Question

How do you use the Henderson-Hasselbalch to calculate the $pH$ of a buffer solution that is $.50{\text{ }}M$ in $N{H_3}$ and $.20{\text{ }}M$ in $N{H_4}Cl$?

Answer 1

Buffer Solution is a water solvent-based solution which consists of a blend containing a weak acid and the form base of the weak acid, or a feeble base and the form acid of the frail base. They resist an adjustment in $pH$ upon weakening or upon the expansion of small amounts of $acid/soluble{\text{ }}base$ to them. That buffer solution has a $pH$ of $9.65$.

Complete step by step answer: Before I present Henderson-Hasselbalch's equation, we should recognize the acid and base. Alkali $\left( {N{H_3}} \right)$ is always a base and the ammonium particle $\left( {N{H_4}^ + } \right)$ is the form acid of smelling salts. A form acid has one more proton $\left( {{H^ + }} \right)$ than the base you started with.
Presently, we can use this equation:
$pH = p{K_a} + \log \dfrac{{[base]}}{{[acid]}}$
As you can see, we are given a $p{K_b}$ instead of a $p{K_a}$. Yet, no worries we can use the accompanying equation that relates the two constants to one another:
$p{K_b} + p{K_a} = 14$
We can solve for the $p{K_a}$ by subtracting the given $p{K_b}$ from$14$:
$14 - 4.75 = 9.25$
Thus, your $p{K_a}$ is $9.25$
Next, we can acquire the $\left[ {base} \right]$ and $\left[ {acid} \right]$ from the question.
$\left[ {N{H_3}} \right] = {\text{ }}.50{\text{ }}M{\text{ }}\left[ {N{H_4}} \right]{\text{ }} = .20{\text{ }}M$
We're not actually worried about the chloride anion that is connected to the ammonium particle because it's a spectator particle and it has no impact on the buffer system.
Presently, we have the entirety of the data to decide the $pH$. How about we plug our values into the equation:
$pH = 9.25 + log\left( {0.500.20} \right)$
$pH = 9.65$

Note:
The $pH$ of Buffer Solutions shows insignificant change upon the expansion of an extremely small amount of strong acid or strong base. They are consequently used to keep the $pH$ at a constant worth.