- #1
4LeafClover
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Problem:
I need 750mL of an acetic acid-sodium acetate buffer with pH = 4.3. Solid sodium acetate (CH3COONa) and glacial acetic acid are available. Glacial acetic acid is 99% CH3COOH by mass and has a density of 1.05 g/mL. If the buffer is to be 0.25M in CH3COOH, how many grams of CH3COONa must be used?
My work:
I looked up the Ka value of CH3COOH and found it to be 1.8*10^-5. I also took the pH and solved for [H+] by using the mathematical process of 10^-4.3 and found [H+] to be 5.0*10^-5. Because I am told that the buffer is 0.25M CH3COOH and I found the concentration of [H+] I went ahead and set up an equilibrium expression which looks like Ka = [H+][CH3COO-]/[CH3COOH] --> 1.8*10^-5 = (5.0*10^-5)(x)/(0.25) and I found x to be 0.09M. Since I only have .750L of solution, I went ahead and multiplied 0.09 by 0.750 and found that there are 0.0675 moles of CH3COONa in the solution. Now, when I multiply the number of moles of CH3COONa in solution by the molecular weight of CH3COONa, I come up with 6 grams, but for some reason this is incorrect. Where did I go wrong?
Additional Question: How many milliliters of glacial acetic acid must be used?2
I need 750mL of an acetic acid-sodium acetate buffer with pH = 4.3. Solid sodium acetate (CH3COONa) and glacial acetic acid are available. Glacial acetic acid is 99% CH3COOH by mass and has a density of 1.05 g/mL. If the buffer is to be 0.25M in CH3COOH, how many grams of CH3COONa must be used?
My work:
I looked up the Ka value of CH3COOH and found it to be 1.8*10^-5. I also took the pH and solved for [H+] by using the mathematical process of 10^-4.3 and found [H+] to be 5.0*10^-5. Because I am told that the buffer is 0.25M CH3COOH and I found the concentration of [H+] I went ahead and set up an equilibrium expression which looks like Ka = [H+][CH3COO-]/[CH3COOH] --> 1.8*10^-5 = (5.0*10^-5)(x)/(0.25) and I found x to be 0.09M. Since I only have .750L of solution, I went ahead and multiplied 0.09 by 0.750 and found that there are 0.0675 moles of CH3COONa in the solution. Now, when I multiply the number of moles of CH3COONa in solution by the molecular weight of CH3COONa, I come up with 6 grams, but for some reason this is incorrect. Where did I go wrong?
Additional Question: How many milliliters of glacial acetic acid must be used?2