Experiment Acid Dissociation Constant by Titration 0.9%2M 2 5m Concentration of

Question

Experiment Acid Dissociation Constant by Titration 0.9%2M 2 5m Concentration of standard NaOH solution Original volume of acetic acid solution in beaker Data: Titration Results Volume (mL) pH Equivalence point ½ equivalence point What is the k, and pk, of you acetic acid solution? Calculate the percent error 0f your measurement. 1. 4.11 n fra 24 10 2. Calculate the concentration of your acid solution? Molarity of acid = MolarityNaoWX EquivalenceVolumeNaOH (L) ÷ Origina!Volume,cid (L) 3. For the reaction of acetic acid and sodium hydroxide, why was the pH at the equivalence point not equal to 7?

Explanation / Answer

Q1

Ka for acetic acid is given at half equivalence point

from your data

pH = pKa + log(A-/HA)

at 50% equiv. A- = HA

then

pH = pKa = 4.77

Ka = 10^-pKa = 10^-4.77

Ka = 1.69824*10^-5

Q2

find concentration of acid

Macid*Vacid = Mbase*Vbase

Veuqivlaence = 2Xequivalence 50% = 2*11.11 mL = 22.22 mL

Vacid = 2.5 mL

0.0952*22.22 = Macid*2.5

Macid =0.0952*22.22 /2.5

Macid = 0.8461 M

Q3

the pH is not 7 since this is a weak acid

then it will hydrolyse in water

CH3COO- + H2O <-> CH3COOH + OH-

since Oh-is present, expect basic medaia, i.e pH >7

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.