Ka value for acetic acid is 1.85x10^-5. Also, I need to calculate the pH of a solution prepared by mixing 15 mL of .1 M NaOH and 30 mL of .1M benzoic acid solution. (benzoic acid is monoprotic, its dissociation constant is 6.5x10^-5.
2007-03-16 06:36:19 · 4 answers · asked by Sarah 1 in Science & Mathematics ➔ Chemistry
Use the Henderson-Hasselbach equation.
pH = pKa + log ([OAc-]/[AcOH]
pH = 4.74 + log (0.50/0.75) = 4.74 + log (0.667)
pH = 4.56
For the second one, since you're ading half of the amount of NaOH as you have benzoic acid, what you'll end up with is a solution that has equal concentrations of benzoic acid and sodium benzoate (the NaOH will react with half of the benzoic acid to produce the benzoate). By Henderson-Hasselbach, since [C6H5CO2-] = [C6H5CO2H], the log term will drop out and pH = pKa = 4.19.
2007-03-16 06:43:47 · answer #1 · answered by TheOnlyBeldin 7 · 1⤊ 0⤋
Sodium acetate is the deprotonated form of acetic acid.
[A-] : NaAcetate and [HA]: Acetate
Henderson-Hasselbach equation:. pH = pKa + log [A-] / [HA]
so pKa = -log Ka and we get pKa = -log(1.85*10^-5)
pKa = 4.732828
and therefore
pH = 4.732828 + log(.5/.75)
pH = 4.732828 - 0.17609
pH = 4.5570
next problem:
find moles of each first
mol = Molarity * Liters
mol(NaOH) = .1M * 0.015L = 0.0015 mol NaOH
mol(Benzoic Acid) = .1M * 0.030 L = 0.0030 mol Benzoic acid
so you have a 1:2 ratio of NaOH to benzoic acid
NaOH is sufficciently basic so we assume 100% dissociation
so lets see 0.0030 mol BA * 6.5*10^-5 = 0.000000195
pH = -log([H+]) or -log([1*10^-14 / [A-])
using the former,
pH = 6.07099
I am not so sure about this answer so you might want to double check this second part, havent done this stuff in a while ;)
rate me the best
2007-03-16 13:54:40 · answer #2 · answered by Anonymous · 1⤊ 3⤋
Use the eqn. pH = pKa + log [Salt]/[Acid]
2007-03-16 14:04:49 · answer #3 · answered by ag_iitkgp 7 · 0⤊ 0⤋
huh?
2007-03-16 13:38:33 · answer #4 · answered by robe_genious1@sbcglobal.net 2 · 0⤊ 3⤋