Question

The \(\mathrm{pH}\) of a \(0.063-M\) solution of hypobromous acid \((\mathrm{HOBr}\) but usually written \(\mathrm{HBrO}\) ) is 4.95. Calculate \(K_{\mathrm{a}}\).

Short answer

The acid dissociation constant (Ka) of hypobromous acid is approximately \(2.0 \times 10^{-9}\).

Step-by-step solution

Write the dissociation reaction for hypobromous acid (HBrO)

HBrO is a weak acid, so it partially dissociates into its ions in water as follows: \[ \mathrm{HBrO} \rightleftharpoons \mathrm{H^+} + \mathrm{BrO^-} \]

Define the Ka expression for the reaction

For this dissociation reaction, we can write the Ka expression as: \[K_\mathrm{a} = \frac{[\mathrm{H^+}][\mathrm{BrO^-}]}{[\mathrm{HBrO}]} \]

Use the pH to find the hydrogen ion concentration [H+]

The pH of the solution is 4.95. We can use the relationship between pH and hydrogen ion concentration to find [H+]: \[ \mathrm{pH} = -\log [\mathrm{H^+}] \] Solving for [H+]: \[ [\mathrm{H^+}] = 10^{-\mathrm{pH}} = 10^{- 4.95} \] \[[\mathrm{H^+}] = 1.122 \times 10^{-5} \, M \]

Set up an ICE table to relate the initial and equilibrium concentrations of all components

Since HBrO is a weak acid, we'll assume that its initial concentration doesn't change significantly during the dissociation process. Here's the ICE table: Initial: \[ [\mathrm{HBrO}]_0 = 0.063 \, M \quad [\mathrm{H^+}]_0 = 0 \, M \quad [\mathrm{BrO^-}]_0 = 0 \, M \] Change: \[ -x \quad +x \quad +x \] Equilibrium: \[ [\mathrm{HBrO}] = 0.063 - x \, M \quad [\mathrm{H^+}] = 1.122 \times 10^{-5} + x \, M \quad [\mathrm{BrO^-}] = x \, M \]

Substitute equilibrium concentrations into the Ka expression and solve for Ka

Now we can substitute the expressions from the ICE table into the Ka expression: \[K_\mathrm{a} = \frac{(1.122 \times 10^{-5} + x)(x)}{(0.063 - x)} \] Since HBrO is a weak acid, we can assume x is much smaller than the initial concentration of HBrO: \[K_\mathrm{a} \approx \frac{(1.122 \times 10^{-5})(x)}{0.063} \] Now, we can solve for x, which is the equilibrium concentration of BrO-: \[ x \approx \frac{[\mathrm{H^+}] K_\mathrm{a}}{K_\mathrm{a} + 1.12 \times 10^{-5}} \] Since the equilibrium concentration of H+ is known, we can solve for Ka: \[ K_\mathrm{a} \approx \frac{[\mathrm{H^+}]^2}{0.063} - [\mathrm{H^+}] \] \[ K_\mathrm{a} \approx \frac{(1.122 \times 10^{-5})^2}{0.063} - 1.122 \times 10^{-5} \] \[ K_\mathrm{a} \approx 2.0 \times 10^{-9} \] Therefore, the acid dissociation constant (Ka) of hypobromous acid is approximately \(2.0 \times 10^{-9}\).

Key concepts

pH Calculation

The pH scale is a measure of acidity or basicity in a solution. It ranges from 0 to 14, with lower values indicating higher acidity. The term pH stands for 'potential of hydrogen.'
It quantifies the concentration of hydrogen ions, \([\mathrm{H^+}]\), in a solution using the formula:

• \[\mathrm{pH} = -\log [\mathrm{H^+}]\]

From this equation, you can see how pH relates logarithmically to hydrogen ion concentration. This means that small changes in pH represent significant changes in hydrogen ion concentration.
For example, knowing the pH of a solution allows us to calculate \([\mathrm{H^+}]\) by rearranging the formula to:

• \[[\mathrm{H^+}] = 10^{-\mathrm{pH}}\]

In our given exercise with a pH of 4.95, plugging into this equation provides \([\mathrm{H^+}]\) as \(1.122 \times 10^{-5} \, M\).
Understanding pH is essential in chemistry for analyzing the acidity and reactivity of substances.

Weak Acids

Weak acids partially dissociate in water, unlike strong acids which dissociate completely.
This partial dissociation means that weak acids establish an equilibrium between their undissociated and dissociated forms.
For weak acids like hypobromous acid (HBrO), the dissociation can be represented by:

• \[\mathrm{HBrO} \rightleftharpoons \mathrm{H^+} + \mathrm{BrO^-}\]

The backward and forward reaction rates are comparable, reaching a balance point.
Because of partial dissociation, the concentration of produced ions is lower compared to the initial concentration of the weak acid.
Understanding weak acids is crucial for calculating the acid dissociation constant \(K_a\), as it quantifies the degree of ionization.
This understanding helps us predict acidity levels and chemical behavior of different solutions.

Equilibrium Expressions

Equilibrium expressions are vital in understanding how reactants and products coexist in a dynamic balance.
For any reversible reaction, such as the dissociation of a weak acid, an equilibrium expression can describe concentrations at equilibrium.

• For the dissociation of \(\mathrm{HBrO}\):\[K_a = \frac{[\mathrm{H^+}][\mathrm{BrO^-}]}{[\mathrm{HBrO}]}\]

Here, \(K_a\) is the acid dissociation constant, measuring the extent of dissociation of the acid.
A higher \(K_a\) suggests a stronger acid, indicating more dissociation, while a lower \(K_a\) indicates a weaker acid.
This equilibrium expression helps chemists understand the balance set up between undissociated acid and its ions.
By knowing \([\mathrm{H^+}]\) from the pH and expressing other concentrations relative to \([\mathrm{H^+}]\) using an ICE table, we can solve for \(K_a\).
This process elucidates how weak acids behave in solution.

ICE Table

The ICE table (Initial, Change, Equilibrium) is a helpful tool for organizing and calculating concentrations during chemical equilibria.
It helps track changes from initial concentrations to equilibrium states in chemical reactions.

• For hypobromous acid:Initial:\([\mathrm{HBrO}]_0 = 0.063 \, M\quad [\mathrm{H^+}]_0 = 0 \, M\quad [\mathrm{BrO^-}]_0 = 0 \, M\)Change:\(-x \quad +x \quad +x\)Equilibrium:\([\mathrm{HBrO}] = 0.063 - x \, M \quad [\mathrm{H^+}] = 1.122 \times 10^{-5} + x \, M \quad [\mathrm{BrO^-}] = x \, M\)

The ICE table simplifies complex calculations by organizing data step-by-step.
In this exercise, since the solution is a weak acid, an assumption is made that \(x\) is very small and can be approximated for easier calculation.
Using the ICE results, we substitute into the equilibrium expression to solve for \(K_a\), helping us understand the degree of dissociation at equilibrium.
Training ourselves to use ICE tables enhances problem-solving skills in equilibrium calculations.