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In an acid-base titration, the equivalence point is the stage at which the amount of titrant added is chemically equivalent to the substance in the solution being titrated. In simple terms, this is the point where the number of moles of acid equals the number of moles of base. This is crucial because it tells us that the acid has been completely neutralized by the base.

For example, in our given problem, the equivalence point is reached when 30.0 mL of NaOH is used. This is because NaOH is added to neutralize the unknown monoprotic acid, meaning that both reactants have reacted in a 1:1 ratio. Identifying this point accurately allows for the correct calculation of the molar mass, which depends on knowing exactly how many moles were involved in the reaction.

Equivalence points are often determined by noting a change in pH, which is commonly monitored using an indicator dye or a pH meter. It is not necessarily the same as the endpoint, which is the point where a visible change confirms the completion of the reaction, though they are ideally as close as possible.

Calculating the molar mass of an unknown substance in a titration involves knowing two things: the mass of the sample and the moles of the substance involved in the reaction. The formula for molar mass is:\[\text{Molar Mass} = \frac{\text{Mass of Sample}}{\text{Moles of Substance}}.\]

In our example, we first need to establish the moles of NaOH added at the equivalence point, which matched the moles of the unknown acid due to their 1:1 reaction ratio. We found the moles of NaOH by multiplying its concentration (0.0950 M) by its volume (0.030 L), resulting in 0.00285 moles.

With a mass of 0.2140 g for the acid, we calculated its molar mass by dividing the mass by the number of moles (0.2140 g / 0.00285 mol), resulting in a molar mass of 75.09 g/mol. This calculation is crucial for determining the identity of the unknown acid by comparing the calculated molar mass with those of known acids.

The dissociation constant, represented as Ka, is a crucial value that shows the strength of an acid in a solution. It measures how well an acid dissociates into its ions at a given concentration. The larger the Ka, the stronger the acid, as it means more acid molecules are ionized.

For the unknown acid, calculating the Ka involved understanding the concentrations of hydrogen ions, the conjugate base, and the undissociated acid at the half-equivalence point. At this point, the concentration of the acid equals the base that has reacted with the titrant, simplifying the calculation drastically.

Given that the pH at the half-equivalence point was 6.50, we could calculate \[ [H_3O^+] = 10^{-pH} = 3.16 \times 10^{-7} \text{ M}. \]
The concentrations for both the conjugate base \( A^- \) and undissociated acid \( HA \) are crucial for calculating the Ka value using:

\[ \text{Ka} = \frac{[H_3O^+][A^-]}{[HA]}. \]

Substituting the values, we found Ka to be \( 3.16 \times 10^{-7} \), demonstrating the relatively weak acidic strength of the unknown monoprotic acid.