Anticipated pH of Buffer Solution ( )

  • Thread starter Thread starter erjkism
  • Start date Start date
  • Tags Tags
    Buffer Ph
Click For Summary
SUMMARY

The discussion centers on calculating the anticipated pH of a buffer solution consisting of 100 mL of acetic acid (CH3COOH) and its conjugate base acetate (CH3CO2-) after the addition of 100 mL of 0.0983 M hydrochloric acid (HCl). The user initially calculated a pH of 1.72 but later revised it to 2.67 using the equilibrium expression with a dissociation constant (Ka) of 1.8x10-5. The calculations involved determining the moles of acetic acid and acetate before and after the addition of HCl, leading to the final pH value. The user expressed confusion over varying pH readings in subsequent experiments.

PREREQUISITES
  • Understanding of buffer solutions and their components (acetic acid and acetate).
  • Knowledge of acid-base equilibrium and the Henderson-Hasselbalch equation.
  • Familiarity with calculating molarity and moles in solution.
  • Proficiency in logarithmic calculations for pH determination.
NEXT STEPS
  • Study the Henderson-Hasselbalch equation for buffer pH calculations.
  • Learn about the effects of strong acids on buffer capacity.
  • Explore the concept of buffer solutions in different pH ranges.
  • Investigate the impact of dilution on pH in buffer systems.
USEFUL FOR

Chemistry students, laboratory technicians, and educators involved in acid-base chemistry and buffer solution analysis will benefit from this discussion.

erjkism
Messages
54
Reaction score
0
[SOLVED] Anticipated pH of Buffer Solution (urgent!)

Homework Statement


Can someone verify my process here? i did a lab where i had 100 mL CH3COOH/CH3CO2- buffer solution. I have to find the pH if 100 mL of 0.0983 M HCl is added to it. In the lab i got pH of 1.72

2. Homework Equations

[H3O+][CH3CO2-]
Ka= ---------------------
[CH3COOH]

Ka : 1.8x10-5
initial moles CH3COOH: 0.00996
Initial moles CH3CO2-: 0.0100
moles H+ in 100 mL 0.0983 M HCl: 0.00983

The Attempt at a Solution



It's my understanding that the mole ratio of H+/CH3CO2- is one to one.
Next i think i do this:

moles CH3COOH = 0.00996 + 0.00983 = 0.0198 moles
moles CH3CO2-= 0.0100- 0.00983= 0.00017 moles

concentration CH3COOH= 0.0198 moles/(0.200 L) = 0.099 M
concentration CH3CO2-= 0.00017/(0.200 L) = 0.00085 M

Then by rearranging the equilibrium expression i can solve for [H3O+].
[H3O+]=(1.8x10-5)(0.099M) / (0.00085) = 0.00212 M

pH= -log(0.00212M) = 2.67

is that how am supposed to go about doing this? because my other lab values are very strange. for instance, i keep getting a pH of 4.7 after adding 5, 10, and 20 mL of HCl to it.
 
Physics news on Phys.org
k i finally got the answer thanks anyone who looked at it tho
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 22 ·
Replies
22
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 6 ·
Replies
6
Views
14K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
2K