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 pH of a \(0.25 \mathrm{M}\) solution of this substance?

Short answer

The pH of a 0.25 M solution of saccharin, a weak acid with a pKa of 2.32, is approximately 1.46.

Step-by-step solution

Write the ionization equilibrium equation and Ka expression

The ionization of saccharin in an aqueous solution can be represented by the following equation: \[ \mathrm{HNC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{NC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}^{-}(a q) \] The equilibrium constant expression for this weak acid ionization is given by: \[K_a = \frac{[\mathrm{H}^+][\mathrm{NC}_7\mathrm{H}_4\mathrm{SO}_3^-]}{[\mathrm{HNC}_7\mathrm{H}_4\mathrm{SO}_3]}\]

Determine the initial concentrations

The initial concentration of saccharin is given as 0.25 M. Since the solution is initially un-ionized, the initial concentrations of the other species (H+ and NC₇H₄SO₃⁻) are both 0: \[[\mathrm{HNC}_7\mathrm{H}_4\mathrm{SO}_3]_0 = 0.25 \mathrm{M}\] \[[\mathrm{H}^+]_0 = 0\] \[[\mathrm{NC}_7\mathrm{H}_4\mathrm{SO}_3^-]_0 = 0\]

Set up an equilibrium table (ICE table)

An ICE (Initial, Change, Equilibrium) table helps us calculate the changes in concentration of the species in the ionization process: ``` [HNC₇H₄SO₃] [H⁺] [NC₇H₄SO₃⁻] Initial 0.25M 0 0 Change -x +x +x Equilibrium 0.25-x x x ```

Set up the Ka equation and solve

Substitute the equilibrium concentrations in the Ka expression: \[K_a = \frac{x \times x}{(0.25 - x)}\] Since the pKa is given as 2.32, the Ka can be calculated as: \[K_a = 10^{-\mathrm{p}K_a} = 10^{-2.32} = 4.82 \times 10^{-3}\] Now we can solve for "x": \[4.82 \times 10^{-3} = \frac{x^2}{(0.25 - x)}\] Simplifying further, we have \(x^2 \approx 4.82 \times 10^{-3} \times 0.25\): \[x \approx \sqrt{1.21 \times 10^{-3}}\] \[x \approx 0.0348\]

Calculate the pH

Since "x" is the concentration of H+ ions at equilibrium, we can now calculate the pH of the solution using the formula: \[\mathrm{pH} = -\log{[\mathrm{H}^+]}\] \[\mathrm{pH} = -\log{(0.0348)}\] \[\mathrm{pH} \approx 1.46\] The pH of the 0.25 M saccharin solution is approximately 1.46.

Key concepts

Equilibrium Constant

In chemistry, an equilibrium constant (Ka) plays a key role in understanding the behavior of acids and bases in solution. When a weak acid, like saccharin, dissolves in water, it partially ionizes into its ions.
Think of the equilibrium constant, Ka, as a measure of how far this ionization process goes.
It is a ratio of the concentration of the products to the reactants, as shown in the equation: \[K_a = \frac{[\mathrm{H}^+][\mathrm{NC}_7\mathrm{H}_4\mathrm{SO}_3^-]}{[\mathrm{HNC}_7\mathrm{H}_4\mathrm{SO}_3]}\]
For saccharin, which is a weak acid, the equilibrium state shows the extent to which it loses an H^+ ion in solution.
The smaller the Ka value, the weaker the acid, indicating fewer ions are produced.

Weak Acid Ionization

Weak acids, such as saccharin, do not fully dissociate into their ions in water.
This is in contrast to strong acids, which completely break down into ions.
The partial ionization of a weak acid is an important concept. It means that in a solution, both the acid and its ionized form exist.
The ionization equation for saccharin is:\[\mathrm{HNC}_7 \mathrm{H}_4 \mathrm{SO}_3(aq) \rightleftharpoons \mathrm{H}^+(aq) + \mathrm{NC}_7 \mathrm{H}_4 \mathrm{SO}_3^-(aq)\]
Weak acids have Ka values which help determine the extent of ionization.
A higher Ka indicates a greater degree of ionization, but for weak acids, this value is typically small. Understanding this concept is crucial for pH calculations and predicting how the acid behaves in different conditions.

ICE Table

An ICE table is a useful tool chemists use to simplify equilibrium problems like this one involving saccharin.
ICE stands for Initial, Change, Equilibrium, and it's a way to organize the starting conditions, changes, and final concentrations of reactants and products.
When set up, the table looks like this:

• **Initial** - shows the initial concentrations before any reaction. For saccharin, it's 0.25 M for the acid, and 0 M for the ions.
• **Change** - shows how much of each substance forms or gets used up. Here, we suppose it changes by an amount x .
• **Equilibrium** - details the concentration of each species at equilibrium. Using sacrine, it's 0.25 - x for the HNC_7H_4SO_3 and x for both ions.

This setup makes it straightforward to plug values into the equilibrium expression and solve for unknowns.

Saccharin

Saccharin is more than just a sweetener. It's an excellent example of a weak acid in teaching chemistry.
As a weak acid, saccharin provides a practical context for understanding concepts like equilibrium and ionization.
It's particularly interesting because of its unique structure and the fact that it partially ionizes, which affects the flavor profile.
In addition to its use in sweetening without calories, understanding its chemical properties helps in predicting its behavior in different chemical reactions.