Question

Limewater is calcium hydroxide solution. In an experiment to find the concentration of calcium hydroxide in limewater, 25.0 cm³ of limewater needed 18.8 cm³ of 0.0400 mol/dm³ hydrochloric acid to neutralise it.

Ca(OH)₂(aq) + 2HCl(aq) → CaCl₂(aq) + 2H₂O(1)
Calculate the concentration of the calcium hydroxide in:
a mol/dm³
b g/dm³
(A: H= 1, O=16, Ca = 40)

Answer 1

The balanced chemical equation for the reaction is:

Ca(OH)₂(aq) + 2HCl(aq) → CaCl₂(aq) + 2H₂O(l)

From the equation, we see that one mole of Ca(OH)₂ reacts with two moles of HCl.

a) We can use the given information to calculate the number of moles of HCl that reacted with the Ca(OH)₂ in the 25.0 cm³ of limewater:

Number of moles of HCl = concentration × volume
Number of moles of HCl = 0.0400 mol/dm³ × 18.8 cm³/1000 cm³
Number of moles of HCl = 0.000752 mol

Since two moles of HCl react with one mole of Ca(OH)₂, we have:

Number of moles of Ca(OH)₂ = 0.000752 mol / 2
Number of moles of Ca(OH)₂ = 0.000376 mol

The volume of the limewater used is 25.0 cm³, which is equivalent to 0.0250 dm³. Therefore, the concentration of calcium hydroxide in the limewater is:

Concentration of Ca(OH)₂ = number of moles / volume
Concentration of Ca(OH)₂ = 0.000376 mol / 0.0250 dm³
Concentration of Ca(OH)₂ = 0.015 mol/dm³

Therefore, the concentration of the calcium hydroxide in the limewater is 0.015 mol/dm³.

b) To calculate the concentration of Ca(OH)₂ in g/dm³, we need to use the molar mass of Ca(OH)₂:

Molar mass of Ca(OH)₂ = 1 × 40.08 g/mol (for Ca) + 2 × 16.00 g/mol (for O) + 2 × 1.01 g/mol (for H)
Molar mass of Ca(OH)₂ = 74.10 g/mol

The mass of Ca(OH)₂ in 25.0 cm³ of limewater can be calculated using the number of moles we previously calculated:

Mass of Ca(OH)₂ = number of moles × molar mass
Mass of Ca(OH)₂ = 0.000376 mol × 74.10 g/mol
Mass of Ca(OH)₂ = 0.0279 g

The volume of the limewater used is 25.0 cm³, which is equivalent to 0.0250 dm³. Therefore, the concentration of calcium hydroxide in the limewater in g/dm³ is:

Concentration of Ca(OH)₂ = mass / volume
Concentration of Ca(OH)₂ = 0.0279 g / 0.0250 dm³
Concentration of Ca(OH)₂ = 1.12 g/dm³

Therefore, the concentration of the calcium hydroxide in the limewater is 1.12 g/dm³

Answer 2

To find the concentration of calcium hydroxide in limewater, we need to use the balanced chemical equation and stoichiometry to determine the number of moles of calcium hydroxide that reacted with the hydrochloric acid.

From the balanced equation, we can see that 1 mole of calcium hydroxide reacts with 2 moles of hydrochloric acid. Therefore, the number of moles of calcium hydroxide in the limewater is:

moles of Ca(OH)₂ = (0.0400 mol/dm³) × (18.8 cm³ / 1000 cm³) × (1/2) = 0.00376 mol

a) The concentration of calcium hydroxide in mol/dm³ is:

concentration of Ca(OH)₂ = moles of Ca(OH)₂ / volume of limewater in dm³

We are given the volume of limewater as 25.0 cm³, which is equivalent to 0.025 dm³. Therefore,

concentration of Ca(OH)₂ = 0.00376 mol / 0.025 dm³ = 0.150 mol/dm³

b) The molar mass of calcium hydroxide is:

Ca(OH)₂ = (1 × 40.08 g/mol) + (2 × 16.00 g/mol) + (2 × 1.01 g/mol) = 74.10 g/mol

To convert the concentration from mol/dm³ to g/dm³, we need to multiply by the molar mass. Therefore,

concentration of Ca(OH)₂ = 0.150 mol/dm³ × 74.10 g/mol = 11.1 g/dm³

So, the concentration of calcium hydroxide in limewater is 0.150 mol/dm³ or 11.1 g/dm³.