Datasheet Unknown No. or Letter Trial Mass of weigh boat plus sample (g) Mass of

Question

Datasheet Unknown No. or Letter Trial Mass of weigh boat plus sample (g) Mass of weigh boat g) Mass of sample (g) volume of NaOH to eq point (mL Molarity ofNaOH Equivalents of NaOH Equivalents of unknown acid Equivalent mass of unknown acid (g/eq) Average equivalent mass Literature value of equivalent mass error in equivalent mass Volume of NaOH at X equivalence point (mL) pH at equivalence point pKa of unknown acid Average pKa of unknown acid Average Ka of unknown acid Literature value of pKa Uterature value of Ka error in Ka

Explanation / Answer

From the given data

Fro trial 1

moles of NaOH used = 0.1 M x 24 ml = 2.4 mmol

equivalence of NaOH = 1 equivalent

moles of acid reacted = 2.4 mmol

Equivalent mass of unknown acid = 0.2605 g/0.0024 mol = 108.54 g/Eq

Volume of NaOH at half-equivalence point = 12 ml

pH at half-equivalence point = pKa of unknown acid

Compare the mass 108.54 g/mol with given table of monoprotic acid to find identity of acid.

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

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