Flag Physical Chemistry> ionic equilibrium...
question mark

a solution contains 2M CH3COOH and 1M CH3COONa. (Ka= 1.8*10^-5).a) calculate the change in pH of the buffer after the addition of 0.01 mol NaOH. b) the pH of the solution after addition of 0.01 mol HCL? please solve the ques completely

Parth Malik , 14 Years ago
Grade 11
anser 1 Answers
Askiitians Tutor Team

To tackle this problem, we need to understand how buffers work and how to apply the Henderson-Hasselbalch equation. A buffer solution, like the one you have with acetic acid (CH3COOH) and sodium acetate (CH3COONa), resists changes in pH when small amounts of acid or base are added. Let's break down the calculations for both scenarios: adding NaOH and adding HCl.

Part A: Change in pH After Adding NaOH

First, we will calculate the initial pH of the buffer solution using the Henderson-Hasselbalch equation:

Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Where:

  • [A-] = concentration of the conjugate base (CH3COONa)
  • [HA] = concentration of the weak acid (CH3COOH)
  • pKa = -log(Ka)

Given:

  • [HA] = 2 M (CH3COOH)
  • [A-] = 1 M (CH3COONa)
  • Ka = 1.8 x 10^-5

First, calculate pKa:

pKa = -log(1.8 x 10^-5) ≈ 4.74

Now, substitute the values into the Henderson-Hasselbalch equation:

pH = 4.74 + log(1/2) = 4.74 - 0.301 = 4.44

Next, we add 0.01 mol of NaOH to the buffer. NaOH will react with the acetic acid (HA) to form more acetate (A-):

CH3COOH + NaOH → CH3COONa + H2O

Initially, we have:

  • 0.01 mol NaOH will consume 0.01 mol CH3COOH.
  • New [HA] = 2 M - (0.01 mol / total volume) = 2 M - 0.01 mol/L (assuming 1 L for simplicity) = 1.99 M
  • New [A-] = 1 M + 0.01 mol = 1.01 M

Now, recalculate the pH:

pH = 4.74 + log(1.01/1.99)

pH = 4.74 + log(0.508) ≈ 4.74 - 0.295 = 4.45

The change in pH after adding NaOH is:

ΔpH = 4.45 - 4.44 = 0.01

Part B: pH After Adding HCl

Now, let's find the pH after adding 0.01 mol of HCl. HCl will react with the acetate (A-) to form acetic acid (HA):

CH3COO- + HCl → CH3COOH + Cl-

In this case:

  • 0.01 mol HCl will consume 0.01 mol CH3COO-.
  • New [A-] = 1 M - 0.01 mol = 0.99 M
  • New [HA] = 2 M + 0.01 mol = 2.01 M

Now, we can recalculate the pH using the Henderson-Hasselbalch equation:

pH = 4.74 + log(0.99/2.01)

pH = 4.74 + log(0.492) ≈ 4.74 - 0.31 = 4.43

Thus, the pH after adding HCl is approximately 4.43.

Summary of Changes

  • Change in pH after adding NaOH: ΔpH = 0.01
  • pH after adding HCl: pH ≈ 4.43

In conclusion, the buffer system effectively minimizes changes in pH despite the addition of strong acids or bases, demonstrating its utility in maintaining pH levels in various chemical and biological processes.

ApprovedApproved
Last Activity: 7 Months ago
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments