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

Hi Cole,

To start this buffer problem, let's look at the species that are involved in making the buffer. Recall that a buffer is defined as a solution of a weak conjugate acid/base pair (some HA and A^-), that are in near-equal proportions (a molar ratio between 1:10 and 10:1 with respect to HA and A^-), within 1 pH unit of the conjugate acid's pKa.

This will help us answer part 1. Our acid will be propanoic acid, HC₃H₅O₂, and the conjugate base will be C₃H₅O₂^-. We know they constitute a conjugate pair because the difference in chemical formula is one H⁺, the acidic proton. It is important for the acid/base conjugate pair to have near-equal molar portions because we know the buffer region in a titration is where [HA] and [A^-] are near-equal. The next step is confirming their near-equal proportions:

We find the molar mass of sodium propanoate to be 96.07 g/mol.

0.496 g NaC₃H₅O₂ / (96.07 g/mol) = 0.00516 mol

Using Molarity = n/V (with n being moles and V being volume in liters), we can find the moles of acid:

First converting the 50.0 mL into Liters by dividing by 1000 mL/L yields 0.0500 L. We can multiply 0.0500 L * 0.265 M = 0.0133 mol.

So, looking at our two numbers, we have close to a 2:1 ratio of HA to A^-, certainly within our desired range of 1:10 to 10:1. We can conclude that our solution makes a good buffer.

Part 2 wants us to calculate the pH of our buffer. The go-to equation for calculating the pH of a buffer is the Henderson-Hasselbach equation, pH = pKa + log( [A^-] / [HA] ). Our acid dissociation constant is in terms of Ka, as 1.34x10^-5. To find pKa, we take the -log(Ka) = -log(1.34x10^-5) = 4.87.

Putting it all together:

pH = 4.87 + log(0.00516/0.0133) = 4.46

(Note that since the concentrations are both in the same solution, the 0.050 L cancels out and moles can simply be used).

Let's make sure this answer makes sense. Relative to the pKa, our pH is lower and therefore more acidic. Looking back at our moles, we did have more of the acid, so this answer checks out!

Let me know if you have any follow-up questions or need clarification on anything.