What is the pH after adding HCl to the buffer solution?

  • Thread starter Thread starter ahhppull
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around calculating the pH of a buffer solution after the addition of hydrochloric acid (HCl). The context includes a specific homework problem involving the concentrations of acetic acid and sodium acetate in the buffer solution.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant states the initial pH of the buffer solution is 4.74 and attempts to calculate the new pH after adding HCl, expressing uncertainty about significant figures and the correctness of their answer.
  • The participant proposes using the equation pH = pKa + log[base]/[acid] to find the new pH after the addition of HCl, adjusting concentrations of acetic acid and sodium acetate accordingly.
  • Another participant requests clarification on the initial participant's problem-solving process.
  • The initial participant later corrects their approach, recalculating the moles of acetic acid and sodium acetate after the addition of HCl and arriving at a new pH of 4.67.

Areas of Agreement / Disagreement

The discussion reflects a progression of thought from an initial calculation to a corrected approach, but there is no explicit consensus on the final answer as participants focus on the process rather than confirming the result.

Contextual Notes

Participants express uncertainty regarding significant figures and the proper application of the buffer equation, indicating potential limitations in their calculations.

ahhppull
Messages
59
Reaction score
0

Homework Statement



A 200.0 mL buffer solution is 0.280 M in acetic acid and 0.280 M in sodium acetate. What is the pH after addition of 0.0050 mol of HCl?

Homework Equations



pH = pKa + log[base]/[acid]

The Attempt at a Solution



I have a problem with the significant figures. The website won't accept my answer. Or maybe I'm just wrong.

The initial ph is 4.74 and when HCl is added, the concentration of sodium acetate decreases by .0050 and the concentration of acetic acid increases by .0050

Therefore the equation becomes pH = 4.74 + log [.275]/[.285]

My answers I have inputted were 4.72 and 4.73
 
Physics news on Phys.org
Nvm, no help needed. I figured out what the problem is.
 
Would be nice if you can tell what it was.
 
Oh, I did the problem wrong. The right was to do it was:

.28 M acetic acid x .200 L = .056 moles acetic acid
.056 moles acetic acid + .005 moles (from HCl) = .061 moles acetic acid

.28 M sodium acetate x .200 L = .056 moles sodium acetate
.056 moles sodium acetate - .005 moles = .051 moles sodium acetate

pH = 4.74 + log [.051]/[.061]
pH = 4.67
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K