Consider a different titrant for this exercise. Suppose Ca(OH) 2 were used as th

Question

Consider a different titrant for this exercise. Suppose Ca(OH)2 were used as the titrant, instead of NaOH. This will make the titrant twice as concentrated in hydroxide ion. The analyte will still be HC2H3O2.

(a) What is the stoichiometry of HC2H3O2 to Ca(OH)2? (2:1)

Consider a different titrant for this exercise. Suppose Ca(OH)2 were used as the titrant, instead of NaOH. This will make the titrant twice as concentrated in hydroxide ion. The analyte will still be HC2H3O2. (a) What is the stoichiometry of HC2H3O2 to Ca(OH)2? (2:1) (b) Complete the following table for this titration. Data Table P1: Titration of acetic acid with calcium hydroxide

Explanation / Answer

The equation of this reaction is :

. . . . . . . . . Ca(OH)2 . . . + . . . 2 CH3-COOH . . . -----> . . . 2 CH3-COO(-) + Ca(2+) + 2 H2O

. . . or more simply :

. . . . . . . . . 2 OH(-) . . . + . . . 2 CH3-COOH . . . -----> . . . 2 CH3-COO(-) . . . + . . . 2 H2O

. . . . . . . . .2 moles . . . . . . . . . 1 mole



n(Ca(OH)2) ) = Concentration(Ca(OH)2) x V(Ca(OH)2)
n(Ca(OH)2) ) = 0.365 x 12.63x10-3 = 4.60x10-3 mole
n(Ca(OH)2) ) = 4.6 mmol of Ca(OH)2


. . . according the equation :

n(CH3-COOH) = n(Ca(OH)2) x 2
n(CH3-COOH) = 4.6 x2
n(CH3-COOH) = 9.2 mmol of CH3-COOH


. . . then :


M(CH3-COOH) = 2M(C) + 4M(H) + 2M(O)
M(CH3-COOH) = (2 x 12) + (4 x 1) + (2 x 16)
M(CH3-COOH) = 60 g/mole


Mass of CH3-COOH = M(CH3-COOH) x n(CH3-COOH)
Mass of CH3-COOH = 60 x 9.2x10-3
Mass of CH3-COOH = 0.552 g of CH3-COOH


Molarity of CH3-COOH in the original sample = n(CH3-COOH) / V
Molarity of CH3-COOH in the original sample = 9.2x10-3 / 12.04x10-3
Molarity of CH3-COOH in the original sample = 0.764 mol/Liter


Mass% of CH3-COOH in the original sample = m(CH3-COOH) x 100 / m(solution of CH3-COOH)
Mass% of CH3-COOH in the original sample = 0.552x 100 / 12.10
Mass% of CH3-COOH in the original sample = 4.56% of CH3-COOH


final answers are :

1.mmol of Ca(OH)2 = 4.6 mmol
2.mmol of HC2H3O2 = 9.2 mmol
3.mass of HC2H3O2 = 0.552g
4.mass % of HC2H3O2 in the original sample = 4.56 %
5.molarity of HC2H3O2 in the original sample = 0.764 M

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

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