Question

A student dissolves 0.0100 mol of an unknown weak base in 100.0 \(\mathrm{mL}\) water and titrates the solution with 0.100 \(\mathrm{M} \mathrm{HNO}_{3}\) After 40.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{HNO}_{3}\) was added, the \(\mathrm{pH}\) of the resulting solution was \(8.00 .\) Calculate the \(K_{\mathrm{b}}\) value for the weak base.

Short answer

The Kb value for the unknown weak base is approximately 4.29 x 10⁻⁵.

Step-by-step solution

Determine moles of weak base and HNO3 added

Initially, we have 0.0100 mol of the weak base dissolved in 100.0 mL of water. During titration, we add 40.0 mL of 0.100 M HNO3. To determine the moles of HNO3 added, multiply its volume and concentration: Moles of HNO3 = volume x concentration Moles of HNO3 = 40.0 mL x 0.100 M = 0.0040 mol

Calculate the moles of unreacted base

Since the weak base and HNO3 react in a 1:1 ratio during titration, the moles of unreacted base can be found by subtracting the moles of HNO3 added from initial moles of the weak base: Moles of unreacted base = initial moles of base - moles of HNO3 Moles of unreacted base = 0.0100 mol - 0.0040 mol = 0.0060 mol

Calculate moles of OH- present after titration

We are given the pH of the solution after adding 40 mL of HNO3, which is 8.00. Using the relation between pH and pOH: pOH = 14 - pH pOH = 14 - 8.00 = 6.00 Now, using the pOH value, we can calculate the concentration of OH- ions present in the solution: [OH-] = \(10^{-pOH}\) = \(10^{-6.00}\) = 1.0 x 10⁻⁶ M To find the moles of OH-, we will multiply this concentration by the total volume of the solution, which is the initial 100 mL of water plus the 40 mL of HNO3 added: Total volume = 100.0 mL + 40.0 mL = 140.0 mL Moles of OH- = [OH-] x total volume Moles of OH- = 1.0 x 10⁻⁶ M x 140.0 mL = 1.4 x 10⁻⁴ mol

Determine moles of reacted base

Since each mole of reacted base produces one mole of OH-, we can determine the moles of reacted base: Moles of reacted base = moles of OH- Moles of reacted base = 1.4 x 10⁻⁴ mol

Calculate the Kb value for the weak base

Now, we can calculate the Kb value for the weak base using the relation: Kb = \(\frac{[OH-][unreacted\ base]}{[reacted\ base]}\) First, divide the moles by the total volume to get the concentrations: [OH-] = \(\frac{1.4 \times 10^{-4}\ mol}{140.0\ mL}\) = 1.0 x 10⁻⁶ M [unreacted base] = \(\frac{0.0060\ mol}{140.0\ mL}\) = 4.29 x 10⁻⁵ M [reacted base] = \(\frac{1.4 \times 10^{-4}\ mol}{140.0\ mL}\) = 1.0 x 10⁻⁶ M Now, substitute these concentrations into the Kb expression: Kb = \(\frac{(1.0 \times 10^{-6})(4.29 \times 10^{-5})}{(1.0 \times 10^{-6})}\) = 4.29 x 10⁻⁵ The Kb value for the weak base is approximately 4.29 x 10⁻⁵.

Key concepts

Acid-Base Equilibrium

In chemistry, acid-base equilibrium describes the balance between acidic and basic species in a solution. This concept is crucial when studying titrations, which involve adding a known concentration of an acid to determine the concentration of a base, or vice versa. During titrations, acids and bases react to form water and potentially other products, reaching an equilibrium point. This equilibrium is governed by reversible reactions where weak acids and bases partially ionize. In weak base titrations, the focus is on how weak bases equilibrate with their conjugate acids after reaction with the titrant (strong acid). Such systems demonstrate dynamic equilibrium, continuously adjusting to changes until a stable concentration of products and reactants is achieved. Understanding this equilibrium allows chemists to calculate various parameters, such as the strength of the base involved in the reaction. This process is crucial to determining the base's ion product constant, or Kb, as illustrated in the example calculation.

Kb Constant

The Kb constant, or base dissociation constant, measures the strength of a base in solution. It represents how efficiently a base accepts protons and forms ionic species with water. To calculate Kb, we use the formula: \[ Kb = \frac{[OH^-][B^+]}{[B]} \] where \( [OH^-] \) is the hydroxide ion concentration, \( [B^+] \) is the concentration of the conjugate acid, and \( [B] \) is the concentration of the base. High values of Kb indicate a stronger base, as more base molecules dissociate into ions, while lower values denote weaker bases. During titrations, by knowing the pH or pOH, and subsequently the concentration of ions, one can determine the Kb value. For example, having the determined moles of unreacted and reacted base after a titration allows the calculation of Kb by considering how much of the weak base remains versus what has dissociated. This constant is essential for predicting the behavior of bases in various chemical contexts.

pH and pOH

The concepts of pH and pOH are fundamental in understanding the acidity and basicity of solutions. pH (\[ \text{pH} = -\log [H^+] \]) measures the concentration of hydrogen ions, whereas pOH (\[ \text{pOH} = -\log [OH^-] \]) measures hydroxide ions. For a solution at 25°C, pH and pOH are related by the equation \( \text{pH} + \text{pOH} = 14 \). Knowing either allows the determination of the other, which is particularly valuable during titration processes. For instance, using the pH of 8.00 in our example, the pOH is 6.00 by subtraction from 14. This relationship helps in determining the concentration of hydroxide ions in solution, which was crucial in calculating the moles of OH⁻ present after the titration, and consequently, the degree of base dissociation. Understanding these values aids in comprehending the overall dynamics of acid-base reactions and their equilibria.

Molarity Calculations

Molarity is a measure of the concentration of solutes in a solution, given in moles per liter (mol/L). It is used to express concentration levels, allowing chemists to perform precise calculations when mixing and reacting chemical components. For titrations, knowing the molarity of titrants (e.g., HNO3) is key to determining how much of it reacts with the base and how much remains. This is done by considering the volume of the solution and its molarity, as shown in the example where 0.100 M of sulfuric acid was used. Calculations of molarity help in assessing both the initial and equilibrium states of reaction components. When the pH of a solution changes, the resulting concentration of ions can be reflected back to these molarity values, facilitating calculations like those involving Kb. Through stoichiometry, and the known reaction ratios, you can calculate unknown concentrations in the reaction, thereby thoroughly understanding the chemical interplay. Molarity calculations shape understanding in titrations by giving quantifiable meaning to chemical changes.