Question

Two buffers are prepared by adding an equal number of moles of formic acid (HCOOH) and sodium formate (HCOONA) to enough water to make 1.00 L of solution. Buffer A is prepared using 1.00 mol each of formic acid and sodium formate. Buffer B is prepared by using 0.010 mol of each. (Ka(HCOOH) = 1.8×10⁻⁴. What is the pH of buffer A?

Answer 1

The pH of buffer A, with 1.00 M concentrations of formic acid and sodium formate, is 3.74, calculated using the Henderson-Hasselbalch equation.

Step-by-step explanation:

The pH of buffer A, which contains equal moles of formic acid (HCOOH) and sodium formate (HCOONa), can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the base form (sodium formate) and [HA] is the concentration of the acid form (formic acid). Given that the concentrations of formic acid and sodium formate are both 1.00 M, the equation simplifies to pH = pKa + log(1/1), which then simplifies to pH = pKa since log(1) is 0. Using the provided acid dissociation constant (Ka) for formic acid (1.8×10⁻⁴), the pKa can be calculated as pKa = -log(Ka), which gives pKa = 3.74. As a result, the pH of buffer A is also 3.74.