Question

Determine the pH of each of the following solutions ( \(K_{a}\) and \(K_{b}\) values are given in Appendix D): (a) \(0.095 M\) hypochlorous acid, (b) \(0.0085 \mathrm{M}\) hydrazine, (c) \(0.165 \mathrm{M}\) hydroxylamine.

Short answer

(a) The pH of a \(0.095\mathrm{M}\) hypochlorous acid solution is approximately \(4.24\). (b) The pH of a \(0.0085\mathrm{M}\) hydrazine solution is approximately \(10.18\). (c) The pH of a \(0.165\mathrm{M}\) hydroxylamine solution is approximately \(9.91\).

Step-by-step solution

Write the ionization equation and the \(K_a\) expression for hypochlorous acid (HClO)

The ionization equation for hypochlorous acid is \[HClO \rightleftharpoons H^+ + ClO^-\] The \(K_a\) expression for this reaction is: \[K_a = \frac{[H^+][ClO^-]}{[HClO]}\]

Set up and solve for x, which represents the change in concentration due to ionization

Let \(x\) represent the change in concentration due to ionization. Initially, we have \(0.095 M\) of HClO, and no \(H^+\) or \(ClO^-\) ions. After ionization, we have: \[HClO = 0.095-x\] \[H^+ = x\] \[ClO^- = x\] Substitute these values into the \(K_a\) expression: \[K_a = \frac{x(x)}{0.095-x}\] According to Appendix D, the \(K_{a}\) is 3.5 × 10^-8 for hypochlorous acid. So, we have: \[3.5 \times 10^{-8} = \frac{x^2}{0.095-x}\] Assuming that \(x\) is very small compared to \(0.095\), solve for x: \[x^2 \approx 3.5 \times 10^{-8} × 0.095\]

Find the concentration of \(H^+\) ions and calculate the pH

Solve for x: \[x = \sqrt{3.5 \times 10^{-8} × 0.095} \approx 5.8 \times 10^{-5}\] Since \(x\) corresponds to the \([H^+]\) concentration, we can calculate the pH as: \[pH = -\log [H^+] = -\log 5.8 \times 10^{-5} \approx 4.24\] (b) For hydrazine (N2H4), we have a \(0.0085M\) solution:

Write the ionization equation and the \(K_b\) expression for hydrazine (N2H4)

The ionization equation for hydrazine is \[N2H4 + H2O \rightleftharpoons N2H5^+ + OH^-\] The \(K_b\) expression for this reaction is: \[K_b = \frac{[N2H5^+][OH^-]}{[N2H4]}\]

Set up and solve for x, which represents the change in concentration due to ionization

Let \(x\) represent the change in concentration due to ionization. Initially, we have \(0.0085 M\) of N2H4, and no \(N2H5^+\) or \(OH^-\) ions. After ionization, we have: \[N2H4 = 0.0085-x\] \[N2H5^+ = x\] \[OH^-= x\] Substitute these values into the \(K_b\) expression: \[K_b = \frac{x(x)}{0.0085-x}\] According to Appendix D, the \(K_{b}\) is 3.1 × 10^-6 for hydrazine. So, we have: \[3.1 \times 10^{-6} = \frac{x^2}{0.0085-x}\] Assuming that \(x\) is very small compared to \(0.0085\), solve for x: \[x^2 \approx 3.1 \times 10^{-6} × 0.0085\]

Find the concentration of \(OH^-\) ions and calculate the pH

Solve for x: \[x = \sqrt{3.1 \times 10^{-6} × 0.0085}\approx 1.5 \times 10^{-4}\] Since \(x\) corresponds to the \([OH^-]\) concentration, we can calculate the pOH as: \[pOH = -\log [OH^-] = -\log 1.5 \times 10^{-4} \approx 3.82\] Now, we can find the pH using the relationship between pH and pOH: \[pH = 14 - pOH = 14 - 3.82 \approx 10.18\] (c) For hydroxylamine (NH2OH), we have a \(0.165 M\) solution:

Write the ionization equation and the \(K_a\) expression for hydroxylamine (NH2OH)

