Question

Consider an aqueous solution containing a weak acid $C{H}_{3}COOH$ and a strong base $NaOH$ at ${25}^{\circ }C$
This can act as an acidic buffer when :

• A. Molar concentration of weak acid is more than the concentration of strong base i.e $\left[C{H}_{3}COOH\right]>\left[NaOH\right]$
• B. Molar concentration of weak acid is less than the concentration of strong base i.e $\left[C{H}_{3}COOH\right]<\left[NaOH\right]$
• C. Molar concentration of weak acid is equal to the concentration of strong base i.e $\left[C{H}_{3}COOH\right]=\left[NaOH\right]$
• D. An acidic buffer is always formed by $C{H}_{3}COOH$ and $NaOH$

Answer 1

The correct option is A Molar concentration of weak acid is more than the concentration of strong base i.e $\left[C{H}_{3}COOH\right]>\left[NaOH\right]$

Theory:
Condition-1 where $\left[WA\right]>\left[SB\right]$
$C{H}_{3}COOH\left(aq\right)+NaOH\left(aq\right)\to C{H}_{3}COONa\left(aq\right)+H_{2}O\left(l\right)$
$43--$
$1033$
This can act as acidic buffer

Condition-2 where $\left[WA\right]=\left[SB\right]$
$C{H}_{3}COOH\left(aq\right)+NaOH\left(aq\right)\to C{H}_{3}COONa\left(aq\right)+H_{2}O\left(l\right)$
$33--$
$0033$
(Complete consumption of acid and base, this is salt hydrolysis case. This is not a buffer)

Condition-3 where $\left[WA\right]<\left[SB\right]$
$C{H}_{3}COOH\left(aq\right)+NaOH\left(aq\right)\to C{H}_{3}COONa\left(aq\right)+H_{2}O\left(l\right)$
$34--$
$0133$
(This is not a buffer as in this case due to high $O{H}^{-}$ we can assume that hydrolysis is not occurring. $\therefore \left(pOH\right)=-log\left[1\right]$

In condition-1 where $\left[WA\right]>\left[SB\right]$ is a buffer solution.
$C{H}_{3}COOH\left(aq\right)+NaOH\left(aq\right)\to C{H}_{3}COONa\left(aq\right)+H_{2}O\left(l\right)$
$43--$
$1033$
3 moles of $C{H}_{3}COONa$ are at equilibrium:
The reaction here is $C{H}_{3}COONa\left(aq\right)\to C{H}_{3}CO{O}^{-}\left(aq\right)+N{a}^{+}\left(aq\right)$
$3--$
$-33$
For, $C{H}_{3}COOH$
At equilibrium, $C{H}_{3}COOH\left(aq\right)\rightleftharpoons C{H}_{3}CO{O}^{-}\left(aq\right)+H^{+}\left(aq\right)$
$1(1-\alpha )3+\alpha \alpha$

Due to common ion effect, this equilibrium is pushed backward. Therefore $\alpha$ can be ignored.
So, $1-\alpha \approx 1$
$\therefore$ By addition of a small amount of $H^{+}$ there is no significant change in the $pH$ value as the equilibrium is moved backward and $H^{+}$ is reduced.
$K_a=\frac{\left[H^{+}\right]\left[X^{-}\right]}{HX}$
$K_a$ is constant.
$\left[X^{-}\right]$ and $\left[HX\right]$ almost constant as $\alpha$ can be ignored.
$\left[H^{+}\right]$ has no significant change.
$C{H}_{3}COOH\left(aq\right)\rightleftharpoons C{H}_{3}CO{O}^{-}\left(aq\right)+H^{+}\left(aq\right)$
In simple terms
If extra $H^{+}$ is added to buffer (from a strong acid)
$H^{+}\left(aq\right)+C{H}_{3}CO{O}^{-}\rightleftharpoons C{H}_{3}COOH\left(aq\right)$
Weak conjugate base $\left(C{H}_{3}CO{O}^{-}\right)$ neutralises the strong acid
If extra $O{H}^{-}$ is added to the buffer (from a strong base)
$O{H}^{-}\left(aq\right)+C{H}_{2}COOH\left(aq\right)\to C{H}_{3}CO{O}^{-}\left(aq\right)+H_{2}O$
Water is formed preventing large increase in $\left[O{H}^{-}\right]$