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: $$ \mathrm{HNC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}(a q)=\mathrm{H}^{+}(a q)+\mathrm{NC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}^{-}(a q) $$ What is the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution of this substance?

Short answer

The pH of a 0.10 M solution of saccharin is approximately 1.16.

Step-by-step solution

Write the ionization equation and equilibrium constant expression for saccharin

Write the ionization equation for saccharin: \( \mathrm{HNC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}(a q) \rightleftharpoons \mathrm{H}^{+}(a q) + \mathrm{NC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}^{-}(a q) \) Now, we can write the equilibrium constant expression for this reaction as: \(K_a = \frac{[\mathrm{H}^{+}] [\mathrm{NC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}^{-}]}{[\mathrm{HNC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}]} \) The given value for PKa is 2.32. To find the Ka, we can use the equation: \(K_a = 10^{-\mathrm{P} K_{a}}\)

Calculate the value of Ka

Using the given PKa value, we can calculate Ka as follows: \(K_a = 10^{-2.32} \approx 4.79 \times 10^{-3} \)

Set up an ICE table

An ICE (Initial, Change, Equilibrium) table will help us find the equilibrium concentrations of the ions involved in the reaction. | | HNC7H4SO3 | H+ | NC7H4SO3- | |-----|----------:|------:|-----------| | I | 0.10 | 0 | 0 | | C | -x | +x | +x | | E | 0.10 - x | x | x | Here, x represents the change in concentration as the reaction proceeds to equilibrium.

Use the equilibrium concentrations to solve for x

Now, we will substitute the equilibrium concentrations from the ICE table into the equilibrium expression for Ka: \(4.79 \times 10^{-3} = \frac{(x) (x)}{0.10 - x} \) Since saccharin is a weak acid and doesn't ionize completely, we can assume that x is small enough compared to the initial concentration, which results in: \(4.79 \times 10^{-3} = \frac{x^2}{0.10}\) Solve for x: \(x^2 = 4.79 \times 10^{-3} \times 0.10\) \(x \approx 6.92 \times 10^{-2}\)

Calculate the pH of the solution

Now that we have the value of x, which represents the equilibrium concentration of \(\mathrm{H}^+\) ions, we can calculate the pH of the solution using the definition of pH: \( \mathrm{pH} = -\log_{10} [\mathrm{H}^{+}] \) Plug in the value of x for the concentration of \(\mathrm{H}^+\) ions: \( \mathrm{pH} = -\log_{10} (6.92 \times 10^{-2}) \approx 1.16 \) The pH of a 0.10 M solution of saccharin is approximately 1.16.

Key concepts

Equilibrium Constant

The equilibrium constant (\( K_a \)) is a crucial concept in understanding chemical reactions, especially weak acid ionization. It represents the ratio of the concentrations of the products to the reactants when a reaction is at equilibrium. For acid dissociation, \( K_a \) specifically refers to how well an acid ionizes in water.
Understanding \( K_a \) involves:

• Writing the expression for the dissociation of the acid. For saccharin, the reaction is \( \mathrm{HNC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{NC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}^{-}(aq) \)
• From this, the equilibrium constant expression is \( K_a = \frac{[\mathrm{H}^{+}] [\mathrm{NC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}^{-}]}{[\mathrm{HNC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}]} \)
• This tells you how much the acid ionizes in equilibrium.

A low \( K_a \) value indicates a weak acid, which doesn’t fully ionize in water.

Weak Acid

A weak acid is one that doesn’t completely dissociate in water. This partial ionization is characterized by a \( K_a \) value that is less than that of strong acids, meaning it produces fewer ions in solution.
Key points about weak acids include:

• They have larger \( pK_a \) values since \( pK_a = -\log_{10} K_a \).
• For saccharin, with a \( pK_a \) of 2.32, it’s evident that it has a moderate acidity, much weaker than strong acids like hydrochloric acid.
• Weak acids are in equilibrium in their aqueous solutions, meaning the acid and its ions are present together.

Understanding weak acids helps when calculating \( \text{pH} \) and determining acid strength.

Ionization Equation

Ionization equations depict the balance between an acid and its ionized form. For a weak acid, this involves equilibrium between the uncharged acid and its ions.
In the case of saccharin, the ionization is:\[ \mathrm{HNC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{NC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}^{-}(aq) \]

• This equation shows how saccharin releases \( \mathrm{H}^{+} \)
• The equilibrium highlights that not all saccharin molecules donate protons.
• The double arrow indicates that the reaction can proceed in both directions until equilibrium is reached.

Understanding this helps simplify the calculation of concentrations at equilibrium and aids in calculating \( \text{pH} \).

ICE Table

An ICE table (Initial, Change, Equilibrium) is a tool used to track changes in concentrations throughout the reaction process. It is especially useful for reactions like weak acid ionization.
Steps for using an ICE table:

• List initial concentrations of all species involved before any reaction takes place. For saccharin, this means beginning with 0.10 M of the acid and 0 M of its ions.
• Define the change in terms of variable x, which represents the amount ionized, such as \(-x\) for the acid and \(+x\) for the ions.
• List equilibrium concentrations as expressions involving x (e.g., \(0.10 - x \) for the acid).

The ICE table simplifies solving the equilibrium expression by providing clear values for substitution into the \( K_a \) expression.