# If "25 cm"^3 of a solution of calcium hydroxide with a concentration of "1 g/dm"^3 takes a titre of 25 cm^3 of hydrochloric acid, what is the molar concentration of the "HCl"?

Jan 2, 2015

I suspect that you diluted a stock solution of HCl and used this to titrate the limewater. If that is is correct, there are 3 × 10⁻⁴ mol of Ca(OH)₂ in a 25 cm³ pipette.

If a 25 mL of the Ca(OH)₂ solution takes a titre of 25 cm³ of HCl, the concentration of HCl must be 0.03 mol·dm⁻³.

Step 1. Calculate the molar concentration of the limewater.

("1 g Ca(OH)"_2)/("1 dm"^3"soln") × ("1 mol Ca(OH)"_2)/("74.09 g Ca(OH)"_2) = "0.013 mol·dm"^-3
(1 significant figure + 1 guard digit)

Step 2. Calculate the moles of Ca(OH)₂ in your sample.

${\text{0.025 dm"^3"soln" × ("0.013 mol Ca(OH)"_2)/("1 dm"^3"soln") = 3 × 10^-4"mol Ca(OH)}}_{2}$
(1 significant figure)

Step 3. Write the balanced chemical equation for the neutralization.

Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O

Step 4. Calculate the moles of HCl required.

3.4× 10^-4"mol Ca(OH)"_2 × "2 mol HCl"/("1 mol Ca(OH)"_2) = 6.7 ×10^-4"mol HCl"
(1 significant figure + 1 guard digit)

Step 5. Calculate the molarity of the HCl.

Molarity = (6.7 ×10^-4"mol")/("0.025 dm"^3) = "0.03 mol·dm"^-3
(1 significant figure)

Note: The answer can have only 1 significant figure, because that is all you gave for the concentration of the limewater. If you need more precision, you will have to recalculate.