Answer:
The pH of the buffer solution is 4.60.
Explanation:
Concentration of acid = [tex][HC_2H_3O_2]=0.225 M[/tex]
Concentration of salt = [tex][KC_2H_3O_2]=0.162 M[/tex]
Dissociation constant = [tex] K_a=1.8 \times 10^{-5}[/tex]
The pH of the buffer can be determined by Henderson-Hasselbalch equation:
[tex]pH=pK_a+\log\frac{[salt]}{[acid]}[/tex]
[tex]pH=-\log[1.8 \times 10^{-5}]+\log\frac{0.162 M}{0.225 M}[/tex]
pH = 4.60
The pH of the buffer solution is 4.60.
4.602
Given:
A buffer system consisting of 0.225 M HC₂H₃O₂ and 0.162 M KC₂H₃O₂.
The Ka for HC₂H₃O₂ is 1.8 x 10⁻⁵.
Question:
Calculate the pH of this buffer.
The Process:
Let us first observe the ionization reaction of the KC₂H₃O₂ salt below.
[tex]\boxed{ \ KC_2H_3O_2 \rightleftharpoons K^+ + C_2H_3O_2^- \ }[/tex]
To calculate the specific pH of a given buffer, we need using The Henderson-Hasselbalch equation for acidic buffers:
[tex]\boxed{ \ pH = pK_a + log\frac{[A^-]}{[HA]} \ }[/tex]
where,
[tex]\boxed{ \ pH = pK_a + log\frac{[C_2H_3O_2^-]}{[HC_2H_3O_2]} \ }[/tex]
[tex]\boxed{ \ pH = -log(1.8 \times 10^{-5}) + log\frac{[0.162]}{[0.225]} \ }[/tex]
[tex]\boxed{ \ pH = 5-log \ 1.8 - 0.1427 \ }[/tex]
[tex]\boxed{ \ pH = 5 - 0.2553 - 0.1427 \ }[/tex]
[tex]\boxed{ \ pH = 4.602 \ }[/tex]
Thus, the pH of this buffer equal to 4.602.
