Question

Which of the following represents a correct relation for a basic buffer formed by $N{H}_{4}OH$ and $N{H}_{4}Cl$ ?

• A. $pOH=p{K}_b+log\left[\frac{\left[N{H}_4^{+}\right]}{\left[N{H}_{4}OH\right]}\right]$
  
• B. $pH=p{K}_b+log\left[\frac{\left[N{H}_4^{+}\right]}{\left[N{H}_{4}OH\right]}\right]$
  
  
• C. $pOH=p{K}_b+log\left[\frac{\left[N{H}_{4}OH\right]}{\left[N{H}_4^{+}\right]}\right]$
  
• D. $pH=p{K}_a+log\left[\frac{\left[N{H}_{4}OH\right]}{\left[N{H}_4^{+}\right]}\right]$

Answer 1

The correct option is A $pOH=p{K}_b+log\left[\frac{\left[N{H}_4^{+}\right]}{\left[N{H}_{4}OH\right]}\right]$


Basic buffers:
Solutions of a mixture of weak base and the salt of the conjugate acid of that base with a strong acid
For example:
Ammonium Hydroxide + Ammonium Chloride The salt's ammonium ion serves as the conjugate acid of weak base.
$N{H}_{4}OH\left(aq\right)+HCl\left(aq\right)\to N{H}_{4}Cl\left(aq\right)+H_{2}O\left(l\right)$
$43--$
$1033$
We have 3 for $\left[N{H}_{4}Cl\right]$
We have 1 for $\left[N{H}_{4}OH\right]$
The reaction here is $N{H}_{4}OH\left(aq\right)\rightleftharpoons N{H}_4^{+}\left(aq\right)+O{H}^{-}\left(aq\right))$
$1(1-\alpha )1\alpha +31\alpha$
$\approx 3$
$\alpha$ is too small as compared to the concentrations of $\left[N{H}_4^{+}\right]$ and $\left[N{H}_{4}OH\right]$ so for a small addition of acid solution $\left[O{H}^{-}\right]$ is not going to change. So pOH and pH remain the same.
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 $H^{+}$ there is no significant change in the $pH$ value as the equilibrium is moved backward and $H^{+}$ is reduced.

Taking logarithm and multiplying by (-1) on both sides, we get.
$log\left[O{H}^{-}\right]=-log{K}_b-log\left[N{H}_{4}OH\right]+log\left[N{H}_4^{+}\right]$
$pOH=p{K}_b+log\left[\frac{\left[N{H}_4^{+}\right]}{\left[N{H}_{4}OH\right]}\right]$

The quantity $\frac{\left[N{H}_4^{+}\right]}{\left[N{H}_{4}OH\right]}$ is the ratio of concentration of cation of the salt and the base present in the mixture.
$pOH=p{K}_b+log\left[\frac{\left[Cationofsalt\right]}{\left[Base\right]}\right]$