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# The pH of buffer of $N{{H}_{4}}OH+N{{H}_{4}}Cl$ type is given by:[A] $pH-p{{K}_{b}}$[B] $pH=\dfrac{1}{2}{{K}_{b}}-\log \dfrac{\left[ salt \right]}{\left[ base \right]}$[C] $pH=14-p{{K}_{b}}-\log \dfrac{\left[ salt \right]}{\left[ base \right]}$[D] $pH=pOH-p{{K}_{b}}+\log \dfrac{\left[ salt \right]}{\left[ base \right]}$  Verified
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Hint: pH of a solution is the negative log of the concentration of hydrogen ions present in the solution. To find the acidity of a basic buffer, we have to find the pOH of the buffer and then put it in the equation: pH + pOH=14 to get the correct solution.

Generally, we define buffer solution as a solution which is resistant to change in pH upon addition of a small amount of base or acid. Buffer solution usually consists of a weak acid and salt of its conjugate base or of a weak base and a salt of its conjugate acid.

Here, we have$N{{H}_{4}}OH+N{{H}_{4}}Cl$ which is a basic buffer as $N{{H}_{4}}OH$ is a weak base and $N{{H}_{4}}Cl$ is a salt of its conjugate acid which is $N{{H}_{4}}^{+}$.
The pH of a buffer solution is calculated by the Henderson-Hasselbalch equation. We can derive this equation as follows.

The weak base will dissociate as-
$N{{H}_{4}}OH\rightleftharpoons N{{H}_{4}}^{+}+O{{H}^{-}}$

Therefore, we can write the base dissociation constant${{K}_{b}}$as-
${{K}_{b}}=\dfrac{\left[ N{{H}_{4}}^{+} \right]\left[ O{{H}^{-}} \right]}{\left[ N{{H}_{4}}OH \right]}$, which is the product of concentration of the dissociated ions divided by the concentration of the base.
From the above equation, we can write that$\left[ O{{H}^{-}} \right]=\dfrac{{{K}_{b}}\left[ N{{H}_{4}}OH \right]}{\left[ N{{H}_{4}}^{+} \right]}$
Taking negative log on both sides, we will get
$-\log \left[ O{{H}^{-}} \right]=-\log \left\{ \dfrac{{{K}_{b}}\left[ N{{H}_{4}}OH \right]}{\left[ N{{H}_{4}}^{+} \right]} \right\}$
We know that,$pH=-\log \left[ {{H}^{+}} \right]$
Therefore, we can write that $pOH=-\log \left[ O{{H}^{-}} \right]$
Therefore, the equation becomes $pOH=-\log {{K}_{b}}-\log \left\{ \dfrac{\left[ N{{H}_{4}}OH \right]}{\left[ N{{H}_{4}}^{+} \right]} \right\}$
Or, we can rewrite this as $pOH=p{{K}_{b}}+\log \left\{ \dfrac{\left[ N{{H}_{4}}^{+} \right]}{\left[ N{{H}_{4}}OH \right]} \right\}$
We know that$pH=14-pOH$
Therefore, in the equation we can put the value of pOH to get the pH as-
$pH=14-\left( p{{K}_{b}}+\log \left\{ \dfrac{\left[ N{{H}_{4}}^{+} \right]}{\left[ N{{H}_{4}}OH \right]} \right\} \right)$
Or,$pH=14-p{{K}_{b}}-\log \left\{ \dfrac{\left[ N{{H}_{4}}^{+} \right]}{\left[ N{{H}_{4}}OH \right]} \right\}$
In the above equation, $N{{H}_{4}}^{+}$is the salt and$N{{H}_{4}}OH$is the base. Therefore, we can

write the equation as-
$pH=14-p{{K}_{b}}-\log \left\{ \dfrac{\left[ salt \right]}{\left[ base \right]} \right\}$
Therefore, the correct answer is option [C] $pH=14-p{{K}_{b}}-\log \dfrac{\left[ salt \right]}{\left[ base \right]}$.

Note: Here, we cannot calculate the acidity directly as the given buffer is a basic buffer. The pH of a buffer solution depends only on the ratio of concentration of the salt and base (or salt and acid in case of acidic buffer) and not on the individual concentrations of the salt and the base. If we make the solution dilute the ratio will remain unchanged thus, the pH will also remain unchanged. The pH will change only on the addition of ${{H}^{+}}$ or $O{{H}^{-}}$ ions.