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"#?

Answer 1

I believe you titrated the limewater by diluting a stock solution of HCl; if this is accurate, a 25 cm³ pipette contains 3 × 10⁻⁴ mol of Ca(OH)₂.

When 25 cm³ of HCl is added to a 25 mL Ca(OH)₂ solution, the HCl concentration needs to be 0.03 mol·dm⁻³.

Step 1: Determine the limewater's molar concentration.

#("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)

Determine the amount of Ca(OH)₂ in your sample in step two.

#"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: Compose the neutralization's balanced chemical equation.

2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O

Step 4: Determine how many moles of HCl are needed.

#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: Determine the HCl's molarity.

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

Note: You only provided one significant figure for the limewater concentration, so the answer can only have one. You will need to recalculate if you require more accuracy.

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Answer 2

The molar concentration of the hydrochloric acid (HCl) can be calculated using the given information and the principle of stoichiometry. Since the volume of calcium hydroxide solution is equal to the volume of the hydrochloric acid solution, the number of moles of HCl is equal to the number of moles of calcium hydroxide.

Given: Volume of calcium hydroxide solution = 25 cm^3 Concentration of calcium hydroxide solution = 1 g/dm^3 Volume of hydrochloric acid solution = 25 cm^3

To find the number of moles of calcium hydroxide:

  1. Convert the volume of calcium hydroxide solution to dm^3: 25 cm^3 = 25/1000 dm^3 = 0.025 dm^3
  2. Calculate the number of moles of calcium hydroxide using its concentration and volume: Moles = concentration (in mol/dm^3) × volume (in dm^3) Moles = 1 g/dm^3 × 0.025 dm^3 = 0.025 mol

Since the volume and the number of moles of HCl are the same as those of calcium hydroxide, the molar concentration of HCl is also 0.025 mol/dm^3.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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