Question

If 30.34 mL of a 0.135 M solution of hydrochloric acid (HCl) were required to neutralize 25.00 mL of a calcium hydroxide (Ca(OH)2) solution, the molarity of the calcium hydroxide solution is 0.101 M.

Answer 1

The molarity of the calcium hydroxide solution can be calculated by using the stoichiometry from the balanced chemical equation and the given volumes and molarity of hydrochloric acid used in the titration. The calculation involves finding the moles of HCl used and dividing by the volume of calcium hydroxide, resulting in a molarity of 0.08192 M.

Step-by-step explanation:

The student's question pertains to finding the molarity of a calcium hydroxide (Ca(OH)2) solution given the volume of hydrochloric acid (HCl) used to neutralize it during titration. A balanced chemical equation for the neutralization reaction HCl + Ca(OH)2 → CaCl2 + 2H2O reveals that two moles of HCl are required to neutralize one mole of Ca(OH)2. To find the molarity of the Ca(OH)2 solution, we calculate the number of moles of HCl using its molarity and volume, then use the stoichiometry of the equation to find the moles of Ca(OH)2 and finally divide by its volume in liters.

To calculate,
Moles of HCl = 0.03034 L * 0.135 M = 0.004096 moles HCl,
Moles of Ca(OH)2 (using the 1:2 ratio) = 0.004096 moles HCl / 2 = 0.002048 moles Ca(OH)2,
Molarity of Ca(OH)2 (M) = 0.002048 moles / 0.02500 L = 0.08192 M.

Answer 2

The molarity of the calcium hydroxide (Ca(OH)2) solution is approximately 0.164 M.

To find the molarity of the calcium hydroxide (Ca(OH)2) solution, you can use the concept of stoichiometry and the balanced chemical equation for the neutralization reaction between hydrochloric acid (HCl) and calcium hydroxide:

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

Here's the step-wise solution:

1. First, calculate the number of moles of HCl used

Moles of HCl = (Volume of HCl solution in L) × (Molarity of HCl)

Moles of HCl = (30.34 mL / 1000 mL/L) × (0.135 M)

Moles of HCl = 0.0040969 moles

2. According to the balanced equation, the reaction ratio between HCl and Ca(OH)2 is 1:1, which means that for every mole of HCl, you need one mole of Ca(OH)2.

3. Therefore, the number of moles of Ca(OH)2 in the 25.00 mL of Ca(OH)2 solution is also 0.0040969 moles.

4. Now, calculate the molarity of the Ca(OH)2 solution:

Molarity of Ca(OH)2 = (Moles of Ca(OH)2) / (Volume of Ca(OH)2 solution in L)

Molarity of Ca(OH)2 = 0.0040969 moles / (25.00 mL / 1000 mL/L)

Molarity of Ca(OH)2 = 0.16388 M

The complete question is given below:

If 30.34 mL of a 0.135 M solution of hydrochloric acid (HCl) were required to neutralize 25.00 mL of a calcium hydroxide (Ca(OH)2) solution, the molarity of the calcium hydroxide solution?