Question

Calculate the pH of a solution prepared by mixing \(2.50 \mathrm{~g}\) of hypobromous acid \((\mathrm{HOBr})\) and \(0.750 \mathrm{~g}\) of \(\mathrm{KOH}\) in water. \(\left(K_{\mathrm{a}} \mathrm{HOBr}=2.5 \times 10^{-9}\right)\).

Short answer

Answer: The pH of the solution is 8.56.

Step-by-step solution

Calculate initial concentrations of HOBr and KOH

First, let's find the initial concentration of hypobromous acid (HOBr) in the solution: To do this, we'll divide the mass of HOBr by its molar mass and divide it by the total volume of the solution (assumed to be 1 Liter). Given \(2.50~g\) of HOBr, and considering the molar mass of HOBr \((88.91~\frac{g}{mol})\), we can find the moles of HOBr: \(\frac{2.50}{88.91} = 0.0281~mol\) Now, let's assume the solution is prepared in \(1~L\) of water, so we can find the initial concentration of HOBr: \([\mathrm{HOBr}]_i = \frac{0.0281~mol}{1~L} = 0.0281~M\) Next, let's find the initial concentration of potassium hydroxide (KOH) in the solution: Given \(0.750~g\) of KOH, and considering the molar mass of KOH \((56.11~\frac{g}{mol})\), we can find the moles of KOH: \(\frac{0.750}{56.11} = 0.0134~mol\) Now, let's assume the solution is prepared in \(1~L\) of water, so we can find the initial concentration of KOH: \([\mathrm{KOH}]_i = \frac{0.0134~mol}{1~L} = 0.0134~M\)

Calculate the neutralization reaction

Since KOH is a strong base and HOBr is a weak acid, they will react in a neutralization reaction: \(\mathrm{HOBr} + \mathrm{OH^-} \rightarrow \mathrm{H_2O} + \mathrm{OBr^-}\) The limiting reactant in this reaction is KOH, so let's calculate the amount of substances reacting and the resulting concentrations in the solution after the reaction: \([\mathrm{OH^-}]_{reacted} = [\mathrm{KOH}]_i = 0.0134~M\) The new concentration of HOBr will be: \([\mathrm{HOBr}]_{f} = [\mathrm{HOBr}]_i - [\mathrm{OH^-}]_{reacted} = 0.0281 - 0.0134 = 0.0147~M\) The concentration of NaOBr will be: \([\mathrm{OBr^-}]_{f} = [\mathrm{OH^-}]_{reacted} = 0.0134~M\)

Use Ka equation to find pH

Now, let's use the Ka equation for HOBr to find the concentration of H3O+ ions in the solution: \(K_a = \frac{[\mathrm{H_3O^+}][\mathrm{OBr^-}]}{[\mathrm{HOBr}]}\) \(2.5 \times 10^{-9} = \frac{[\mathrm{H_3O^+}](0.0134)}{(0.0147)}\) Solving for \([\mathrm{H_3O^+}]\), we get: \([\mathrm{H_3O^+}] = 2.74 \times 10^{-9}~M\) Finally, let's calculate the pH of the solution: \(pH = -\log{[\mathrm{H_3O^+}]} = -\log{(2.74 \times 10^{-9})} = 8.56\)

Final Answer

The pH of the solution prepared by mixing \(2.50 \mathrm{~g}\) of \(\mathrm{HOBr}\) and \(0.750 \mathrm{~g}\) of \(\mathrm{KOH}\) is \(8.56\).

Key concepts

Neutralization Reaction

A neutralization reaction is a type of chemical reaction where an acid and a base react together to form water and a salt. In our exercise, hypobromous acid (HOBr), which is a weak acid, reacts with potassium hydroxide (KOH), a strong base. This results in the formation of water (H2O) and potassium hypobromite (KOBr), a salt.
In any neutralization reaction, the acid donates a proton (H+) to the base. This can be seen in the chemical equation:
\[\mathrm{HOBr} + \mathrm{OH^-} \rightarrow \mathrm{H_2O} + \mathrm{OBr^-}\]
Paying attention to the moles of reactants is crucial as it allows us to determine the limiting reactant - the reactant that gets fully consumed in the reaction. In the given problem, KOH is the limiting reactant. Thus, the amount of OH- from the KOH dictates the extent to which the HOBr is neutralized. Understanding the stoichiometry – the relationship between the quantities of reactants – is key to tackling such problems.

Acid-Base Equilibrium

After the initial neutralization reaction, we're left with a solution of the weak acid HOBr and its conjugate base, OBr-. This is a classic example of an acid-base equilibrium situation. The concept of acid-base equilibrium revolves around the dissociation of a weak acid in water and the corresponding formation of its conjugate base.
For HOBr, this equilibrium can be represented by the equation:
\[\mathrm{HOBr} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{OBr^-}\]
To understand how this equilibrium affects pH, we use the acid dissociation constant (Ka), which provides a measure of the strength of an acid in solution. The lower the value of Ka, the weaker the acid. In the problem, we calculate the concentration of H3O+ ions with the given Ka and the concentrations of HOBr and OBr- after neutralization. This equilibrium calculation allows us to determine the pH of the solution, which gives us insight into the acidity or basicity of the resulting solution.

Molar Concentration

Molar concentration, also known as molarity, is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute divided by the volume of the solution in liters. This concept is fundamental in the calculation of pH since the pH is inherently linked to the molar concentration of hydrogen ions, H3O+, in the solution.
In the step by step solution, we determine the initial molar concentrations of both HOBr and KOH by dividing the number of moles of each substance by the total volume of the solution. For instance, the molar concentration of HOBr is calculated as:
\([\mathrm{HOBr}]_i = \frac{0.0281~\text{mol}}{1~L} = 0.0281~M\)
After completing the neutralization reaction, recalculating molar concentrations is necessary to reflect the changes in the number of moles of the remaining acid and the formed salt. Understanding molar concentration is vital for students to follow the changes that occur during the reaction and to accurately compute the pH value.