Question

What are \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in a \(2.86 \mathrm{M}\) solution of \(\mathrm{HBr}\) ?

Short answer

\([\mathrm{H}^{+}] = 2.86 \, \text{M}\) and \([\mathrm{OH}^{-}] \approx 3.50 \times 10^{-15} \, \text{M}\).

Step-by-step solution

Understanding the Problem

We need to find the concentrations of hydrogen ions (\(\left[\mathrm{H}^{+}\right]\)) and hydroxide ions (\(\left[\mathrm{OH}^{-}\right]\)) in a solution that contains 2.86 M of hydrobromic acid (HBr). HBr is a strong acid, so it dissociates completely in solution.

Dissociation of HBr in Water

Since HBr is a strong acid, it dissociates completely into its ions in water: \(\mathrm{HBr} \rightarrow \mathrm{H}^{+} + \mathrm{Br}^{-}\). This means the concentration of \(\mathrm{H}^{+}\) ions will be equal to the concentration of HBr, which is 2.86 M.

Calculating the Concentration of \(\left[\mathrm{H}^{+}\right]\)

The concentration of hydrogen ions \(\left[\mathrm{H}^{+}\right]\) is equal to the initial concentration of HBr because it dissociates fully. Therefore, \(\left[\mathrm{H}^{+}\right] = 2.86\, \text{M}\).

Using the Water Ion Product to Find \(\left[\mathrm{OH}^{-}\right]\)

In water at 25°C, the ion product \(K_w\) is always \(1.0 \times 10^{-14}\). We can use this to find the \(\left[\mathrm{OH}^{-}\right]\) concentration using the formula: \([\mathrm{H}^{+}][\mathrm{OH}^{-}] = K_w\). Replacing \(\left[\mathrm{H}^{+}\right]\) with 2.86 M:\[ [\mathrm{OH}^{-}] = \frac{K_w}{[\mathrm{H}^{+}]} = \frac{1.0 \times 10^{-14}}{2.86}\\]

Calculating \(\left[\mathrm{OH}^{-}\right]\)

Compute the concentration of \([\mathrm{OH}^{-}]\) using the formula: \[ [\mathrm{OH}^{-}] = \frac{1.0 \times 10^{-14}}{2.86} \approx 3.50 \times 10^{-15} \, \text{M}\]Thus, the concentration of hydroxide ions in the solution is approximately \(3.50 \times 10^{-15}\, \text{M}\).

Key concepts

Hydrogen Ion Concentration

Hydrogen ion concentration, denoted as \([\mathrm{H}^{+}]\), is a measure of the acidity in a solution. It tells us how many hydrogen ions are present in the solution at a given time. Understanding hydrogen ion concentration is essential when studying acids and bases, as it directly relates to the pH level of a solution.

For strong acids like hydrobromic acid (HBr), this becomes quite straightforward. HBr is classified as a strong acid because it dissociates completely when dissolved in water, meaning every molecule of HBr breaks down into one hydrogen ion \((\mathrm{H}^{+})\) and one bromide ion \((\mathrm{Br}^{-})\). This dissociation can be represented by the equation:

\[ \mathrm{HBr} \rightarrow \mathrm{H}^{+} + \mathrm{Br}^{-} \]

• Strong acids lead to direct hydrogen ion concentration.
• For a 2.86 M HBr solution, \([\mathrm{H}^{+}]\) is 2.86 M.

Knowing the hydrogen ion concentration helps calculate other characteristics of the solution, such as its pH, which is given by \(-\log[\mathrm{H}^{+}]\). This forms the basis for understanding the solution's acidic properties.

Dissociation of Strong Acids

Strong acids like hydrobromic acid have a unique characteristic. They dissociate entirely in water, meaning they separate into their constituent ions without leaving any undissociated molecules behind. This complete dissociation is what makes strong acids particularly effective at increasing the hydrogen ion concentration in a solution.
When HBr is dissolved, it fully dissociates according to: \[ \mathrm{HBr} \rightarrow \mathrm{H}^{+} + \mathrm{Br}^{-} \]Here, every mole of HBr contributes one mole of \(\mathrm{H}^{+}\).

• Complete dissociation implies all acid molecules split into ions.
• The concentration of \(\mathrm{H}^{+}\) equals the initial concentration of the acid.
• This property applies to other strong acids like HCl and HNO3.

Understanding this concept allows us to predict the ionic composition of solutions formed by strong acids and helps in further calculations involving chemical equilibria.

Water Ion Product (Kw)

The water ion product, or \(K_w\), is a fundamental concept when exploring acid-base equilibria. \(K_w\) explains the equilibrium condition of the self-ionization of water, where water molecules produce equal concentrations of hydrogen ions \(\mathrm{H}^{+}\) and hydroxide ions \(\mathrm{OH}^{-}\):
\[\mathrm{H}_2\mathrm{O} \leftrightarrow \mathrm{H}^{+} + \mathrm{OH}^{-}\]
At 25°C, \(K_w\) is defined as \(1.0 \times 10^{-14}\). This forms the basis for deducing the concentration of hydroxide ions when hydrogen ion concentration is known. For example, in an HBr solution, knowing \([\mathrm{H}^{+}]\) helps us find \([\mathrm{OH}^{-}]\) using:

\[[\mathrm{H}^{+}][\mathrm{OH}^{-}] = K_w\]Plugging in the known hydrogen ion concentration:\[[\mathrm{OH}^{-}] = \frac{1.0 \times 10^{-14}}{2.86}\]
These calculations show why \(K_w\) is a constant value central to understanding water's role in balancing acid-base reactions.

• Explains the self-ionization of water leading to ionic product.
• Used to calculate unknown ion concentrations in solutions.
• Central to determining solution pH and pOH.

Recognizing \(K_w\) helps predict how acidic or basic a solution is, driving home the importance of equilibrium in aqueous solutions.