Problem 61 - Chemistry The Central Science In Si Units

Question

Saccharin, a sugar substitute, is a weak acid with \(\mathrm{p} K_{a}=2.32\) at \(25^{\circ} \mathrm{C}\). It ionizes in aqueous solution as follows: $$ \mathrm{HNC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{NC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}^{-}(a q) $$ What is the \(\mathrm{pH}\) of a \(0.25 \mathrm{M}\) solution of this substance?

Step-by-Step Solution

Answer

The pH of the 0.25 M saccharin solution is approximately 1.46.

Step 1: Understand the Equilibrium Expression

The given ionization reaction of saccharin can be represented as \( \mathrm{HNC}_7\mathrm{H}_4\mathrm{SO}_3 \rightleftharpoons \mathrm{H}^+ + \mathrm{NC}_7\mathrm{H}_4\mathrm{SO}_3^- \). The equilibrium constant expression \(K_a\) for this reaction is given by \[ K_a = \frac{[\mathrm{H}^+][\mathrm{NC}_7\mathrm{H}_4\mathrm{SO}_3^-]}{[\mathrm{HNC}_7\mathrm{H}_4\mathrm{SO}_3]} \] Given the \(\mathrm{p} K_a = 2.32\), we can find \(K_a\) using \( K_a = 10^{-\mathrm{p} K_a} \).

Step 2: Calculate the Ionization Constant \(K_a\)

Using the formula \( K_a = 10^{-\mathrm{p} K_a} \), plug in the given value:\[ K_a = 10^{-2.32} \approx 4.79 \times 10^{-3} \] This is the equilibrium constant for the ionization of saccharin.

Step 3: Assume Initial Conditions and Changes at Equilibrium

As the initial concentration of saccharin is 0.25 M and it has not ionized initially, we consider:- Initial: \([\mathrm{HNC}_7\mathrm{H}_4\mathrm{SO}_3] = 0.25\) M, \([\mathrm{H}^+] = 0\), \( [\mathrm{NC}_7\mathrm{H}_4\mathrm{SO}_3^-] = 0\)- Change: \([-x, +x, +x]\)- Equilibrium: \([\mathrm{HNC}_7\mathrm{H}_4\mathrm{SO}_3] = 0.25 - x\), \([\mathrm{H}^+] = x\), \([\mathrm{NC}_7\mathrm{H}_4\mathrm{SO}_3^-] = x\)

Step 4: Set up the Equilibrium Expression

Substitute the equilibrium concentrations into the \(K_a\) expression:\[ 4.79 \times 10^{-3} = \frac{x \cdot x}{0.25 - x} \approx \frac{x^2}{0.25} \]Since \(x\) is expected to be small, we approximate \(0.25 - x \approx 0.25\).

Step 5: Solve for \(x\), the Concentration of \(\mathrm{H}^+\)

Now solve for \(x\):\[ x^2 = 4.79 \times 10^{-3} \times 0.25 \]\[ x^2 = 1.1975 \times 10^{-3} \]\[ x = \sqrt{1.1975 \times 10^{-3}} \approx 0.0346 \]Thus, \([\mathrm{H}^+] = 0.0346\) M.

Step 6: Calculate the \(\mathrm{pH}\) of the Solution

Using the hydrogen ion concentration, calculate \(\mathrm{pH}\):\[ \mathrm{pH} = -\log [\mathrm{H}^+] \]\[ \mathrm{pH} = -\log(0.0346) \approx 1.46 \]Thus, the \(\mathrm{pH}\) of the 0.25 M saccharin solution is approximately 1.46.

Key Concepts

Weak Acids

Acids can be categorized into strong and weak, depending on their ability to dissociate in water. Unlike strong acids, weak acids do not completely ionize in solution. Instead, they reach an equilibrium state where both the undissociated acid and its ions are present.
This equilibrium state is essential to understand as it forms the basis for calculating the pH of the solution.
Weak acids have a specific ionization reaction, such as the reaction of saccharin:

• \[\mathrm{HNC}_7\mathrm{H}_4\mathrm{SO}_3 \rightleftharpoons \mathrm{H}^+ + \mathrm{NC}_7\mathrm{H}_4\mathrm{SO}_3^-\]

In this reaction, not all the saccharin molecules release hydrogen ions; only a fraction does. This makes weak acids interesting since their pH is not an immediate reflection of their concentration.

pH Calculation

Knowing how to calculate the pH of a weak acid is essential. pH is a measure of the acidity of a solution, closely linked to the concentration of hydrogen ions present.For weak acids, the pH calculation generally involves using the ionization constant and the initial concentration.
With saccharin, after finding the concentration of hydrogen ions, indicated as

• \[ [\mathrm{H}^+] \approx 0.0346\ \text{M} \]

we can calculate the pH as follows:

• \[ \mathrm{pH} = -\log(0.0346) \]
• \[ \mathrm{pH} \approx 1.46 \]

By using the mathematical calculation above, pH shows that saccharin in a 0.25 M solution yields a highly acidic environment.

Equilibrium Constant

The equilibrium constant, denoted as \(K_a\) for acids, is a crucial factor in determining the extent of ionization of a weak acid. It quantifies the ratio of the concentration of products to reactants at equilibrium.
For saccharin, knowing its \(K_a\) is essential:

• \[ K_a = \frac{[\mathrm{H}^+][\mathrm{NC}_7\mathrm{H}_4\mathrm{SO}_3^-]}{[\mathrm{HNC}_7\mathrm{H}_4\mathrm{SO}_3]} \]

This small numerical value of \(K_a\) for weak acids like saccharin indicates low ionization in solution.
In this problem, saccharin's \(K_a\) is computed from its \(\mathrm{p} K_a = 2.32\) by the formula:

• \[ K_a = 10^{-2.32} \approx 4.79 \times 10^{-3} \]

Thus, an equilibrium calculation allows one to predict how much of the acid is dissociated into ions in a given solution.

Ionization

Ionization is the process by which a molecule splits into ions when dissolved in water. For weak acids, this only happens partially. In the specific case of saccharin, the ionization reaction is depicted by:

• \[ \mathrm{HNC}_7\mathrm{H}_4\mathrm{SO}_3 \rightleftharpoons \mathrm{H}^+ + \mathrm{NC}_7\mathrm{H}_4\mathrm{SO}_3^- \]

The equilibrium amounts of ions illustrate how not all saccharin molecules are transformed into ions.
Only a small fraction of the molecules contributes to the

• hydrogen ion concentration, \([\mathrm{H}^+]\)
• and the conjugate base, \([\mathrm{NC}_7\mathrm{H}_4\mathrm{SO}_3^-]\).

Calculations concerning weak acid ionization often require considering initial concentrations and changes upon reaching equilibrium, simplifying the understanding of its behavior in aqueous solutions.