Question

One liter of a buffer contains 40g of $N{{H}_{4}}Cl$ and 20g of $N{{H}_{3}}$. Calculate the pH of the solution. ${{K}_{b}}(N{{H}_{3}})=1.84\times {{10}^{-5}}$ at 298 K.

Answer 1

Hint: The pOH of a solution is the negative logarithm of the concentration of hydroxide-ion. In the pOH scale, 7 indicates a neutral solution, pOH less than 7 indicates a basic solution, and the pOH greater than 7 indicates an acidic solution.

Complete step by step answer:
-Ammonium chloride shows acidic behavior when dissolved in acidic. Ammonium chloride when dissolved in water liberates $N{{H}_{4}}^{+}$and $C{{l}^{-}}$ ions. Ammonium ion combines with water to form ammonium hydroxide and hydrogen cations. Ammonium hydroxide is a weak base that remains almost unionized in solution. As a result of an increase in the hydrogen, cations make the solution acidic.
 -pOH is calculated based on pH value or the hydrogen ion concentration. As we know, hydrogen ion concentration and hydroxide ion concentration are interrelated, so pOH can be calculated using both the concentrations.

$pOH=-\log [O{{H}^{-}}]$
$[O{{H}^{-}}]=\dfrac{{{K}_{W}}}{[{{H}^{+}}]}$ ; $p{{K}_{W}}=pH+pOH$ and also, $p{{K}_{W}}=p{{K}_{A}}+p{{K}_{B}}$
Where $p{{K}_{A}}$ is the acidity constant and $p{{K}_{B}}$ is the base constant.
Since our question is concerned with $N{{H}_{4}}Cl$,
$\Rightarrow pOH=p{{K}_{b}}+\log \dfrac{[Salt]}{[Base]}$

-Calculating the concentration of the salt,
$[Salt]=\dfrac{40}{53.5}=0.747moles$

-Calculating the concentration of the base,
$[Base]=\dfrac{20}{17}=1.176moles$
$pOH=5-\log (1.85)+\log \dfrac{0.747}{1.176}$
$\Rightarrow pOH=4.548$
-Since, $pH=-\log [{{H}^{+}}]$
And from the definition of pOH, $pOH=-\log [O{{H}^{-}}]$
-At $25{}^\circ C$ , Ph + pOH = 14
$\Rightarrow pH=14-pOH=14-4.548$
Therefore the pH of the solution is 9.452.

Note: ${{K}_{A}},{{K}_{B,}}p{{K}_{A}}\text{ and }p{{K}_{B}}$ are the terms used in predicting whether a solution will donate or accept protons at a specific pH value. \[{{K}_{B}}\text{ and }p{{K}_{B}}\] are related to bases while ${{K}_{A}}\text{ and }p{{K}_{A}}$ are related to bases. ${{K}_{A}}\text{ and }p{{K}_{A}}$ accounts for the concentration of hydrogen ion, while \[{{K}_{B}}\text{ and }p{{K}_{B}}\] accounts for the concentration of hydroxide ion. ${{K}_{A}}$ is the acid dissociation constant while ${{K}_{B}}$ is the base dissociation constant. $p{{K}_{A}}$ is the negative logarithm of the acid dissociation constant while $p{{K}_{B}}$ is the negative logarithm of the base dissociation constant. The acid and base dissociation constants are expressed in $mol/L$ . A large value of${{K}_{A}}$indicates a strong acid because it means that the acid is readily dissociated into ions. The smaller value of $p{{K}_{A}}$ indicated the stronger acid. A larger value of ${{K}_{B}}$ indicates the high level of dissociation of a strong base. A lower value of $p{{K}_{B}}$ indicates a stronger base.