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 \(K_b\) value for the unknown weak base is approximately \(6.67 \times 10^{-7}\).

Step-by-step solution

Calculate the initial concentration of the weak base and HNO3

First, we find the initial concentration of weak base and HNO3. The weak base is dissolved in 100.0 mL of water, so we can calculate its initial concentration by dividing the given moles by the volume in liters: \[Initial \ concentration \ of \ weak \ base = \frac{0.0100 \ mol}{0.100 \ L} = 0.100 \ M\] The concentration of HNO3 is given as 0.100 M.

Calculate the moles of HNO3 added during titration

Next, we will find the moles of HNO3 added during titration. We are given the volume of HNO3 used (40.0 mL) and its concentration (0.100 M). To find the moles of HNO3 added, multiply the volume in liters by the concentration: \[Moles \ of \ HNO_3 = 0.100 \ M \times 0.0400 \ L = 0.00400 \ mol\]

Find the moles of the weak base and its conjugate acid after titration

During the titration, the weak base reacts with HNO3 to form its conjugate acid. The moles of weak base and HNO3 reacted are equal. Therefore, since 0.00400 moles of HNO3 were added, 0.00400 moles of the weak base have reacted: \[Moles \ of \ weak \ base \ after \ titration = 0.0100 \ mol - 0.00400 \ mol = 0.00600 \ mol\] Since the weak base reacts 1:1 with HNO3, the moles of the conjugate acid will be the same as the moles of reacted HNO3: \[Moles \ of \ conjugate \ acid = 0.00400 \ mol\]

Calculate the concentrations of the weak base and its conjugate acid after titration

After the titration, the total volume of the solution is 100.0 mL + 40.0 mL = 140.0 mL or 0.140 L. Now, we can find the concentrations of the weak base and its conjugate acid after titration by dividing their moles by 0.140 L: \[Concentration \ of \ weak \ base = \frac{0.00600 \ mol}{0.140 \ L} = 0.0429 \ M\] \[Concentration \ of \ conjugate \ acid = \frac{0.00400 \ mol}{0.140 \ L} = 0.0286 \ M\]

Calculate the Kb value for the weak base

First, we will find the pOH of the solution from the given pH value. The sum of pH and pOH is equal to 14, so: \[pOH = 14 - pH = 14 - 8.00 = 6.00\] Now we can find the hydroxide ion concentration, [OH-], from the pOH: \[[OH^-] = 10^{-pOH} = 10^{-6.00} = 1.00 \times 10^{-6} \ M\] We will use the Kb expression for the weak base dissociation equilibrium: \[K_b = \frac{[OH^-][conjugate \ acid]}{[weak \ base]}\] Plugging in the concentrations we have calculated: \[K_b = \frac{(1.00 \times 10^{-6})(0.0286)}{0.0429} = 6.67 \times 10^{-7}\] So, the Kb value for the unknown weak base is approximately \(6.67 \times 10^{-7}\).

Key concepts

Titration

Titration is a common laboratory method used to determine the concentration of an unknown solution. In our context, it is utilized to establish the characteristics of a weak base by introducing a known amount of acid. This method involves gradually adding a titrant, whose concentration is known, to a solution until the reaction reaches a point where the number of moles of one substance equals the number of moles of the other. This point is often marked by a noticeable change in a property of the solution, most commonly its pH. During titration, we carefully measure the volume of the titrant used, which allows us to calculate the unknown concentration of the solution being tested. This is possible through stoichiometric calculations based on the balanced chemical equation for the reaction involved.

Weak Base

A weak base is a base that does not completely dissociate into its ions in a solution. This means that in water, a weak base exists in equilibrium with its ions. The equilibrium can be described by the base dissociation constant, Kb. Identifying whether a base is weak or strong is crucial, as it influences how the base will react during titration. In the exercise, the unknown weak base reacts partially with the added nitric acid (HNO3), which is a strong acid. This incomplete dissociation influences the changes in pH observed during the titration process. Understanding weak bases is key to predicting how they will behave in different scenarios, enabling us to calculate important values such as the Kb, which describes the base's strength in terms of its ability to accept protons.

pH Calculation

Calculating the pH of a solution during a titration process involves understanding the dynamic changes in the concentration of hydrogen ions. Initially, when the weak base is titrated with a strong acid like HNO3, the solution's pH shifts as the base is progressively neutralized. The pH scale, which ranges from 0 to 14, is a measure of acidity; lower values correspond to more acidic solutions, while higher values are indicative of basic or alkaline solutions. In the given problem, after adding 40.0 mL of HNO3, the solution reaches a pH of 8.00, which is slightly basic, confirming the partial neutralization typical of a weak base. The relation between pH and pOH (since pH + pOH = 14) is used to analyze the solution's effective hydroxide ion concentration, ultimately affecting calculations related to base dissociation.

Kb Value Determination

Determining the Kb value for a weak base is fundamental to understanding its properties. The Kb, or base dissociation constant, quantifies the base's ability to accept protons from water, forming its conjugate acid and hydroxide ions.In our step-by-step solution, we find Kb by considering the equilibrium concentrations of the ions in question after the titration reaches a significant point. Firstly, we derive the pOH from the known pH, translating it into the concentration of hydroxide ions.Then, we employ the formula: \[ K_b = \frac{[OH^-][conjugate\ acid]}{[weak\ base]} \]Substituting the calculated concentrations into this expression allows us to determine the precise value of Kb. In this example, the Kb reveals the weak base's relatively low ionic dissociation, characteristic of weak bases.