Problem 91 - Chemistry and Chemical Reactivity

Question

Saccharin \(\left(\mathrm{HC}_{7} \mathrm{H}_{4} \mathrm{NO}_{3} \mathrm{S}\right)\) 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 \(0.10 \mathrm{M}\) solution of sodium saccharide at \(25^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The pH of the 0.10 M solution of sodium saccharide is 7.24.

Step 1: Determine Kb using the relationship between Ka and Kb

The ionization constant of a base, Kb, can be found from the acid dissociation constant, Ka, using the equation:\[ K_w = K_a \times K_b \]Where \(K_w\) is the ion-product constant of water (\(1.0 \times 10^{-14})\). Rearrange this to find \(K_b\):\[ K_b = \frac{K_w}{K_a} \]Given \(K_a = 10^{-2.32}\), calculate \(K_b\):\[ K_b = \frac{1.0 \times 10^{-14}}{10^{-2.32}} \approx 10^{-11.68} \]

Step 2: Set up the expression for hydrolysis of the base

When sodium saccharide (\( ext{NaC}_7 ext{H}_4 ext{NO}_3 ext{S} \)) dissolves in water, it forms the saccharide ion \(( ext{C}_7 ext{H}_4 ext{NO}_3 ext{S}^- )\), which acts as a weak base. The hydrolysis reaction is:\[ ext{C}_7 ext{H}_4 ext{NO}_3 ext{S}^- + ext{H}_2 ext{O} \rightleftharpoons ext{HC}_7 ext{H}_4 ext{NO}_3 ext{S} + ext{OH}^- \]

Step 3: Use the Kb expression to calculate [OH-]

Use the hydrolysis equilibrium expression:\[ K_b = \frac{[ ext{HC}_7 ext{H}_4 ext{NO}_3 ext{S}][ ext{OH}^-]}{[ ext{C}_7 ext{H}_4 ext{NO}_3 ext{S}^-]} \]Assuming the initial concentration of \( ext{C}_7 ext{H}_4 ext{NO}_3 ext{S}^-\) is 0.10 M and letting \(x\) be the change in concentration, the equation becomes:\[ K_b = \frac{x^2}{0.10} \approx 10^{-11.68} \]Solve for \(x\):\[ x = \sqrt{10^{-11.68} \times 0.10} \approx 1.74 \times 10^{-7} \]

Step 4: Convert [OH-] to pOH

Calculate \( ext{pOH}\) using the equation:\[ ext{pOH} = -\log[ ext{OH}^-] \]\[ ext{pOH} = -\log(1.74 \times 10^{-7}) \approx 6.76 \]

Step 5: Find pH from pOH

Since \( ext{pH} + ext{pOH} = 14 \), we can find the \( ext{pH}\) by:\[ ext{pH} = 14 - ext{pOH} = 14 - 6.76 = 7.24 \]

Key Concepts

Weak Acids

Weak acids, like saccharin, partially dissociate in water. This means not all acid molecules donate hydrogen ions (\( ext{H}^+ \)) to the solution. Instead, there's a balance between ionized and non-ionized forms. The strength of a weak acid is measured by its acid dissociation constant (\(K_a\)). The smaller the \(K_a\), the weaker the acid is considered. Because weak acids don’t completely ionize, calculating properties like \(pH\) requires understanding this equilibrium. Saccharin, with a \(pK_a\) of 2.32, is a relatively weak acid compared to stronger acids like hydrochloric acid, which almost completely dissociate in water.

pH Calculation

To find the \(pH\) of a solution containing saccharin in salt form, such as sodium saccharide, we first need to know about the \(OH^-\) concentration. Sodium saccharide hydrolyzes in water to produce saccharin (the conjugate acid) and hydroxide ions (\(OH^-\)).

• Begin with the given molarity of the solution.
• Calculate the \(K_b\) for the base using the relationship between \(K_a\) and \(K_b\).
• Use the calculated \(K_b\) to find the concentration of \(OH^-\).

After finding \([OH^-]\), the next step is to calculate the \(pOH\) using the formula \( \text{pOH} = -\log[OH^-] \). Finally, convert to \(pH\) using the relation \( \text{pH} + \text{pOH} = 14 \). This full process gives us the \(pH\) of the solution.

Equilibrium Constants

Equilibrium constants are crucial in understanding acid and base reactions. For weak acids, the equilibrium constant is expressed as \(K_a\) (acid dissociation constant). Conversely, for bases, like the saccharide ion, it’s \(K_b\) (base ionization constant). These constants tell us how much a compound likes to ionize in solution:

• High \(K_a\) and \(K_b\) suggest more complete ionization.
• Low values indicate weak ionization.

The relationship \(K_w = K_a \times K_b\) links acids and their conjugate bases, reflecting their tendency to produce \(H^+\) and \(OH^-\) ions, respectively. By knowing \(K_a\), \(K_b\) can be calculated, which is vital in determining the equilibrium position of hydrolysis reactions.

Chemical Reactions

Chemical reactions involving weak acids and bases, like saccharin dissociating in water, rely on equilibrium principles. In these reactions:

• Reactants and products are continuously being formed and consumed.
• The system reaches a state where the concentration of reactants and products remain constant over time.

This is known as dynamic equilibrium. The hydrolysis of sodium saccharide in water represents an equilibrium between the saccharide ion and saccharin, producing \(OH^-\). Understanding these reactions helps in predicting whether a solution will be acidic or basic and is crucial for many practical applications, such as in pharmaceuticals and food chemistry, where saccharin is commonly used.