Answer:
pH = 3.97
Explanation:
It looks like HC₇H₅O₂ is benzoic acid C ₆ H ₅ C O O H
For short - hand I'll call this H A
Here we have to determine the pH of a buffer solution. The Henderson-Hasselbalch give us the pH through the equation:
pH = pKa + log ((A⁻) / (HA))
where pKa is the -log Ka,
(A⁻) is the concentration of the conjugate base of the weak acid
(HA) is the concentration of the weak acid
By adding the strong acid HCl to the buffer, some of the conjugate base of the weak acid will be consumed and some of the weak acid will produced through the chemical equation:
HCl + LiC₇H₅O₂ ⇒ HC₇H₅O₂ + LiCl
Therefore we need to account for this reaction to answer the question:
Vol HCl = 100 mL = 0.100 L
mol HCl reacted = 0.100 L x 1.00 mol/L = 0.100 mol
mol LiC₇H₅O₂ originally present = 0.250 mol/L x 1.00 L = 0.250 mol
mol LiC₇H₅O₂ after reaction with HCl = 0.250 -0.100 M = 0.250 mol
mol HC₇H₅O₂ originally present = 0.150 mol/L x 1.00 L = 0.150 mol
mol HC₇H₅O₂ produced in reaction = 0.100 mol
mol HC₇H₅O₂ present after reaction = 0.150 mol + 0.100 mol = 0.250 mol
Now we are in position to calculate the pH by plugging our values into the equation:
pH = pKa + log ((A⁻) / (HA)) = 4.20 + log (0.15/0.25) = 4.20 + log 0.60
= 3.97
pH = 3.97
