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

HC3H5O2 is the weak acid and C3H5O2-1 is the conjugate base.

HC3H5O2 will react with added base to resist pH increase.

C3H5O2-1 will react with added acid to resist pH decrease.

2.

pH=pKa+log([conj. base]/[acid])

moles C3H5O2-1 is needed to calculate [conj. base]

0.496g NaC3H5O2/96.06g=0.00516 mol NaC3H5O2

1 mol NaC3H5O2 = 1 mol C3H5O2-1, so

0.00516 mol NaC3H5O2=0.00516 mol C3H5O2-1

50mL/1000=0.050 L

[conj. base]=[C3H5O2-1]=0.00516mol/0.050 L

[C3H5O2-1]=0.103269

pH=pKa+log([conj. base]/[acid])

pH=pKa+log([C3H5O2-1]/[HC3H5O2])

pH=-log(1.34E-5)+log(0.103269/0.265)

pH=4.46