Question

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

Short answer

The \(K_{a}\) of hypobromous acid (HBrO) can be calculated using the formula \[K_{a}=\frac{([\mathrm{H}^+])([\mathrm{BrO^-}])}{0.063}\] where \([\mathrm{H}^+]\) and \([\mathrm{BrO^-}]\) are the concentrations of hydrogen and bromate ions, respectively. After calculating these concentrations from the given pH value, we find that the \(K_{a}\) of HBrO is approximately \(1.79 \times 10^{-9}\).

Step-by-step solution

Find the concentration of hydrogen ions [H+] from the given pH

To find the concentration of H+ ions, we can use the formula: \[ \mathrm{pH} = -\log_{10} [\mathrm{H}^+]\] We can rearrange the formula to find the concentration of H+ ions: \[ [\mathrm{H}^+] = 10^{-\mathrm{pH}}\] Now, we have the pH value (4.95). Let's plug that in the formula to find the concentration of hydrogen ions: \[ [\mathrm{H}^+] = 10^{-4.95}\]

Calculate the Concentration of BrO-

Since the hypobromous acid (HBrO) dissociates as follows: \[ \mathrm{HBrO} \rightleftharpoons \mathrm{H}^+ + \mathrm{BrO}^-\] We can assume that the initial concentration of BrO- ions is the same as the concentration of hydrogen ions formed during dissociation since one molecule of HBrO generates one molecule of BrO-. Therefore, the concentration of BrO- ions is equal to the concentration of H+ ions. \[ [\mathrm{BrO}^-] = [\mathrm{H}^+]\] Using the value of [H+] we found in step 1, we can determine the concentration of BrO- ions.

Calculate the Ka of Hypobromous Acid

Now, we have all the necessary values for calculating the Ka of hypobromous acid. The equilibrium expression for the dissociation of HBrO in water is: \[K_a = \frac{[\mathrm{H}^+] [\mathrm{BrO}^-]}{[\mathrm{HBrO}]}\] Note that we were given the initial concentration of HBrO (0.063 M). As the dissociation takes place, it is assumed that the change in the concentration of HBrO is -x, while H+ and BrO- concentration increases by x. Therefore, the concentration of HBrO will be 0.063 - x. Using the equilibrium concentrations, plug in the values for H+ ions, BrO- ions, and the initial concentration of HBrO: \[K_{a}=\frac{([\mathrm{H}^+])([\mathrm{BrO^-}])}{0.063-x}\] Since x is very small compared to 0.063, we can consider (0.063 - x) to be approximately equal to 0.063: \[K_{a}=\frac{([\mathrm{H}^+])([\mathrm{BrO^-}])}{0.063}\] Now just insert the concentration values for H+ and BrO- ions calculated earlier and find the value of Ka for hypobromous acid.

Key concepts

pH Calculation

pH is a measure of acidity or alkalinity of a solution. It is calculated by taking the negative logarithm (base 10) of the hydrogen ion concentration \[\mathrm{[H^+]}\]. The formula is written as:

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

This equation tells us how the concentration of hydrogen ions affects the pH value, with lower pH indicating a higher concentration of hydrogen ions. For instance, if the pH of a solution is 4.95, that means \([\mathrm{H}^+]\) is calculated using \(10^{-4.95}\). Understanding this relationship is crucial as it forms the basis for calculating the acid dissociation constant (\(K_a\)). Whenever you know the pH, it's straightforward to find out the hydrogen ion concentration. This is an essential step in solving equilibrium problems related to weak acids.

Hypobromous Acid

Hypobromous acid, denoted as HBrO or HOBr, is a weak acid that partially dissociates in aqueous solutions. It can be represented by the following equilibrium reaction:

• \( \mathrm{HBrO} \rightleftharpoons \mathrm{H}^+ + \mathrm{BrO}^- \)

This expression shows us that hypobromous acid breaks down into hydrogen ions (\([\mathrm{H}^+]\)) and bromate ions (\([\mathrm{BrO}^-]\)). As a weak acid, HBrO does not completely dissociate, so its equilibrium constants and ion concentrations are needed to determine its behavior in solution. Knowing its initial concentration and the pH allows us to find how much it dissociates and hence, helps in calculating \(K_a\), the acid dissociation constant, which quantifies the strength of the acid in solution.

Equilibrium Expressions

An equilibrium expression in the context of acid dissociation is an equation that represents the relative concentrations of the acid and its dissociated ions in a solution at equilibrium. For hypobromous acid, it is given by:

• \(K_a = \frac{[\mathrm{H}^+][\mathrm{BrO}^-]}{[\mathrm{HBrO}]}\)

The equilibrium constant (\(K_a\)) reflects the extent to which HBrO dissociates into \(\mathrm{H}^+\) and \(\mathrm{BrO}^-\) ions. At equilibrium, the concentration of undissociated HBrO is adjusted by subtracting the concentration of the hydrogen ions already determined using the pH. If a solution starts with 0.063 M of HBrO and assuming minimal change in its concentration due to dissociation (since HBrO is a weak acid), we approximate the equilibrium as:

• \(K_a = \frac{[\mathrm{H}^+]\cdot[\mathrm{BrO}^-]}{0.063}\)

This framework allows us to calculate \(K_a\) precisely, providing insight into the acid's reactivity and properties in the solution.