PH and the Acid Dissociation Constant

Click For Summary
SUMMARY

The discussion focuses on calculating the hydronium ion concentration and pH for an aqueous solution of ammonia (NH3) with a hydroxide ion concentration of 2.25E-3 M, resulting in a hydronium ion concentration of 4.44 x 10-12 M and a pH of 11.35. Additionally, it addresses the determination of the acid dissociation constant (K_a) for a 0.010 M nitrous acid solution with a pH of 2.70, yielding a K_a value of 4.97E-4. The calculations utilize the formulas for pH and K_a, demonstrating the relationship between hydronium ion concentration and pH in weak acid solutions.

PREREQUISITES
  • Understanding of pH and pOH calculations
  • Knowledge of weak acid dissociation and equilibrium expressions
  • Familiarity with logarithmic functions and antilogarithms
  • Basic principles of acid-base chemistry
NEXT STEPS
  • Study the concept of weak acids and their dissociation constants
  • Learn about the Henderson-Hasselbalch equation for buffer solutions
  • Explore the relationship between pH, pOH, and ion concentrations in aqueous solutions
  • Investigate the properties and applications of ammonia in acid-base reactions
USEFUL FOR

Chemistry students, educators, and professionals involved in acid-base chemistry, particularly those focusing on pH calculations and acid dissociation constants.

Soaring Crane
Messages
461
Reaction score
0
1. What is the hydronium ion concentration and the pH for an aqueous solution of NH3 that has a hydroxide ion concentration of 2.25E-3 M?

a. 4.44 x 10-11 M, 3.65
b. 4.44 x 10-11 M, 10.35
c. 4.44 x 10-12 M, 2.65
d. 4.44 x 10-12 M, 11.35

[OH] = 2.25E-3
pOH = -log(2.25E-3) = 2.6478
-log[H3O+] = 14 - 2.6478 = 11.352 = pH
[H3O+] = antilog(-11.352] = 4.44E-12


2. Determine the acid dissociation constant for a 0.010 M nitrous acid solution that has a pH of 2.70. Nitrous acid is a weak monoprotic acid and the equilibrium equation of interest is HNO2 + H2O <-> H3O+ + NO2-.

a. 8.0 x 10-3
b. 2.0 x 10-3
c. 5.0 x 10-4
d. 4.0 x 10-4

K_a = [NO2-][H30+]/[HNO2]

pH = -log[H3O+]
2.70 = -log[H3O+]
[H3O+] = antilog(-2.70) = 0.001995 M

K_a = [0.001995 M]^2/[0.010 M - 0.001995 M] = 4.97E-4

Thanks.
 
Physics news on Phys.org
2xOK
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
2K
  • · Replies 48 ·
2
Replies
48
Views
9K
Replies
4
Views
4K