ {{A}^{-}} \right]$in equation $(ii)$is taken to be almost equal to the concentration of the salt in equation $(i)$ because the acetate ions coming from fully dissociated salt suppress the ionization of the weak acid. If concentration of salt is equal to that of the acid, then in such case $pH=pKa$. While deriving the above expression always be careful while performing the logarithms calculations and logarithms formulas of addition and subtraction.