Problem 99 - Chemistry and Chemical Reactivity

Question

Saccharin (HC \(_{7} \mathrm{H}_{4} \mathrm{NO}_{3} \mathrm{S}\) ) is a weak acid with \(\mathrm{p} K_{\mathrm{a}}=2.32\) at \(25^{\circ} \mathrm{C} .\) It is used in the form of sodium saccharide, \(\mathrm{NaC}_{7} \mathrm{H}_{4} \mathrm{NO}_{3} \mathrm{S}\). What is the \(\mathrm{pH}\) of a 0.10 M solution of sodium saccharide at \(25^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The pH of the solution is approximately 6.16.

Step 1: Understand the Problem

We're given sodium saccharide, which is the salt form of saccharin. Sodium saccharide in water will dissociate, producing the saccharin ion, a weak base. We need to find the pH of its 0.10 M solution using the given \(\text{p}K_a\).

Step 2: Find Kb from Ka

The equilibrium constant \( K_a \) is given as \( 2.32 \) for saccharin. We can find the base dissociation constant \( K_b \) using the relation\[ K_w = K_a \times K_b \]where \( K_w = 1.0 \times 10^{-14} \). Thus, \[K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{10^{-2.32}}\] Calculate the \( K_b \).

Step 3: Calculate Kb Value

Continue with the calculation:\[K_b = \frac{1.0 \times 10^{-14}}{4.79 \times 10^{-3}} \approx 2.09 \times 10^{-12}\]

Step 4: Set up the Equilibrium Expression

The saccharin ion \((C_7H_4NO_3S^- )\) will react with water to produce \(OH^-\) and \(HC_7H_4NO_3S\). The equilibrium expression is:\[ C_7H_4NO_3S^- + H_2O \rightleftharpoons OH^- + HC_7H_4NO_3S \]Initially, the concentration of \( C_7H_4NO_3S^- \) is \(0.10 \) M, and \([OH^-] = [HC_7H_4NO_3S] = 0\).

Step 5: Calculate OH- Concentration

Using the expression:\[K_b = \frac{[OH^-][HC_7H_4NO_3S]}{[C_7H_4NO_3S^-]}\]Substitute:\[2.09 \times 10^{-12} = \frac{x^2}{0.10 - x}\]With \(x\) being small, simplify to:\[x = \sqrt{2.09 \times 10^{-13}} \approx 1.45 \times 10^{-7}\]So, \([OH^-] = 1.45 \times 10^{-7} \text{M}\).

Step 6: Calculate pOH and pH

Find \( \text{pOH} \) using:\[\text{pOH} = -\log([OH^-]) \approx 7.84\]Next, calculate \(\text{pH}\):\[\text{pH} = 14 - \text{pOH} = 14 - 7.84 = 6.16\]

Step 7: Conclude the Solution

The pH of the 0.10 M sodium saccharide solution is approximately \( 6.16 \).

Key Concepts

Weak Acids

A weak acid is a type of acid that does not completely dissociate into its ions in a solution. Instead of releasing all its hydrogen ions, a weak acid will only partly ionize in water. This partial dissociation is characterized by an equilibrium state between the undissociated acid molecules and the ions formed. The equilibrium is represented by an equilibrium constant known as the acid dissociation constant, denoted as \( K_a \).
This constant, \( K_a \), provides a quantitative measure of the acid's strength in solution—the larger the \( K_a \), the stronger the acid, as more hydrogen ions are released. Weak acids like saccharin have low \( K_a \) values, indicating they do not easily give up their hydrogen ions, thus having only a modest effect on the pH of a solution.
In the given exercise, saccharin, represented by the chemical formula \( ext{HC}_7 ext{H}_4 ext{NO}_3 ext{S}\), is identified as a weak acid with a \( ext{p}K_a \) of 2.32. This is indicative of its tendency to remain largely un-ionized in solution, establishing an equilibrium that must be considered during pH calculations.

Sodium Saccharide

Sodium saccharide is actually the sodium salt form of saccharin. In water, sodium saccharide dissociates into sodium ions and saccharin ions. These saccharin ions are what interact in the solution, potentially affecting pH by forming a weak conjugate base.
The dissociation can be thought of as a straightforward ionization process:

• The sodium ion, \( ext{Na}^+\), is a spectator ion and does not affect the acid-base equilibrium.
• The saccharin ion, \( ext{C}_7 ext{H}_4 ext{NO}_3 ext{S}^- \), acts as a weak base, meaning it will partially accept protons from water, influencing the solution's basicity and thus its pH.

A key understanding here is that salts formed from weak acids like saccharin often result in slightly basic solutions, as their conjugate bases tend to accept hydrogen ions slowly, raising the pH past neutrality.

Acid-Base Equilibrium

Acid-base equilibrium refers to the balance that occurs in a solution containing weak acids and their conjugate bases. This concept is crucial for understanding how solutions maintain their pH level and how to calculate changes in pH.
In the equilibrium of sodium saccharide solution, the saccharin ion (which is a weak base) reacts with water to establish the following equilibrium:

• \( ext{C}_7 ext{H}_4 ext{NO}_3 ext{S}^- + ext{H}_2 ext{O} \rightleftharpoons ext{OH}^- + ext{HC}_7 ext{H}_4 ext{NO}_3 ext{S} \)

The equilibrium constant for this reaction, \( K_b \), is derived from the \( K_a \) of saccharin using the relationship \( K_w = K_a \times K_b \), where \( K_w \) is the ion product of water. By calculating \( K_b \) and using an equilibrium expression for the formation of \( ext{OH}^- \), we can estimate the ion concentrations, thus determining the pH and \( ext{pOH} \) of the solution.
This equilibrium reflects how the weak conjugate base in sodium saccharide makes the solution slightly basic, causing the pH to rise above 7.

Saccharin

Saccharin, known for being a sugar substitute, is not just a sweetener—its chemistry involves interesting acid-base behavior. The formula for saccharin is \( ext{HC}_7 ext{H}_4 ext{NO}_3 ext{S}\), and it is known to be a weak acid.
In its pure form, saccharin is part of dietary products, but in solutions like sodium saccharide, its role transitions into affecting pH. Being a weak acid, saccharin partially dissociates: its conjugate base \( ext{C}_7 ext{H}_4 ext{NO}_3 ext{S}^- \) forms and establishes an equilibrium with water that innovates the calculated pH.
While saccharin itself is not largely soluble in water, its sodium salt form dissociates quite readily, making it possible for the weak acid characteristics to prevail and influence the chemistry of solutions. Understanding these interactions is key for processes in food science and pharmaceuticals where saccharin is used.