Asked by Tumi
Describe the full preparation of 1.0L 0.2mol/L pH 4.4 acetate buffer. (pka=4.76, Mr Na acetate 82, glacial acetic acid is 17.4mol/L)
Answers
Answered by
DrBob222
For 1 L of 0.2 M buffer, you want
acid + base = 0.2
pH = pKa + log(base)/(acid)
4.40 = 4.76 + log(base/acid)
base/acid = 0.436 but you should confirm that. I estimated here and there.
That gives you two equations; solve them simultaneously for acid and base concnc.
eqn 1 is base + acid = 0.2 mol
eqn 2 is base/acid = 0.436 mol
Solve for (base) = ??
Solve for (acid) = ??
Then M acid = moles/L. You know moles and M, solve for L of the 17.4 stuff you want.
For the base, I would solve for grams = moles x molar mass.
I assume you can handle the details.
acid + base = 0.2
pH = pKa + log(base)/(acid)
4.40 = 4.76 + log(base/acid)
base/acid = 0.436 but you should confirm that. I estimated here and there.
That gives you two equations; solve them simultaneously for acid and base concnc.
eqn 1 is base + acid = 0.2 mol
eqn 2 is base/acid = 0.436 mol
Solve for (base) = ??
Solve for (acid) = ??
Then M acid = moles/L. You know moles and M, solve for L of the 17.4 stuff you want.
For the base, I would solve for grams = moles x molar mass.
I assume you can handle the details.
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