Question

Calculate the molar concentration of \(\mathrm{OH}^{-}\) in a \(0.724 \mathrm{M}\) solution of hypobromite ion \(\left(\mathrm{BrO}^{-} ; K_{b}=4.0 \times 10^{-6}\right) .\) What is the \(\mathrm{pH}\) of this solution?

Short answer

The molar concentration of \(\mathrm{OH}^{-}\) in the \(0.724\ \mathrm{M}\) solution of hypobromite ion \(\mathrm{BrO}^{-}\) is approximately \(1.702 \times 10^{-3}\ \mathrm{M}\). The \(\mathrm{pH}\) of this solution is approximately 11.23.

Step-by-step solution

Write the equilibrium expression for the reaction of \(\mathrm{BrO}^{-}\) with water

The reaction of the hypobromite ion (\(\mathrm{BrO}^{-}\)) with water can be written as: \[\mathrm{BrO}^{-} + \mathrm{H_{2}O} \rightleftharpoons \mathrm{OH}^{-} + \mathrm{HBrO}\] Now, we can write the expression for the \(K_{b}\) constant: \[K_{b} = \frac{[\mathrm{OH}^{-}][\mathrm{HBrO}]}{[\mathrm{BrO}^{-}]}\]

Define the initial and equilibrium concentrations

Let's assume that the change in the concentration of \(\mathrm{BrO}^{-}\) is represented by \(x\). At equilibrium, the concentrations will be: \([\mathrm{BrO}^{-}] = 0.724 - x\) \([\mathrm{OH}^{-}] = x\) \([\mathrm{HBrO}] = x\)

Substitute the equilibrium concentrations into the \(K_{b}\) expression

Plugging in the equilibrium concentrations into the \(K_{b}\) expression, we get: \[4.0 \times 10^{-6} = \frac{x\cdot x}{0.724 - x}\]

Solve for x

Since \(K_{b}\) is very small, we can make the assumption that \(x << 0.724\). In that case, we can approximate the denominator: \[0.724 - x \approx 0.724\] Now, solving for \(x\), we have: \[x^2 = 4.0 \times 10^{-6} \times 0.724\] \[x^2 = 2.896 \times 10^{-6}\] \[x \approx 1.702 \times 10^{-3}\] As \(x\) represents the concentration of \(\mathrm{OH}^{-}\), we have: \([\mathrm{OH}^{-}] \approx 1.702 \times 10^{-3}\ \mathrm{M}\)

Calculate the \(\mathrm{pOH}\) and then the \(\mathrm{pH}\)

Now that we have the concentration of \(\mathrm{OH}^{-}\), we can calculate the \(\mathrm{pOH}\) as: \[\mathrm{pOH} = -\log{[\mathrm{OH}^{-}]}\] \[\mathrm{pOH} = -\log{(1.702 \times 10^{-3})} \approx 2.77\] And finally, we can determine the \(\mathrm{pH}\) as: \[\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 2.77 \approx 11.23\] Therefore, the \(\mathrm{pH}\) of the given solution is approximately 11.23.

Key concepts

Molar Concentration

The concept of molar concentration is essential for understanding how much of a particular solute is present in a given volume of solution. In this context, we are interested in the molar concentration of the hydroxide ion (\(\mathrm{OH}^{-} \)) in a solution of hypobromite ions (\(\mathrm{BrO}^{-} \)). Knowing the initial molar concentration of the \(\mathrm{BrO}^{-} \), which is 0.724 M, helps us start our analysis of how the ions interact in the solution and achieve equilibrium.To determine the equilibrium molar concentration of a product like \(\mathrm{OH}^{-} \), we set up the equilibrium reaction of the \(\mathrm{BrO}^{-} \) ion with water. This helps us apply the basic principles of chemistry to calculate the changes in concentrations, which are elementary to reaching equilibrium. By using known equilibrium constants and the assumption that changes in concentrations are small relative to the initial values, we can effectively simplify and calculate the concentration of interest in the solution.

pH Calculation

Calculating the \(\mathrm{pH} \) of a solution helps us understand its acidity or basicity. In this exercise, we calculate the \(\mathrm{pH} \) from the concentration of hydroxide ions (\(\mathrm{OH}^{-} \)) in a basic solution. The \(\mathrm{pH} \) scale ranges from 0 to 14, with values below 7 indicating acidic solutions, values above 7 indicating basic solutions, and a value of 7 representing neutrality.To find the \(\mathrm{pH} \), we first determine the \(\mathrm{pOH} \) by taking the negative logarithm of the hydroxide ion concentration. This value reflects the basicity directly. Given that \(\mathrm{pH} + \mathrm{pOH} = 14 \), we can then easily solve for the \(\mathrm{pH} \). Knowing how to calculate \(\mathrm{pH} \) from \(\mathrm{pOH} \) allows us to understand the overall behavior of the solution in question in terms of acidity and basicity.

Equilibrium Constant (K_b)

The equilibrium constant, \(K_b\), is fundamental to understanding the behavior of weak bases in solution, such as the hypobromite ion \(\mathrm{BrO}^{-} \). \(K_b\) is a measure of the strength of a base: the larger the \(K_b\), the stronger the base, and the more it tends to form \(\mathrm{OH}^{-} \) ions when dissolved in water.In this exercise, we use the \(K_b\) value of \(4.0 \times 10^{-6}\) to set up an expression that represents the equilibrium state of the system. When setting up the \(K_b\) formula, we compare the concentrations of the products (\(\mathrm{OH}^{-} \) and \(\mathrm{HBrO} \)) to the concentration of the reactants (\(\mathrm{BrO}^{-} \)).Using the small \(K_b\) approximation, which assumes minimal change in the initial concentration due to the small equilibrium constant, simplifies the solving process. This approach helps students focus on the primary influences of equilibrium, improving conceptual understanding by reducing computational errors. Understanding \(K_b\) allows us to grasp how equilibrium in weak base reactions profoundly affects the \(\mathrm{pH} \) and general properties of the solution.