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

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SUMMARY

To prepare a buffer solution with a pH of 4.50 using 1.00L of 1.80 mol/L acetic acid (HC2H3O2) and sodium acetate (NaC2H3O2), the Henderson-Hasselbalch equation is utilized. Given the pKa of acetic acid is 4.74, the concentration of acetate ions (C2H3O2-) must be calculated to achieve the desired pH. The calculation involves determining the required moles of sodium acetate to compensate for the hydrolysis of acetate ions, which affects the final concentration. The molar mass of sodium acetate is 82.0 g/mol, and the final amount to be added can be derived from the difference between the required and existing concentrations of acetate ions.

PREREQUISITES
  • Understanding of the Henderson-Hasselbalch equation
  • Knowledge of buffer solutions and their components
  • Familiarity with molarity and molar mass calculations
  • Basic concepts of acid-base chemistry, particularly weak acids
NEXT STEPS
  • Calculate the concentration of C2H3O2- ions using the Henderson-Hasselbalch equation
  • Learn about the hydrolysis of acetate ions in buffer solutions
  • Explore the impact of weak acid dissociation on buffer capacity
  • Investigate other buffer preparation methods and their applications
USEFUL FOR

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

linuxux
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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.
 
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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
 
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