Question

consider the chemical equation 2HCl(aq) + Ca(OH)2(aq) 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.

Answer 1

I don’t have your experimental numbers, so I’ll give the method and a worked example — then you can either plug in your values or send them and I’ll do the arithmetic.

Reaction: 2 HCl + Ca(OH)2 → CaCl2 + 2 H2O

Steps and formula
- Moles HCl = C_HCl × V_HCl (use V in dm3 = L).
- Stoichiometry: 2 mol HCl react with 1 mol Ca(OH)2, so moles Ca(OH)2 = moles HCl / 2.
- Concentration of Ca(OH)2 (mol/dm3) = (C_HCl × V_HCl) / (2 × V_Ca(OH)2) with all volumes in dm3.
- Convert to g/dm3 by multiplying mol/dm3 by molar mass of Ca(OH)2. M(Ca(OH)2) ≈ 74.09 g/mol.

Worked example
Suppose 25.00 mL (0.02500 dm3) of Q was titrated and used 20.00 mL (0.02000 dm3) of 0.1000 mol/dm3 HCl.
- moles HCl = 0.1000 × 0.02000 = 0.002000 mol
- moles Ca(OH)2 = 0.002000 / 2 = 0.001000 mol
- [Ca(OH)2] = 0.001000 / 0.02500 = 0.04000 mol/dm3
- in g/dm3 = 0.04000 × 74.09 = 2.964 g/dm3

If you give me your V_Ca(OH)2, V_HCl and C_HCl (and whether volumes were mL), I’ll calculate i) and ii) with your actual data.