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

Explanation

In this question, we are looking at a simple acid-base titration. The acid in this case is hydrochloric acid (HCl) and the base is sodium hydroxide (NaOH). The concept of molarity (M) is fundamental here, which is a measure of the concentration of a solute in a solution, or of any chemical species, in terms of amount of substance in a given volume.

The balanced chemical equation for the reaction between NaOH and HCl is:

\(NaOH + HCl \rightarrow NaCl + H_2O\)

From the balanced equation, we can see that the ratio of NaOH to HCl is 1:1. This means that one mole of NaOH reacts with one mole of HCl.

Since molarity (M) is defined as the moles of solute per liter of solution, we can use the formula:

\[M = \frac{moles}{volume}\]

In a titration, the point where the reaction is complete is called the endpoint. At this point, the moles of the acid will be equal to the moles of the base. So, we can say that:

\[moles_{NaOH} = moles_{HCl}\]

We know the molarity and volume of the NaOH solution, so we can calculate the moles of NaOH using the formula:

\[moles = Molarity \times Volume\]

Substituting the given values:

\[moles_{NaOH} = 0.15M \times 24.83cm^3\]

Note that the volume should be in liters, so we convert from cm^3 to L by dividing by 1000:

\[moles_{NaOH} = 0.15M \times \frac{24.83cm^3}{1000} = 0.003725 moles\]

Since \(moles_{NaOH} = moles_{HCl}\), we also have 0.003725 moles of HCl. We can now find the molarity of the HCl solution using the formula \(M = \frac{moles}{volume}\). The volume of HCl is given as 39.45 cm^3, which we convert to liters:

\[M_{HCl} = \frac{0.003725 moles}{\frac{39.45cm^3}{1000}} = 0.094M\]

So, the molarity of the HCl solution is 0.094M, which corresponds to Option A.