I'm new to chemistry and I need help with this

Question

A chemist starts with a 50.0 mL sample of 0.265 M propanoic acid, HC3H5O2, then adds 0.496 g of sodium propanoate, NaC3H5O2.

1.Explain how this creates an effective buffer.

2.Propanoic acid has an acid dissociation constant of 1.34 x 10-5. Assuming that no change in the volume of the solution occurs, calculate the pH of the buffer solution.

Answer 1

1). This creates an effective buffer because you have a weak acid (propanoic acid) and the salt/conjugate base of that acid (propanoate). They way this works is ... if you add acid (H+) it will react with the conjugate base (C3H5O2-) to produce the weak acid. If you add base (OH-), it will react with the weak acid (HC3H5O2) to produce H2O and propanoate. Thus, it will resist changes in pH.

2). Use the Henderson Hasselbalch equation

pH = pKa + log [propanoate]/[propanoic acid]

pKa = -log Ka = -log 1.34x10^-5 = 4.87

[propanoate] = 0.496 g NaC3H5O2 x 1 mol NaC3H5O2/96.1 g / 0.05 L = 0.10 M

pH = 4.87 + log (0.1 M/0.265 M)

pH = 4.87 + log 0.377

pH = 4.87 + (-0.42)

pH = 4.45