Problem 187 - Chemistry An Atoms First Approach

Question

A \(2.14-\mathrm{g}\) sample of sodium hypoiodite is dissolved in water to make 1.25 L of solution. The solution \(\mathrm{pH}\) is \(11.32 .\) What is \(K_{\mathrm{b}}\) for the hypoiodite ion?

Step-by-Step Solution

Answer

The value of \(K_b\) for the hypoiodite ion is \(4.29 \times 10^{-4}\), which is calculated using the given mass of sodium hypoiodite, volume of the solution, and pH. Concentration of sodium hypoiodite and OH- ions are calculated, and finally, the Kb formula is used to determine the value.

Step 1: Identify important information

Given: Mass of sodium hypoiodite = 2.14 g and volume of the solution = 1.25 L, pH = 11.32.

Step 2: Calculate the concentration of sodium hypoiodite

First, we need to find the molar mass of sodium hypoiodite (NaOI). The molecular weight of Na = 22.99 g/mol, O = 16.00 g/mol, and I = 126.90 g/mol. Therefore, molar mass of NaOI can be calculated as: Molar mass of NaOI = 22.99 g/mol + 16.00 g/mol + 126.90 g/mol = 165.89 g/mol Now, we will find the moles of sodium hypoiodite using the given mass and molar mass: moles of NaOI = (2.14 g) / (165.89 g/mol) = 0.0129 mol The concentration of sodium hypoiodite in the solution can be calculated by dividing the moles by the volume: [NaOI] = (0.0129 mol) / (1.25 L) = 0.01032 M

Step 3: Calculate the concentration of OH- ions

We're given the pH of the solution which is 11.32. We can find the concentration of OH- ions using the following relationship: \(pH + pOH = 14\) \(pOH = 14 - pH\) \(pOH = 14 - 11.32 = 2.68\) To find the concentration of OH- ions, we'll use the formula: [OH-] = \(10^{-pOH}\) [OH-] = \(10^{-2.68}\) = 0.002095 M

Step 4: Determine the Kb value for the hypoiodite ion

Since sodium hypoiodite dissociates into sodium cations (Na+) and hypoiodite anions (OI-), we can assume that the concentrations of OI- and Na+ are equal. Hence, the concentration of OH- ions is equal to the concentration of OI- ions. Now we can use the Kb formula to calculate the Kb value: \(K_b = \frac{[\text{OI}^-][\text{OH}^-]}{[\text{NaOI}]}\) \(K_b = \frac{(0.002095)^2}{0.01032}\) \(K_b = 4.29 \times 10^{-4}\) Therefore, the value of Kb for the hypoiodite ion is \(4.29 \times 10^{-4}\).

Key Concepts

Understanding Sodium Hypoiodite

Sodium hypoiodite is a chemical compound composed of sodium (Na), oxygen (O), and iodine (I) with the molecular formula NaOI. It is a type of salt that can dissolve in water, leading to the dissociation of its ions. When sodium hypoiodite is dissolved in water, it separates into sodium ions (Na\(^+\)) and hypoiodite ions (OI\(^-\)). This dissociation is crucial for understanding reactions and equilibrium processes in solution, particularly when assessing the compound's basicity.Sodium hypoiodite is typically used in applications that involve oxidation and disinfection due to its oxidative properties. Its utility in educational chemistry exercises often revolves around understanding how it behaves in solution, such as calculating its equilibrium constants like Kb, which reflects the extent to which it can donate hydroxide ions (OH\(^-\)) in water. Understanding these reactions can help you grasp general chemical behavior and equilibrium in reactions involving weak bases.

Molarity Calculation Simplified

Molarity is a measure of concentration that expresses the number of moles of a solute present in one liter of solution. Understanding how to calculate molarity is essential in chemistry as it helps quantify substances in reactions and predict the outcomes of chemical processes.To calculate the molarity of sodium hypoiodite in solution, follow these steps:

• Determine the molar mass: For sodium hypoiodite, you add the atomic masses of sodium (22.99 g/mol), oxygen (16.00 g/mol), and iodine (126.90 g/mol), which equals 165.89 g/mol.
• Calculate moles of solute: Use the formula \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\). For a 2.14 g sample of NaOI, the moles would be \(\frac{2.14}{165.89} = 0.0129\) mol.
• Find molarity: Divide moles by the volume of the solution in liters. With a solution volume of 1.25 L, the molarity is \(\frac{0.0129}{1.25} = 0.01032\) M.

This simple methodological calculation allows you to find the concentration of any substance when similar data is available.

Interconnection of pH and pOH

In chemistry, the concept of pH and pOH provides insight into the acidic or basic nature of a solution. pH is a measure of the hydrogen ion concentration \(\text{H}^{+}\) in a solution, while pOH measures the hydroxide ion concentration \[OH^-\].One important relationship that connects them is: \[\text{pH} + \text{pOH} = 14\] at 25°C. This means if you know the pH, you can directly calculate the pOH and vice versa.For the given problem, since the pH is 11.32:

• Calculate pOH: \(\text{pOH} = 14 - 11.32 = 2.68\)
• Convert to concentration: Use \([OH^-] = 10^{-\text{pOH}}\) to find \([OH^-] = 10^{-2.68} = 0.002095 \) M.

This interdependence simplifies the calculations for determining concentrations and helps make predictions about reactions occurring in solution.

Understanding Chemical Equilibrium

Chemical equilibrium occurs in a reversible reaction when the rate of the forward reaction equals the rate of the backward reaction, leading to a steady state where the concentrations of reactants and products remain constant over time.For weak bases like the hypoiodite ion \([\text{OI}^-]\), calculating the base dissociation constant (Kb) involves using equilibrium concepts.Here's how you can use equilibrium constants to find Kb for the hypoiodite ion:

• Recognize that in a sodium hypoiodite solution: \([\text{NaOI}]\) dissociates to \([\text{Na}^+]\) and \([\text{OI}^-]\).
• Establish equilibrium expressions: \(K_b = \frac{[OI^-][OH^-]}{[NaOI]}\).
• Use given concentrations: From the problem, \([OH^-] = 0.002095 \text{ M}\), and assume \([OI^-] = [OH^-]\), thus calculate \(K_b = \frac{(0.002095)^2}{0.01032}\).
• Result: \(K_b = 4.29 \times 10^{-4}\).

Understanding chemical equilibrium concepts is crucial for predicting the direction of reactions and calculating the strength or weakness of acids and bases in solution.