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. 
   
2. To make a buffer you will need either a weak acid and the conjugate base or a weak base and the conjugate acid. Here we have the weak acid (propanoic acid) HC3H5O2 and sodium propanoate is the conjugate base (C3H5O2^-1). So since we have a weak acid and the conjugate base in this solution, this is an effective buffer.
3. To find the pH of the buffer solution, we can use the Henderson Hasselbalch Equation.

So we will need to compute the concentration of CH3H5O2^-1 in this solution. We are given that the concentration of HC3H5O2 = 0.265 M = [HA]

1. Concentration of NaC3H5O2 calculation
2. Molar Mass NaC3H5O2 = 96.07 g/mol
3. Mole of NaC3H5O2 = 0.496g / 96.07g/mol = 0.005mol
4. Concentration of NaC3H5O2 = 0.005 mol / 0.050L = 0.103 M
5. So the concentration of C3H5O2^-1 = 0.103M since there is a 1:1 mol ratio between NaC3H5O2 ==> Na+ C3H5O2^-1
6. the pKa = -log Ka = -log(1.34*10^-5) = 4.873
7. pH = 4.873 + log (0.103 / 0.265) = 4.46