If 24.83cm 3 ?of 0.15M NaOH is titrated to its end point with 39.45cm 3

Explanation

Let's start by understanding what the question is asking us to do. We are tasked with finding the molarity of HCl given the molarity and volume of NaOH and the volume of HCl. This is a typical titration problem that can be solved using the concept of molarity and the principle of stoichiometry.

Molarity (M) is a measure of the concentration of a solute in a solution, or of any chemical species in terms of volume. It is defined as the amount of a substance (in moles) divided by the volume of the solution (in liters).

Stoichiometry is the calculation of relative quantities of reactants and products in chemical reactions. In this case, the reaction between NaOH and HCl is a one-to-one reaction: one mole of NaOH reacts with one mole of HCl to produce one mole of NaCl and one mole of water.

Here is the balanced chemical equation for this reaction: \(NaOH + HCl \rightarrow NaCl + H_2O\)

Given that the reaction is one-to-one, we can say that the moles of NaOH are equal to the moles of HCl at the endpoint of the titration. We can calculate the moles of NaOH using the formula: \(Molarity = \frac{Moles}{Volume}\). Rearranging this, we have: \(Moles = Molarity \times Volume\).

Substituting the given values, the moles of NaOH is \(0.15M \times 24.83cm^3 = 3.725moles\). Remember to convert the volume from cm\(^3\) to litres by dividing by 1000. Hence, the volume in litres is \(24.83cm^3/1000 = 0.02483L\).

Now since the moles of NaOH is equal to the moles of HCl, we can find the molarity of the HCl using the formula: \(Molarity = \frac{Moles}{Volume}\), substituting the values, we have: \(Molarity = \frac{3.725moles}{39.45cm^3} = 0.094M\).

Therefore, the molarity of the HCl is 0.094M, which is the correct answer (Option A).