Question

A $1L$ aqueous solution is made by mixing $N{H}_{4}OH$ and $HCl$ solution at ${25}^{\circ }C$.
This can act as a basic buffer when :

• A. Molar concentration of weak base is more than the molar concentration of strong acid i.e $\left[N{H}_{4}OH\right]>\left[HCl\right]$
• B. Molar concentration of weak base is less than the molar concentration of strong acid i.e $\left[N{H}_{4}OH\right]<\left[HCl\right]$
• C. Molar concentration of weak base is equal to the molar concentration of strong acid i.e $\left[N{H}_{4}OH\right]=\left[HCl\right]$
• D. A basic buffer is always formed by $N{H}_{4}OH$ and $HCl$

Answer 1

The correct option is A Molar concentration of weak base is more than the molar concentration of strong acid i.e $\left[N{H}_{4}OH\right]>\left[HCl\right]$

Weak Base (WB) and Strong Acid (SA)
Basic buffer (pH >7) : It is the mixture solution of a weak base and a salt of this weak base with a strong acid.
Consider the combination of $N{H}_{4}OH$ and $HCl$
Condition-1 where$\left[WB\right]>\left[SA\right]$
$N{H}_{4}OH\left(aq\right)+HCl\left(aq\right)\to N{H}_{4}Cl\left(aq\right)+H_{2}O\left(l\right)$
$43--$
$1033$

As $1mol$ of $N{H}_{4}OH$ is left,
The reaction here is $N{H}_{4}OH\left(aq\right)\rightleftharpoons N{H}_4^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$
$1(1-\alpha )\alpha +31\alpha$
$\approx 3$
$\alpha$ is too small as compared to the concentrations of $\left[N{H}_4^{+}\right]$ and $\left[N{H}_{4}OH\right]$
We get $K_b=\frac{\left[N{H}_4^{+}\right]\left[O{H}^{-}\right]}{\left[N{H}_{4}OH\right]}$
$\left[O{H}^{-}\right]=\frac{K_b\left[N{H}_{4}OH\right]}{\left[N{H}_4^{+}\right]}$
$\therefore$ By addition of a small amount of $H^{+}$ ions equilibrium will move in forward direction by the same amount and same number of $O{H}^{-}$ ions are produced. Hence there is no significant change in the $pH$ is observed.
This acts as a buffer solution.

Condition-2 where$\left[WB\right]=\left[SA\right]$
$N{H}_{4}OH\left(aq\right)+HCl\left(aq\right)\to N{H}_{4}Cl\left(aq\right)+H_{2}O\left(l\right)$
$33--$
$0033$
(Hydrolysis occurs here as there is complete consumption of acid and base as a result salt is only left in solution. This is not a buffer)

Condition-3 where $\left[WB\right]<\left[SA\right]$
$N{H}_{4}OH\left(aq\right)+HCl\left(aq\right)\to N{H}_{4}Cl\left(aq\right)+H_{2}O\left(l\right)$
$34--$
$0133$
Hydrolysis of $N{H}_{4}Cl$ can be neglected as degree of hydrolysis was already small and common ion effect due to $H^{+}$ coming from $HCl$ will further decrease it.
So, The $[H^{+}{]}_{total}$ can be assumed to be from $HCl$ only and$pH$ can be directly found from this $\left[H^{+}\right]$.
This is not buffer solution