22.5 ml of a 0.200 m HBr solution was added to a 50.00 ml sample of 0.250 m sodium benzoate (NaC7H5O2). Calculate the pH. Benzoic acid has a Ka 6.30 x 10-5.

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:

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:

pH = 4.45

Learn more about Henderson-Hasselbalch equation in:

c is the answer because when something is left after something evaporates that shows the chemical change.