Problem 76 - Chemistry: The Central Science

Question

Calculate the molar concentration of \(\mathrm{OH}^{-}\) ions in a \(0.550 \mathrm{M}\) solution of hypobromite ion \(\left(\mathrm{BrO}^{-} ;\right.\) \(K_{b}=4.0 \times 10^{-6}\) ). What is the \(\mathrm{pH}\) of this solution?

Step-by-Step Solution

Answer

The molar concentration of OH- ions in a 0.550 M solution of hypobromite ion (BrO-) with a base dissociation constant Kb of 4.0 × 10^(-6) is approximately \(3.57 \times 10^{-4}\) M. The pH of this solution is approximately 10.55.

Step 1: Identify the base dissociation reaction

In order to calculate the molar concentration of OH- ions, we must first understand the base dissociation reaction of the hypobromite ion (BrO-). For a weak base (B) such as hypobromite ion, its dissociation in water can be written as: B + H2O ⇌ BH+ + OH- In our case: \(\ce{BrO- + H2O <=> HOBr + OH-}\)

Step 2: Set up the base dissociation constant equation

Now that we have the base dissociation reaction, we can write the corresponding base dissociation constant (Kb) equation as: \(K_b = \frac{ [\ce{HOBr}] [\ce{OH-}] }{ [\ce{BrO-}] }\) We are given Kb = 4.0 × 10^(-6) and [BrO-] = 0.550 M.

Step 3: Construct the ICE table

An ICE (Initial, Change, Equilibrium) table helps us keep track of the changes in concentrations for each species involved in the reaction. Set up the ICE table for the base dissociation reaction: | | BrO- | H2O | HOBr | OH- | |--------|------|------|------|------| | Initial| 0.550| - | 0 | 0 | | Change | -x | - | +x | +x | | Eqbm | 0.550-x| - | x | x | Note that we don't include H2O in [ ] as it is a liquid and the concentration does not change. x represents the change in concentration of each ion.

Step 4: Calculate the concentration of OH- ions

We can now substitute the values from the equilibrium row of the ICE table into the base dissociation constant equation and solve for x, which represents the molar concentration of OH- ions: \(4.0 \times 10^{-6} = \frac{(x)(x)}{(0.550 - x)}\) As Kb is relatively small, we can make the approximation that x is much smaller compared to 0.550, so we can ignore x in the denominator: \(4.0 \times 10^{-6} = \frac{(x^2)}{0.550}\) Solve for x: \(x = [\ce{OH-}] \approx \sqrt{(4.0 \times 10^{-6})(0.550)} ≈ 3.57 \times 10^{-4} M\)

Step 5: Calculate pOH and pH

Now that we have the concentration of OH- ions, we can use the following relationship to find the pOH: \(pOH = - \log{[\ce{OH-}]}\) \(pOH ≈ - \log{(3.57 \times 10^{-4})} ≈ 3.45\) Finally, the relationship between pH and pOH is given by: \(pH + pOH = 14\) Now, calculate the pH: \(pH = 14 - pOH ≈ 14 - 3.45 ≈ 10.55\) Therefore, the pH of the solution is approximately 10.55.

Key Concepts

Base dissociation

When a base enters an aqueous solution, it undergoes a chemical process called base dissociation. In this process, the base reacts with water molecules to produce hydroxide ions (OH\(^-\)) and a corresponding conjugate acid. The base we are discussing here is the hypobromite ion (BrO\(^-\)).

The reaction can be represented as:

• BrO\(^-\) + H\(_2\)O ⇌ HOBr + OH\(^-\)

This is a reversible reaction, and it indicates that as BrO\(^-\) dissociates, it creates a balance between the reactants and products. The equilibrium state is essential for understanding the concentration of hydroxide ions produced in the solution.

To determine how well a base dissociates, we use the base dissociation constant, known as \(K_b\). This constant is specific to each base and provides insight into the base's strength. A lower \(K_b\) value indicates a weaker base, meaning it dissociates less in solution.

ICE table

An ICE table helps track concentration changes during a chemical reaction. This tool is invaluable in chemical equilibrium calculations, especially those involving weak acids or bases. ICE stands for Initial, Change, and Equilibrium. Here's how it works for our base dissociation reaction:

- **Initial:** Start with the initial concentrations. For BrO\(^-\), we have 0.550 M, while HOBr and OH\(^-\) are initially 0.- **Change:** As the reaction progresses, the concentration of BrO\(^-\) decreases by \(x\), while the concentrations of HOBr and OH\(^-\) increase by \(x\).- **Equilibrium:** At equilibrium, the concentration of BrO\(^-\) is \(0.550 - x\), while for both HOBr and OH\(^-\), it is \(x\).

Using the ICE table helps visualize and set up equations needed to solve for the unknown concentration changes. It aligns perfectly with applying \(K_b\) to find \(x\), which aids in calculating the hydroxide ion concentration.

pH calculation

After determining the concentration of hydroxide ions (\([ ext{OH}^-]\)), we can use this to calculate the pOH and ultimately the pH of the solution. These calculations are key in acid-base chemistry for understanding the basicity or acidity of a solution.

The pOH is calculated using the formula:

- \(pOH = -\log{[ ext{OH}^-]}\)

After calculating the pOH, you can find pH because of the simple relationship between pH and pOH in water:

- \(pH + pOH = 14\)

Using this formula allows you to find the pH, representing how acidic or basic the solution is. For instance, a basic solution, such as the one in our example with \( ext{pH} \approx 10.55 \), indicates its basic nature, as its pH is above 7.

Molar concentration

Molar concentration, also known as molarity, measures the amount of a solute in a given volume of solution. It’s expressed in moles per liter (M). In this exercise, \( ext{molar concentration} \) is vital for determining both the concentration of the initial base, BrO\(^-\), and the resulting OH\(^-\) ions.

Here's how it works:

• The initial molar concentration of BrO\(^-\) was provided as 0.550 M, which means there were 0.550 moles of BrO\(^-\) in one liter of solution.
• When BrO\(^-\) dissociates, we use its change in concentration, represented by \(x\) in the ICE table, to determine the molar concentration of the hydroxide ions (OH\(^-\)).
• From our calculation, OH\(^-\) ends up having a concentration of approximately 3.57 \(\times\) 10\(^{-4}\) M.

Understanding molar concentration is crucial in chemistry as it provides the means to relate chemical quantities to reaction extents and equilibrium, aiding in computations like pH calculation.