Question

Consider an unknown base, RNH. One experiment titrates a \(50.0-\mathrm{mL}\) 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 a \(50.0-\mathrm{mL}\) solution of the unknown base identical to what was used in the first experiment. To this solution is added \(29.95 \mathrm{~mL}\) of \(0.925 \mathrm{M}\) HCl. The 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 \(K_{\mathrm{a}}\) for \(\mathrm{RNH}_{2}^{+} ?\)

Short answer

Question: Calculate the molar mass of the unknown base RNH, its Kb value, and the Ka value of its conjugate acid RNH2+. Answer: (a) The molar mass of the unknown base RNH is 45.13 g/mol. (b) The Kb value for the unknown base RNH is 1.75 × 10^-3. (c) The Ka value for the conjugate acid RNH2+ is 5.71 × 10^-12.

Step-by-step solution

Calculate moles of HCl at the equivalence point

At equivalence point, moles of HCl and moles of RNH are equal. So, let's calculate moles of HCl used. Moles of HCl = (volume of HCl) × (concentration of HCl) Moles of HCl = 59.90 mL × 0.925 M Convert mL to L: Moles of HCl = 0.05990 L × 0.925 mol/L = 0.0554 mol

Calculate molar mass of the unknown base, RNH

Since moles of HCl and RNH are equal at the equivalence point, we have 0.0554 mol of RNH. Now, we can use the mass of RNH and the number of moles to find the molar mass. Molar mass of RNH = (mass of RNH) / (moles of RNH) Molar mass of RNH = 2.500 g / 0.0554 mol = 45.13 g/mol

Calculate the concentration of RNH, RNH2+, and OH- after adding 29.95 mL of HCl

The initial moles of RNH are equal to the moles of HCl at the equivalence point: 0.0554 mol. Initial moles of RNH = 0.0554 mol Initial volume of solution = 50.0 mL + 29.95 mL = 79.95 mL = 0.07995 L New concentration of RNH = moles of RNH/volume = 0.0554 mol / 0.07995 L = 0.693 M Moles of HCl added = 0.925 M × 0.02995 L = 0.0277 mol New concentration of RNH2+ = moles of HCl added/volume = 0.0277 mol / 0.07995 L = 0.346 M Since RNH2+ donates a proton to water molecule to form OH- ions and RNH, at the equilibrium, the concentration of OH- = 0.346 M.

Calculate the concentration of H3O+ and pOH using the given pH

We are given the pH of the solution after adding 29.95mL of HCl. We can calculate the concentrations of H3O+ and OH- from the pH and pOH. pH = 10.77, so pOH = 14 - pH = 3.23 [H3O+] = 10^-pH = 10^-10.77 = 1.69 × 10^-11 M [OH-] = 10^-pOH = 10^-3.23 = 5.88 × 10^-4 M

Calculate Kb for the unknown base

Now, we can use the relationship between the concentrations to find Kb for RNH. Kb = [RNH2+][OH-] / [RNH] Kb = (0.346 M)( 5.88 × 10^-4 M) / (0.693 M) Kb = 1.75 × 10^-3

Calculate Ka for RNH2+

We have found Kb for RNH. To find Ka for its conjugate acid RNH2+, we can use the relationship between Ka and Kb. Ka × Kb = Kw = 1.0 × 10^-14 Ka = (1.0 × 10^-14) / Kb Ka = (1.0 × 10^-14) / (1.75 × 10^-3) Ka = 5.71 × 10^-12 (a) The molar mass of the unknown base RNH is 45.13 g/mol (b) The Kb for the unknown base RNH is 1.75 × 10^-3 (c) The Ka for RNH2+ is 5.71 × 10^-12

Key concepts

Molar Mass Calculation

Understanding the concept of molar mass is crucial when working with chemical compounds. Molar mass represents the weight of one mole of a substance and is expressed in grams per mole (g/mol). In the given exercise, the molar mass of an unknown base RNH is determined by titrating it with a known volume and concentration of hydrochloric acid (HCl). The equivalence point of the titration is used to find the number of moles of RNH, as it is the point where the number of moles of HCl equals the number of moles of RNH.

To calculate the molar mass, we first find the mass of RNH from the experiment, in this case, 2.500 grams. Using the moles of HCl (which are equivalent to the moles of RNH at the equivalence point), we can then use the formula:\[\text{Molar mass of RNH} = \frac{\text{mass of RNH}}{\text{moles of RNH}}\]
Through this calculation, we find the molar mass of the unknown base, giving us vital information to further understand the compound's characteristics and behavior in reactions.

Equivalence Point

The equivalence point in a titration is a key concept to grasp; it is the point at which the number of moles of acid equals the number of moles of base in a solution. This point signifies that the acid has been completely neutralized by the base or vice versa. At the equivalence point, the solution contains the salt of the acid and base, and the number of hydrogen ions equals the number of hydroxide ions.

In the context of the exercise, finding the equivalence point allows us to determine the molar mass of the unknown base by equating the moles of HCl used to the moles of RNH. This is possible because at the equivalence point, we ensure the reaction has gone to completion and all of the RNH has reacted with HCl, allowing for an accurate molar mass calculation.

pH and pOH Relationship

In any aqueous solution at 25°C, the pH and pOH values are inversely related and their sum is a constant: 14. This principle is based on the self-ionization of water where the ion-product constant for water (\(K_w\)) is\(1.0 \times 10^{-14}\). The formula to connect pH and pOH is:\[\text{pH} + \text{pOH} = 14\]The exercise gives us the pH of a solution, from which we can calculate the concentration of hydronium ions (\([H_3O^+]\)) using the formula \(\text{pH} = -\log[H_3O^+]\), and likewise, calculate the concentration of hydroxide ions (\([OH^-]\)) using the pOH by the formula \(\text{pOH} = -\log[OH^-]\). These calculations are instrumental when addressing reactions in an aqueous solution, particularly in determining the concentrations of all species at equilibrium.

Acid-Base Equilibrium Constants

The equilibrium constants for acid-base reactions, Ka and Kb, provide insights into the strength of an acid or a base. The acid dissociation constant (Ka) is a measure of how completely an acid releases hydrogen ions in water, whereas the base dissociation constant (Kb) measures the degree to which a base accepts hydrogen ions. These constants are determined through equilibrium concentrations of the products and reactants in a dissociation reaction.

For the base RNH in our exercise, the calculation of Kb involves using the concentration of RNH that has accepted a proton to become RNH2+, and the concentration of hydroxide ions (\([OH^-]\)) formed in the reaction. The relationship between Ka and Kb is given by the equation:\[Ka \times Kb = Kw\]where \(Kw\) is the ion-product constant of water, which is always \(1.0 \times 10^{-14}\) at 25°C. Using the calculated Kb and the formula above, we can then find Ka for the conjugate acid RNH2+. These calculations are fundamental to understanding the behavior of solutions in acid-base chemistry and making predictions about reaction direction and extent.