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

Hint: 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.