Consider an unknown base, RNH. One experiment titrates a 50.0-mL aqueous solution containing \(2.500 \mathrm{~g}\) of the base. This titration requires \(59.90 \mathrm{~mL}\) of \(0.925 \mathrm{M} \mathrm{HCl}\) to reach the equivalence point. A second experiment uses an identical 50.0-mL solution of the unknown base that was used in the first experiment. To this solution is added \(29.95 \mathrm{~mL}\) of \(0.925 \mathrm{M} \mathrm{HCl}\). The \(\mathrm{pH}\) after the HCl addition is \(10.77 .\) (a) What is the molar mass of the unknown base? (b) What is \(K_{\mathrm{b}}\) for the unknown base? (c) What is \(K_{n}\) for \(\mathrm{RNH}_{2}{ }^{+} ?\)

Understanding the molar mass of a substance is a fundamental concept in chemistry that ties into various areas such as stoichiometry, thermodynamics, and physical chemistry. The molar mass is the weight of one mole of a substance, typically expressed in grams per mole (g/mol). When conducting an acid-base titration, as in the textbook example, molar mass calculation is essential to identify the unknown substance.

To calculate the molar mass, one must divide the mass of the substance (in grams) by the number of moles of the substance. Let's simplify the equation from the example: \[\text{Molar mass} = \frac{\text{Mass (g)}}{\text{Moles (mol)}}\].

By knowing the amount in grams and the stoichiometric equivalence at the titration equivalence point, we can solve for the molar mass, which gives us insight into the molecular structure and composition of the unknown base.

In acid-base chemistry, the equilibrium condition describes the state where the rates of the forward and reverse reactions are equal, leading to constant concentrations of the reactants and products. Given an acid-base system, it is characterized by the equilibrium constant, which for bases is referred to as the base dissociation constant (\(K_{\text{b}}\)).

The \(K_{\text{b}}\) provides a quantitative measure of the strength of a base in solution. It is derived using the concentrations of the hydroxide ion \([\text{OH}^-]\) and the base (\(\text{RNH}\)) versus the conjugate acid (\(\text{RNH}_2^+\)). The typical expression for \(K_{\text{b}}\) is: \[K_{\text{b}} = \frac{[\text{OH}^-][\text{RNH}]}{[\text{RNH}_2^+]}\].

In our example, the \(K_{\text{b}}\) calculation involves the utilization of the \(\text{pH}\) of the solution at the half-equivalence point, reflecting the dynamic nature of acid-base reactions and their tendency to reach an equilibrium state under given conditions.

A critical aspect of any titration is the equivalence point, the juncture at which the amount of titrant added equals the amount of substance in the sample being titrated. In terms of acid-base titrations, it is the point at which the number of moles of acid equals the number of moles of base.

In the given exercise, the equivalence point occurs when the moles of \(HCl\) added from the burette stoichiometrically equal the moles of the unknown base \(RNH\) in solution. This moment allows chemists to calculate the molar mass of the unknown substance and understand its acid-base equilibrium.

In practice, this point can be identified by a sharp change in \(\text{pH}\), or with the help of an indicator that changes color at a particular \(\text{pH}\) level. Understanding the equivalence point is key to many quantitative analytical techniques in chemistry and serves as a foundation for determining the concentration and molar mass of substances in a solution.