Consider the attached graph of a weak diprotic acid. A 0.2658 g samp e titrated

Question

Consider the attached graph of a weak diprotic acid. A 0.2658 g samp e titrated with 0.1000 M NaOH. Based on the data complete the following: ml of NaOH used to reach the first endpoint Mols of the acid present in the sample ' Molecular weight (molar mass) of the unknown acid Consider malonic acid, a typical diprotic weak acid: If 1 mol of the acid is dissolved in water and 0.5 mol of sodium hydroxide is added: How much of the malonic acid left in the solution? The sodium hydroxide reacts with the malonic acid to form what chemical species (show any charges) along with water. What is the relationship between malonic acid and the species from B)? When mols of a weak acid in solution are equal to mols of its conjugate the pH of the solution is equal to the of the weak acid

Explanation / Answer

1- Titration

Volume of NaOH used to reach first equivalence point = 16.0 ml

moles of acid present = 0.1 M x 16 ml = 1.6 mmols

Molecular mass of acid = 0.2658 g/0.0016 mols = 166.125 g/mol

pKa1 of acid = 5.23

pKa2 of acid = 8.00

2- 1 mole of malonic acid reacts with 0.5 mol of NaOH

C3H4O4 + NaOH ----> NaC3H3O4 + H2O

A) malonic acid is a diprotic acid, so when 0.5 mol of NaOH is added, it reacts with 1 H+ from acid and remaining is another 1.0 mol of H+ in malonic acid.

B) Reaction : C3H4O4 + NaOH ----> NaC3H3O4 + H2O

C) The species formed by the reaction of malonic acid with the base is the conjugate base of malonic acid.

D) When moles of a weak acid are equal to the moles of its conjugate base, the pH of the solution is equal to the pKa of the weak acid.

This is the half equivalence point.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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