Solve Buffer Problem: 1.00L 1.80 mol/L, pH 4.50, NaC2H3O2, pKa 4.74

  • Thread starter Thread starter linuxux
  • Start date Start date
  • Tags Tags
    Buffer
Click For Summary

Discussion Overview

The discussion revolves around calculating the amount of Sodium Acetate (NaC2H3O2) needed to achieve a buffer solution with a pH of 4.50, using a 1.00L solution of acetic acid with a concentration of 1.80 mol/L and a known pKa of 4.74. Participants explore the application of the Henderson-Hasselbalch equation and the implications of hydrolysis in the buffer system.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant asks how to calculate the grams of Sodium Acetate needed to achieve the desired pH, referencing the Henderson-Hasselbalch equation.
  • Another participant suggests using the Henderson-Hasselbalch equation but expresses uncertainty about the hydrolysis of the acetate ion (C2H3O2-), questioning how this affects the final concentration after adding NaC2H3O2.
  • A different participant clarifies that the formality of Sodium Acetate comes from the acetic acid present and provides a formula involving hydronium concentration and the formalities of the salt and acid.
  • One participant reflects on their earlier calculations regarding H+ ions, noting the small concentration and recognizing the assumptions made in the Henderson-Hasselbalch equation.
  • A participant shares a website resource related to the problem for others who may be facing similar issues.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the impact of hydrolysis on the calculations and whether the assumptions of the Henderson-Hasselbalch equation hold true in this scenario. There is no consensus on the exact amount of Sodium Acetate to add due to these considerations.

Contextual Notes

Participants note the potential complications arising from the hydrolysis of the acetate ion, which may affect the final concentration of C2H3O2- after adding Sodium Acetate. The discussion also highlights the assumptions inherent in using the Henderson-Hasselbalch equation, particularly regarding the behavior of weak acids.

linuxux
Messages
133
Reaction score
0
Question:
A buffer with pH 4.50 is needed. You have 1.00L of 1.80 mol/L solution. How many grams of Sodium Acetate (NaC2H3O2, molar mass 82.0 g/mol) must be added to give a pH of 4.50? pKa of acetic acid is 4.74.

How to solve this problem? thanks for help.
 
Physics news on Phys.org
linuxux said:
Question:
A buffer with pH 4.50 is needed. You have 1.00L of 1.80 mol/L solution. How many grams of Sodium Acetate (NaC2H3O2, molar mass 82.0 g/mol) must be added to give a pH of 4.50? pKa of acetic acid is 4.74.

How to solve this problem? thanks for help.

henderson hasselbach eq. is your friend.
 
gravenewworld said:
henderson hasselbach eq. is your friend.

thats what i thought since they gave me a pKa value, BUT, could you show how, because I am stuck, won't HC2H3O2's anion hydrolyze?

because in the henderson hasselbach equation, i solve for the concentration of C2H3O2- using 4.50 as the value for pH, correct? and i calculated the concentration of C2H3O2- already present in the solution, and adding NaC2H3O2 is like adding C2H3O2- ions, so i calculate the difference between what concentration of C2H3O2- i have already and what concentration of C2H3O2- i need, and that amount should be that same about of moles of NaC2H3O2 i add, since there is a 1:1 ratio in the dissociation equation for NaC2H3O2, BUT, C2H3O2- hydrolyzes (i think), so then the amount of C2H3O2- i add will react with water to form more HC2H3O2, so the final concentration of C2H3O2- won't be what i calculated! so how much do i add! since the amount i add will not be the amount that stays?
 
Last edited:
Linuxux,
This will be my silly question, but do you say you have 1.00L of 1.8 Molar "Acetic Acid"? So, the formality of the sodium acetate comes directly from that added to the solution, and it does accept some hydronium as you guessed.

Ka = (H)(Fsalt + H)/(Facid - H)

Where H = hydronium concentration, Fsalt=formality of the salt, Facid=formality of the acetic acid;
You already have information to directly determine what H must be; the rest of the variables you find based on the equation and knowledge of the solution values which you have.
 
actually, the amount of H+ ions present in solution i calculated was so small, i had already reasoned to myself why i needed to worry about such a small amount of ions anyway, that was 2-days ago, and then i got thinking that HC2H3O2 is a weak acid, so very few H+ ions will form anyway...oh that what henderson-hasselbach assumes isn't it! I can't believe i wasted a whole weekend with this! anyway, thanks for the help guys.

Here's a website for anyone else with this problem:
http://chimge.unil.ch/En/ph/1ph37.htm
 
Last edited by a moderator:

Similar threads

  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 3 ·
Replies
3
Views
6K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
9K