In this problem, we need to find the pH of a buffer (pH = 4.45) that is produced by the reaction of strong acids as HBr that reacts with weak bases as NaC₇H₅O₂ producing its weak acid and a salt, as follows:
HBr + NaC₇H₅O₂ → HC₇H₅O₂ + NaBr
If in the reaction, the limiting reactant is the strong acid, we will produce a buffer (The aqueous mixture of a weak acid and its strong base).
Using Henderson-Hasselbalch equation, we can find the pH of this buffer:
[tex]pH = pKa + log \frac{[NaC_7H_5O_2]}{[HC_7H_5O_2]}[/tex]
Where:
pKa is -log Ka = 4.20
[] could be taken as the moles of each species
After the reaction, the moles of HBr = Moles of HC₇H₅O₂ and the remaining moles of NaC₇H₅O₂ = Initial moles of NaC₇H₅O₂ - Moles of HBr.
Moles HBr:
22.5mL = 0.0225L * (0.200mol/L) = 4.5x10⁻³ moles HBr
Moles NaC₇H₅O₂:
50.0mL = 0.0500L * (0.250mol/L) = 0.0125 moles NaC₇H₅O₂
That means the moles after the reaction of the species of the buffer are:
Moles HC₇H₅O₂ = Moles HBr = 4.5x10⁻³ moles
Moles NaC₇H₅O₂ = 0.0125 moles - 4.5x10⁻³ moles = 8.0x10⁻³ moles
Replacing in H-H equation:
[tex]pH = 4.20+ log \frac{[8.0x10^{-3}]}{[4.5x10^{-3}]}[/tex]
pH = 4.45
Learn more about Henderson-Hasselbalch equation in:
https://brainly.com/question/13423434
