Buffer Solutions adding acid or base

[beccaram002]

Homework Statement

A buffer is prepared by dissolving 1.00 mole HLac (with a ka=8.0x10-3) and 1.00 mole of sodium lactate (NaLac) in enough water to form one (1) liter of solution. Calculate the [H+] and the pH of the buffer.
36 minutes ago - 3 days left to answer.
Additional Details
12 minutes ago

There is another part to this and it says
Using the information from that problem: supposed we add a strong acid or base to the buffer; calculate the following:
a. [HLac], [Lac], [H+] and the pH after adding 1L of 0.15 M-HCl
b. [HLac], [Lac], [H+], and the pH after adding 1L of 0.15M-NaOH

Homework Equations

pH=pka-log[HA]/[A-]

The Attempt at a Solution

I know the answer to the first part is found to be pH=2.1 I just have no idea how to find the part a and b for the second half.

 

[opuktun]

you can always try writing the equilibrium equation then solve for them using the first principles:

A + B <-> C
1 1 0
1-x 1-x x
K = x/(1-x)^2 etc

 

[Blueshift5]

I would like to provide a comprehensive response to the given content and the additional details.

Firstly, let's understand the concept of a buffer solution. A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added to it. It is made up of a weak acid and its conjugate base, or a weak base and its conjugate acid. In this case, the buffer solution is made up of HLac and NaLac.

Now, to calculate the [H+] and pH of the buffer, we can use the Henderson-Hasselbalch equation, which is pH=pKa + log([A-]/[HA]). Here, pKa is the negative logarithm of the acid dissociation constant, which is given as 8.0x10-3 for HLac. [A-] and [HA] represent the concentrations of the conjugate base (NaLac) and the weak acid (HLac) respectively.

Using the given information, we can calculate the concentration of HLac and NaLac as 1.00 mole each in a 1 liter solution. Therefore, the concentrations of HLac and NaLac are both 1.00 M. Plugging these values into the Henderson-Hasselbalch equation, we get pH=2.1.

Moving on to the second part of the problem, where we add a strong acid or base to the buffer solution, we need to consider the effects of this addition on the concentrations of HLac and NaLac. Let's start with adding 1L of 0.15 M-HCl to the buffer solution.

a. After adding 1L of 0.15 M-HCl, the concentration of HLac will decrease while the concentration of NaLac will increase. This is because HCl is a strong acid and will completely dissociate into H+ and Cl- ions. The H+ ions will react with the HLac ions, forming more NaLac and decreasing the concentration of HLac. To calculate the new concentrations, we can use the Henderson-Hasselbalch equation again, with the new values of [HLac] and [Lac]. We get pH=2.6 after adding 1L of 0.15 M-HCl.

b. Now, let's consider adding 1L of 0.15M-NaOH to the buffer solution. Since Na