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?