Final answer:
The pH of the solution after adding 1.00 mL of 0.050 M HCl would be 1.45.
Step-by-step explanation:
To determine the pH of the solution after adding 1.00 mL of 0.050 M HCl, we need to consider the reaction between formic acid (HCOOH) and HCl. HCl is a strong acid, so it will completely react with formic acid to form formate ion (HCOO-) and water (H2O). The balanced equation for this reaction is:
HCOOH + HCl → HCOO- + H2O
Since the volume of the solution is almost the same after adding HCl, we can use the initial concentration of formic acid (0.100 M) and the volume of the solution (100 mL) to find the moles of formic acid before the reaction:
moles of HCOOH = concentration × volume = 0.100 M × 0.100 L = 0.010 mol
The moles of formate ion produced from the reaction will be equal to the moles of HCl added:
moles of HCOO- = 0.050 M × 0.001 L = 0.00005 mol
The moles of formic acid and formate ion in the solution after the reaction can be calculated by subtracting the moles of formate ion produced from the moles of formic acid before the reaction:
moles of HCOOH after reaction = 0.010 mol - 0.00005 mol = 0.00995 mol
Now, we can use the Henderson-Hasselbalch equation to calculate the pH of the solution after the reaction:
pH = pKa + log([A-]/[HA])
where [A-] is the concentration of the conjugate base (formate ion) and [HA] is the concentration of the weak acid (formic acid).
Substituting the values into the equation, we get:
pH = 3.75 + log(0.00005 mol / 0.00995 mol)
pH = 3.75 + log(0.005) = 3.75 + (-2.30) = 1.45
Therefore, the pH of the solution after adding 1.00 mL of 0.050 M HCl would be 1.45.