Question

m is 0.240 mol/dm3 HCl.contains 22.2g of impure Ca(OH)2 per dm3 of solution. put M into the burette and titrate it against 26.3cm3 portion of Q using phenolphthalein as indicator. repeat the exercise to obtain consistent titre value.1. copy and. complete the table 2. Calculate the average volume of m used.
Burette reading
Final reading 1. 28.10 2. is empty 3. 30.40
Initial reading 1. 0 2. 1.70 3. 5.5
Titre reading 1. 25.30 2. 25.00 3. Is empty
2. 2HCl(aq)+Can(OH)2(at) yield CaCl2(aq)+2H2O(l)
B. From your results and the information given, calculate.
I. The concentration of pure Ca(OH)2 in Q in mol/dm3
ii. The concentration of pure Ca(OH)2 in Q in g/dm3
C. Calculate the percent purity of the Ca(OH)2 user to prepare Q.

Answer 1

1) Completed table (volumes in cm3)

- Run 1: initial = 0.00, final = 25.30, titre = 25.30
- Run 2: initial = 1.70, final = 26.70, titre = 25.00 (final = initial + titre)
- Run 3: initial = 5.50, final = 30.40, titre = 24.90 (titre = final − initial)

(Note: I used the three recorded titre values 25.30 and 25.00, and calculated the third as 30.40 − 5.50 = 24.90.)

2) Average volume of M (HCl) used
Average titre = (25.30 + 25.00 + 24.90) / 3 = 75.20 / 3 = 25.07 cm3 ≈ 25.07 cm3 (≈ 25.1 cm3)

B. Calculations

Given: [HCl] = 0.240 mol·dm−3; aliquot of Q = 26.3 cm3 = 0.0263 dm3. Average HCl volume = 25.07 cm3 = 0.02507 dm3.

I. Moles HCl used = 0.240 mol·dm−3 × 0.02507 dm3 = 0.0060168 mol.
Reaction: 2 HCl + Ca(OH)2 → CaCl2 + 2 H2O, so 2 mol HCl per 1 mol Ca(OH)2.

Moles Ca(OH)2 in aliquot = 0.0060168 / 2 = 0.0030084 mol.

Concentration (mol·dm−3) of Ca(OH)2 in Q = moles / volume of aliquot = 0.0030084 mol / 0.0263 dm3 = 0.11445 mol·dm−3 ≈ 0.114 mol·dm−3.

II. Concentration in g·dm−3:
Molar mass Ca(OH)2 ≈ 74.09 g·mol−1.
Mass per dm3 = 0.11445 mol·dm−3 × 74.09 g·mol−1 = 8.48 g·dm−3 ≈ 8.48 g·dm−3.

C. Percent purity
Impure sample contained 22.2 g impure Ca(OH)2 per dm3. Pure Ca(OH)2 found = 8.48 g·dm−3.

Percent purity = (8.48 / 22.2) × 100% = 38.2% (≈ 38.2%).

Summary of final answers:
- Average titre = 25.07 cm3 (≈ 25.1 cm3)
- [Ca(OH)2] = 0.114 mol·dm−3 ≈ 8.48 g·dm−3
- Percent purity ≈ 38.2%