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 pH of this solution?

Short answer

The molar concentration of the \(\mathrm{OH}^{-}\) ions in the given solution is approximately \(1.702 \times 10^{-3}\mathrm{M}\). The pH value of the 0.724 \(\mathrm{M}\) hypobromite solution is approximately 11.231, indicating that the solution is alkaline or basic.

Step-by-step solution

Write the equilibrium expression for the reaction

For the given hypobromite ion \(\mathrm{(BrO}^{-})\), the equilibrium reaction with water can be written as: \[\mathrm{BrO}^{-} + \mathrm{H_{2}O} \rightleftharpoons \mathrm{BrOH} + \mathrm{OH}^{-}\] Now, using the base dissociation constant \(K_b = 4.0 \times 10^{-6}\), we can write the equilibrium expression: \[K_b = \frac{[\mathrm{BrOH}][\mathrm{OH}^{-}]}{[\mathrm{BrO}^{-}]}\]

Use the initial concentration of hypobromite given in the problem

We are given that the molar concentration of hypobromite ion is 0.724 \(\mathrm{M}\). Let's consider the change in concentrations at the equilibrium state; - Initial concentration of hypobromite ion, \([\mathrm{BrO}^-]_0 =0.724\mathrm{M}\) - At equilibrium, \([\mathrm{BrO}^-]=[0.724-x]\mathrm{M}\) - Let \(x\) be the concentration of \(\mathrm{OH}^{-}\) and \(\mathrm{BrOH}\) ions at equilibrium, so \([\mathrm{OH}^{-}]=[\mathrm{BrOH}]=x\mathrm{M}\)

Substitute equilibrium concentrations in \(K_b\) expression and solve for x

Now, using the \(K_b\) expression, we have: \[K_b = \frac{[\mathrm{BrOH}][\mathrm{OH}^{-}]}{[\mathrm{BrO}^{-}]} = \frac{x^2}{0.724-x}\] Plug in the given value of \(K_b\): \[4.0 \times 10^{-6} = \frac{x^2}{0.724-x}\] Now we will solve this equation for the value of x: \[x^2\approx4.0 \times 10^{-6}(0.724)=2.896\times10^{-6}\] \[x\approx \sqrt{2.896\times10^{-6}}\approx1.702\times 10^{-3}\] Therefore, the molar concentration of the \(\mathrm{OH}^{-}\) ions in the given solution is approximately \(1.702 \times 10^{-3}\mathrm{M}\).

Calculate pOH and pH

We now have the molar concentration of \(\mathrm{OH}^{-}\), and we can calculate the \(\mathrm{pOH}\): \[\mathrm{pOH} = -\log{[\mathrm{OH}^{-}]} = -\log{(1.702 \times 10^{-3})}=2.769\] Now, to find the pH, we'll use the relationship: \[\mathrm{pH} + \mathrm{pOH} = 14\] Finally, calculate the pH: \[\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 2.769 = 11.231\]

Interpret the result

The pH value of the 0.724 \(\mathrm{M}\) hypobromite solution is approximately 11.231. Since the pH is greater than 7, the solution is alkaline or basic.

Key concepts

Base Dissociation Constant (Kb)

Understanding the base dissociation constant (Kb) is critical for studying the properties of bases in solution. In chemistry, the base dissociation constant is a measure of how fully a base dissociates into its constituent ions in water. It is an equilibrium constant which means it provides information on the ratio between the concentrations of the products and the reactants once the reaction has settled into a state of balance.

For a weak base like hypobromite ion \(\mathrm{BrO}^{-}\), which only partially dissociates in water, the equilibrium can be expressed as:\[\mathrm{BrO}^{-} + \mathrm{H_{2}O} \rightleftharpoons \mathrm{BrOH} + \mathrm{OH}^{-}\]Here, the concentration of undissociated hypobromite \(\mathrm{BrO}^{-}\), the product water (though it is often omitted since it is in great excess), the hydroxide ions \(\mathrm{OH}^{-}\), and the hypobromous acid \(\mathrm{BrOH}\) are all part of the equilibrium. The value of Kb is specific to each base and at a given temperature, allows us to predict the extent to which the base will dissociate in solution. A high Kb value indicates a strong base that dissociates easily, whereas a low Kb implies a weak base.

Molar Concentration

Molar concentration, also expressed as molarity, is a measure of the concentration of a solute in a solution. It is denoted by the symbol M and is defined as the moles of solute divided by the liters of solution. When calculating the pH of a solution, understanding molarity is essential because it directly relates to the number of moles of hydroxide ions present in a solution.

For example, in our problem, the molar concentration of hypobromite ion \(\mathrm{BrO^{-}}\) in the solution is given as 0.724 M. This is a starting point to determine the changes in the system as it reaches equilibrium. From this number, we can further calculate the shift in concentration of reactants and products as they establish equilibrium. Knowledge of the molar concentration is fundamental not just for stoichiometric calculations but also for equilibrium and pH calculations in chemistry.

Equilibrium Expression

Writing the Expression

Once the balanced chemical equation for a reaction is known, the equilibrium expression can be written. It is the ratio of the multiply product of the concentrations of the products to the multiply product of the concentrations of the reactants, with each raised to the power of its coefficient in the balanced equation.

For the base dissociation reaction, the equilibrium expression would look like this:\[K_b = \frac{[\mathrm{BrOH}][\mathrm{OH}^{-}]}{[\mathrm{BrO}^{-}]}\]

Understanding Equilibrium

At equilibrium, the rate at which the reactants convert into products is equal to the rate at which the products convert back into reactants. In the provided exercise, the equilibrium expression is crucial for calculating the concentration of hydroxide ions. This is because we need to solve for 'x,' the change in concentration associated with product formation.

pOH Calculation

The concept of pOH is an important part of understanding the chemistry of solutions. The pOH is the negative logarithm of the hydroxide ion concentration:\[\mathrm{pOH} = -\log{[\mathrm{OH}^{-}]}\]. It is directly related to pH through the equation \[\mathrm{pH} + \mathrm{pOH} = 14\] (at 25°C).

Calculating the pOH is a necessary step towards finding the pH of basic solutions, such as the hypobromite solution in our exercise. Once the concentration of hydroxide ions is known from the equilibrium calculations, the pOH can be easily determined. This pOH value then serves as a stepping stone to calculate the pH, giving insight into the acidity or basicity of the solution. A low pOH corresponds to high hydroxide ion concentration and a basic solution, while a high pOH indicates low hydroxide ion concentration and an acidic solution.