Question

Saccharin, a sugar substitute, is a weak acid with $\mathrm{p}{K}_{\mathrm{a}}=2.32$ at ${25}^{\circ }\mathrm{C}$. It ionizes in aqueous solution as follows: ${\mathrm{HNC}}_7{\mathrm{H}}_4{\mathrm{SO}}_3(aq)\rightleftharpoons {\mathrm{H}}^{+}(aq)+{\mathrm{NC}}_7{\mathrm{H}}_4{\mathrm{SO}}_3^{-}(aq)$ What is the $\mathrm{pH}$ of a $0.10\mathrm{M}$ solution of this substance?

Short answer

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

Step-by-step solution

Write the ionization equilibrium expression for saccharin

The ionization equilibrium expression for the weak acid dissociation of saccharin in water is: $$K_a=\frac{[{\mathrm{H}}^{+}][{\mathrm{NC}}_7{\mathrm{H}}_4{\mathrm{SO}}_3^{-}]}{[{\mathrm{HNC}}_7{\mathrm{H}}_4{\mathrm{SO}}_3]}$$

Find the Ka for saccharin using the given pKa

We are given that the pKa of saccharin is 2.32. To find the Ka, use the relationship between pKa and Ka: $$pKa=-\log Ka$$ $$Ka={10}^{-pKa}$$ $$Ka={10}^{-2.32}\approx 4.8\times {10}^{-3}$$

Set up the ICE table for saccharin

Using the provided initial concentration of saccharin, create an ICE table to represent the initial, change, and equilibrium concentrations of the species in the reaction: $$\begin{array}{cccc}\text{Species}& \text{Initial}& \text{Change}& \text{Equilibrium}\\ {\mathrm{HNC}}_7{\mathrm{H}}_4{\mathrm{SO}}_3& 0.10\mathrm{M}& -x& 0.10-x\\ {\mathrm{H}}^{+}& 0\mathrm{M}& +x& x\\ {\mathrm{NC}}_7{\mathrm{H}}_4{\mathrm{SO}}_3^{-}& 0\mathrm{M}& +x& x\end{array}$$

Solve for the equilibrium concentration of H+ ions ([H+])

Plug the equilibrium concentrations from the ICE table into the Ka expression and solve for x: $$K_a=\frac{x\cdot x}{0.10-x}\approx 4.8\times {10}^{-3}$$ $$x^2\approx (0.10-x)(4.8\times {10}^{-3})$$ To simplify problem-solving, we can assume x << 0.10, so equation will be: $$x^2\approx (0.10)(4.8\times {10}^{-3})$$ $$x\approx \sqrt{(0.10)(4.8\times {10}^{-3})}\approx 6.9\times {10}^{-2}$$ Therefore, the equilibrium concentration of H+ ions, [H+], is approximately 6.9 x 10^{-2} M.

Calculate the pH of the solution

Now that we have the equilibrium concentration of H+ ions, we can calculate the pH of the solution using the following formula: $$pH=-\log [H^{+}]$$ $$pH=-\log (6.9\times {10}^{-2})\approx 1.16$$ The pH of the 0.10 M saccharin solution is approximately 1.16.

Key concepts

Weak Acid Dissociation

Weak acids are substances that partially ionize in water, creating a balance between the un-ionized acid and its ions. This equilibrium is essential as it dictates how much of the acid donates its protons (H+) to the solution. Unlike a strong acid that dissociates completely, a weak acid does so incomplete and its strength is quantified by the extent of its dissociation.

For example, saccharin, which is mentioned in the exercise, is a weak acid. When it is dissolved in water, it ionizes but not entirely. The equation representing this is:
$${\mathrm{HNC}}_7{\mathrm{H}}_4{\mathrm{SO}}_3(aq)\rightleftharpoons {\mathrm{H}}^{+}(aq)+{\mathrm{NC}}_7{\mathrm{H}}_4{\mathrm{SO}}_3^{-}(aq)$$
In this equilibrium, not all saccharin molecules turn into ions—only a portion of them. This selective ionization is the key characteristic of weak acid behavior and it significantly impacts the pH calculation.

Acid Dissociation Constant (Ka)

To describe the degree of dissociation and the strength of a weak acid, we reference its acid dissociation constant, denoted as Ka. The Ka value is a numerical representation of the acid's tendency to donate H+ ions to the solution. The formula used to express this equilibrium is:
$$K_a=\frac{[{\mathrm{H}}^{+}][{\mathrm{NC}}_7{\mathrm{H}}_4{\mathrm{SO}}_3^{-}]}{[{\mathrm{HNC}}_7{\mathrm{H}}_4{\mathrm{SO}}_3]}$$
Here, the square brackets indicate the concentration of each species at equilibrium. The higher the Ka value, the stronger the acid is, meaning it dissociates more. As shown in the problem solution, knowing the pKa allows us to find the Ka by using the relationship $Ka={10}^{-pKa}$, which then aids in determining the pH of the solution. This step is fundamental because it bridges the understanding of the acid's inherent strength with the actual hydrogen ion concentration in solution.

Equilibrium Concentration

When a weak acid dissolves in water, the concentrations of the reactants and products reach a state of dynamic equilibrium, meaning that the rate of the forward reaction equals the rate of the reverse reaction. This state is represented by the concentrations of the species involved in the dissociation at equilibrium.

In our calculations, these values are critical because they allow us to determine the pH of the solution. For the saccharin example, we found that the equilibrium concentration of H+ ions was $6.9\times {10}^{-2}$ M. This was achieved by making an informed approximation (assuming x, which represents the change in concentration, is much less than the initial concentration) and then solving for x. The equilibrium concentration of H+ ions is a direct indicator of the acidity of the solution and feeds directly into the pH equation.

ICE Table

The ICE table is an invaluable tool for visualizing the changes in concentration of species in a chemical reaction, where ICE stands for Initial, Change, and Equilibrium. The first column lists the species involved in the reaction, the second shows their initial concentrations, the third indicates the changes during the reaction, usually expressed as ±x, and the last column displays the equilibrium concentrations, which are a combination of the initial concentrations and the changes.

For the saccharin dissociation reaction, setting up an ICE table allows us to track the dissociation process systematically. By filling out the ICE table with the known initial concentrations and the changes represented by x, we can then solve for the unknown change to find the equilibrium concentrations. This systematic approach is conducive to understanding chemical equilibria and essential for accurate pH calculations.