Buffer Question

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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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  • #2
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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.
henderson hasselbach eq. is your friend.
 
  • #3
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henderson hasselbach eq. is your friend.
thats what i thought since they gave me a pKa value, BUT, could you show how, because im stuck, wont 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- wont be what i calculated! so how much do i add! since the amount i add will not be the amount that stays?
 
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  • #4
symbolipoint
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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.
 
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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 isnt it! I cant 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 [Broken]
 
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