The ionization equation for hydroxylamine is \[NH2OH \rightleftharpoons NH2O^+ + OH^-\] The \(K_a\) expression for this reaction is: \[K_a = \frac{[NH2O^+][OH^-]}{[NH2OH]}\]

Set up and solve for x, which represents the change in concentration due to ionization

Let \(x\) represent the change in concentration due to ionization. Initially, we have \(0.165 M\) of NH2OH, and no \(NH2O^+\) or \(OH^-\) ions. After ionization, we have: \[NH2OH = 0.165-x\] \[NH2O^+ = x\] \[OH^-= x\] Substitute these values into the \(K_a\) expression: \[K_a = \frac{x(x)}{0.165-x}\] According to Appendix D, the \(K_{a}\) is 7.0 × 10^-9 for hydroxylamine. So, we have: \[7.0 \times 10^{-9} = \frac{x^2}{0.165-x}\] Assuming that \(x\) is very small compared to \(0.165\), solve for x: \[x^2 \approx 7.0 \times 10^{-9} × 0.165\]

Find the concentration of \(OH^-\) ions and calculate the pH

Solve for x: \[x = \sqrt{7.0 \times 10^{-9} × 0.165} \approx 8.1 \times 10^{-5}\] Since \(x\) corresponds to the \([OH^-]\) concentration, we can calculate the pOH as: \[pOH = -\log [OH^-] = -\log 8.1 \times 10^{-5} \approx 4.09\] Now, we can find the pH using the relationship between pH and pOH: \[pH = 14 - pOH = 14 - 4.09 \approx 9.91\]

Key concepts

Acid-Base Equilibrium

In the world of chemistry, understanding acid-base equilibrium is essential for calculating the pH of solutions. The equilibrium concept revolves around the balance that occurs in a solution when an acid donates protons and a base accepts them.
An acid-base reaction can be represented by an equilibrium equation like \[HA \rightleftharpoons H^+ + A^-\],
where the acid \(HA\) dissociates into a proton \(H^+\) and its conjugate base \(A^-\).
The key idea here is that not all acid or base molecules break apart. A state of equilibrium is reached where the rate of ionization equals the rate of recombination.

• Le Chatelier's Principle: This principle suggests that if changes occur in conditions, such as concentration or temperature, the equilibrium will shift to counteract that change.
• Dynamic Balance: Even though it may appear static, the ions in the solution are continuously being formed and recombined.

Understanding this balance helps in the determination of the pH using the equilibrium constant, \(K_a\) for acids or \(K_b\) for bases.

Ionization Expressions

Ionization expressions are mathematical representations of the degree to which a weak acid or base ionizes in water. These expressions help us calculate unknown concentrations of ions in a chemical solution.
The ionization constant, expressed as \(K_a\) for acids or \(K_b\) for bases, is a crucial part of these expressions.
For a weak acid like hypochlorous acid \(HClO\), the ionization is represented as: \[HClO \rightleftharpoons H^+ + ClO^-\].
The ionization expression is then given by:

• \[K_a = \frac{[H^+][ClO^-]}{[HClO]}\]
• \([H^+] = [ClO^-] = x\)
• \([HClO] = initial - x\)

The expression allows us to solve for \(x\), representing how much the acid ionizes.
These expressions are based on the assumption that ionization is very small compared to the initial concentration, thereby simplifying calculations.

Weak Acids and Bases

Weak acids and bases are distinct because they do not fully dissociate in water. This incomplete ionization means that calculations often involve assumptions to simplify the math.
In contrast to strong acids and bases, which dissociate completely, weak ones reach an equilibrium where both ionized and unionized forms coexist.

• Examples: Hypochlorous acid \(HClO\) and hydrazine \(N2H4\) are classic weak acid and base, respectively.
• Dissociation Constant: Weak acids and bases are characterized by low \(K_a\) and \(K_b\) values, meaning their ionization in solution is minimal.

The pH of such solutions is calculated using the established equilibrium expressions, allowing assumptions like negligible x-terms to approximate ion concentration changes. When calculating pH, these equilibrium expressions provide essential insight into the behavior of weak acids and bases and their effect on the solution's acidity or basicity.