Question

Which of the following represents the correct relation for an acidic buffer at ${25}^{\circ }C?$

• A. $pH=p{K}_a+log\left(\frac{\left[\text{Anion of salt}\right]}{\left[Acid\right]}\right)$
• B. $pH=p{K}_b+log\left(\frac{\left[\text{Cation of salt}\right]}{\left[Acid\right]}\right)$
• C. $pOH=p{K}_a+log\left(\frac{\left[\text{Anion of salt}\right]}{\left[Acid\right]}\right)$
• D. $pH=p{K}_a+log\left(\frac{\left[\text{Cation of salt}\right]}{\left[Acid\right]}\right)$

Answer 1

The correct option is A $pH=p{K}_a+log\left(\frac{\left[\text{Anion of salt}\right]}{\left[Acid\right]}\right)$

$\left[WA\right]>\left[SB\right]$ is a buffer solution.
$C{H}_{3}COOH\left(aq\right)+NaOH\left(aq\right)\to C{H}_{3}COONa\left(aq\right)+H_{2}O\left(l\right)$
$43--$
$1033$
This reaction has weak acid $C{H}_{3}COOH\left(aq\right)$ and the salt (conjugate base) $C{H}_{3}COONa\left(aq\right)$ left in the solution with same anion $C{H}_{3}CO{O}^{-}\left(aq\right)$.
We have 3 for $C{H}_{3}COONa$
The reaction here is $C{H}_{3}COONa\left(aq\right)\to C{H}_{3}CO{O}^{-}\left(aq\right)+N{a}^{+}\left(aq\right)$
$3--$
$-33$
We have 1 for $C{H}_{3}COOH$
The reaction here is $C{H}_{3}COOH\left(aq\right)\rightleftharpoons C{H}_{3}CO{O}^{-}\left(aq\right)+H^{+}\left(aq\right)$
$1(1-\alpha )3+\alpha \alpha$
$-3$
This is a weak dissociation is seen with high concentration at the products and the reaction is pushed backward.
As $C{H}_{3}CO{O}^{-}\left(aq\right)$ is more on the products side dissociation is less and $1-\alpha \approx 1$. Therefore $\alpha$ can be ignored.
$\therefore$ By addition of a small amount of acid or base there is no significant change in the $pH$ value as the equilibrium is maintained.
$C{H}_{3}COOH\left(aq\right)\rightleftharpoons C{H}_{3}CO{O}^{-}\left(aq\right)+H^{+}\left(aq\right)$
$K_a=\frac{\left[H^{+}\right]\left[X^{-}\right]}{\left[HX\right]}$
$K_a$ is constant.
$\left[X^{-}\right]=C{H}_{3}CO{O}^{-}\left(aq\right)$
$\left[HX\right]=C{H}_{3}COOH\left(aq\right)$
$\left[X^{-}\right]$ and $\left[HX\right]$ almost constant and $\alpha$ can be ignored.
So, this is a buffer solution.
$K_a=\frac{\left[H^{+}\right]\left[X^{-}\right]}{\left[HX\right]}$
On rearranging $\frac{K_a\times \left[HX\right]}{\left[X^{-}\right]}=\left[H^{+}\right]$
Taking logarithm on both sides, we get
$log\left[H^{+}\right]=log{K}_a+log\left[HX\right]-log\left[X^{-}\right]$
Multiplying both sides by (-1)
$-log\left[H^{+}\right]=-log{K}_a-log\left[HX\right]+log\left[X^{-}\right]$
$pH=p{K}_a+log\frac{\left[X^{-}\right]}{\left[HX\right]}$
$\frac{\left[X^{-}\right]}{\left[HX\right]}$ this quantity is the ratio of concentration of the anion of the salt and the acid present in the mixture.
Therefore $pH=p{K}_a+log\left(\frac{\left[\text{Anion of salt}\right]}{\left[Acid\right]}\right)$