How do I find the volume of 6M stock solution of acetic acid needed to make a 250 mL buffer at a pH 4.7? pka is 4.74
6 answers
What will you use for the base? Do you intend to add NaOH to acetic acid to make the salt or are you adding an amount of sodium acetate?
Adding 20mL sodium acetate, sorry.
And the concn of the sodium acetate is what?
And it's 0.20M
The Henderson-Hasselbalch equation is
pH = pKa + log[(base)/(acid)]
millimoles base = 20 mL x 0.2M = 4 mmoles.
4.7 = 4.74 + log (4/a)
(4/acid)= 0.91
acid = 4/0.91 = ??
M = mmoles/mL.
You know mmols and you know M, solve for mL and dilute to whatever volume you need. There is a caution that goes along with this. These are small amount in a relatively large volume; therefore, the buffering capacity may be limited. I ran a quick calculation on adding 10 mL of 0.1 M NaOH to the buffer and came out with about 4.35 which is not bad (the buffering capacity is defined as that amount of base or acid to change the pH by +/- 1.0 and 10 mL of 0.1M NaOH doesn't cross that line.)
pH = pKa + log[(base)/(acid)]
millimoles base = 20 mL x 0.2M = 4 mmoles.
4.7 = 4.74 + log (4/a)
(4/acid)= 0.91
acid = 4/0.91 = ??
M = mmoles/mL.
You know mmols and you know M, solve for mL and dilute to whatever volume you need. There is a caution that goes along with this. These are small amount in a relatively large volume; therefore, the buffering capacity may be limited. I ran a quick calculation on adding 10 mL of 0.1 M NaOH to the buffer and came out with about 4.35 which is not bad (the buffering capacity is defined as that amount of base or acid to change the pH by +/- 1.0 and 10 mL of 0.1M NaOH doesn't cross that line.)
Thank you so much!
2 answers