Question

Here is a way to test the validity of the statement made on page 719 in conjunction with the three key ideas governing the ionization of polyprotic acids. Determine the \(\mathrm{pH}\) of \(0.100 \mathrm{M}\) succinic acid in two ways: first by assuming that \(\mathrm{H}_{3} \mathrm{O}^{+}\) is produced only in the first ionization step, and then by allowing for the possibility that some \(\mathrm{H}_{3} \mathrm{O}^{+}\) is also produced in the second ionization step. Compare the results, and discuss the significance of your finding. $$\begin{array}{c} \mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{4}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{HC}_{4} \mathrm{H}_{4} \mathrm{O}_{4}^{-} \\ &K_{\mathrm{a}_{1}}=6.2 \times 10^{-5} \\ \mathrm{HC}_{4} \mathrm{H}_{4} \mathrm{O}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{4}^{2-} \\ &K_{\mathrm{a}_{2}}=2.3 \times 10^{-6} \end{array}$$

Short answer

To provide a short answer, you would need to perform the calculations outlined in the solution steps. This can't be accomplished without specific numerical solutions to the equations provided in the steps.

Step-by-step solution

Calculate the pH assuming the first ionization only

Firstly, we need to set up an equilibrium expression for the first ionization of the acid: \(K_{a1} = \frac {[H_{3} O^{+}] [HC_{4} H_{4} O_{4}^{-}]} {[H_{2} C_{4} H_{4} O_{4}]}\). From the problem statement, we know that \(K_{a1} = 6.2 \times 10^{-5}\) and the initial concentration of the acid was 0.100 M. We can write down the change in concentrations as -x for \(H_{2} C_{4} H_{4} O_{4}\), +x for \(H_{3} O^{+}\) and \(HC_{4} H_{4} O_{4}^{-}\). After substituting the values in equilibrium expression, we can solve it to find out the value of 'x' which will be equal to the \(H_{3} O^{+}\) concentration.

Calculate the pH

Once we have the concentration of \(H_{3} O^{+}\), we can easily calculate the pH of the solution using the formula: \(\mathrm{pH} = -\log([H_{3} O^{+}])\).

Calculation considering both ionizations

Now let's repeat the process considering both ionization steps. But this time, we set up our equilibrium expression for the second ionization step: \(K_{a2} = \frac {[H_{3} O^{+}] [C_{4} H_{4} O_{4}^{2-}]} {[H_{4} C_{4} H_{4} O_{4}^{-}]}\). We can again set up our change in concentrations like we did before and can find out the new concentration for \(H_{3} O^{+}\).

Recalculate the pH

After getting the new concentration of \(H_{3} O^{+}\) considering both the ionizations, we can again calculate the pH of the solution using the formula: \(\mathrm{pH} = -\log([H_{3} O^{+}])\).

Compare the results

Finally, we need to compare the values of pH we got considering only the first ionization and both the ionizations respectively. We need to discuss the significance of these findings.

Key concepts

Ionization Equilibrium

To understand ionization equilibrium in polyprotic acids like succinic acid, we need to grasp the concept of stepwise ionization. Polyprotic acids can donate more than one proton (hydrogen ion) in solution. This occurs in sequential steps, each with its own equilibrium constant. In the case of succinic acid, it undergoes two ionization steps, characterized by the constants \( K_{a1} \) and \( K_{a2} \). The first ionization is generally stronger, with a higher \( K_{a} \) value, as seen with \( K_{a1} = 6.2 \times 10^{-5} \). This represents the conversion of \( H_{2}C_{4}H_{4}O_{4} \) to \( HC_{4}H_{4}O_{4}^{-} \) and \( H_3O^+ \). The second ionization is weaker, highlighted by the lower \( K_{a2} \) value of \( 2.3 \times 10^{-6} \). During this step, \( HC_{4}H_{4}O_{4}^{-} \) ionizes to \( C_{4}H_{4}O_{4}^{2-} \) and additional \( H_3O^+ \).
In equilibrium, both ionic forms coexist with non-ionized molecules. The equilibrium concentrations are calculated by assessing the changes in concentrations from the initial molarity of the acid to the equilibrated state. These calculations are essential to understand the extent of ionization and the resulting \( H_3O^+ \) concentration, which directly influences the pH of the solution.

pH Calculation

Calculating the pH of a solution involves determining the concentration of \( H_3O^+ \) ions produced in the solution. For polyprotic acids such as succinic acid, we often start by looking at the pH impact of the first ionization step alone, due to its typically greater contribution to the \( H_3O^+ \) concentration. We use the formula:

• \[ K_{a1} = \frac{[H_3O^+][HC_4H_4O_4^-]}{[H_2C_4H_4O_4]} \]

This helps us solve for \( x \) where \( x \) represents the \( H_3O^+ \) concentration in mol/L. Once \( x \) is found, the pH can be calculated using the logarithmic formula:

• \( \text{pH} = -\log([H_3O^+]) \)

Considering the second ionization step in the calculation might slightly adjust \( x \), providing a more accurate pH for solutions where both ionizations contribute. For accurate pH, we examine both the first and second ionization steps, find the new \( H_3O^+ \) concentration, and recalculate the pH, giving us insight into the contribution of each ionization step to the overall acidity.

Acid-Base Chemistry

Acid-base chemistry explores the balance and transfer of protons between molecules in solutions and is crucial for understanding how substances like succinic acid behave in solution. In this exercise, succinic acid acts as a Bronsted-Lowry acid, donating protons to water molecules, forming hydronium ions \( H_3O^+ \).
Polyprotic acids, in general, provide multiple protons for donation and have multiple equilibrium constants. This contrasts with monoprotic acids, which only donate one proton. For each ionization step, a distinct \( K_a \) value is assigned that signals the strength of that step. Higher \( K_a \) values imply a stronger acid and greater ionization potential, impacting the pH of the solution dramatically. Understanding these changes within the realm of acid-base chemistry helps predict the behavior of acids like succinic acid, ensuring accurate calculations and interpretations of solution acidity.
This foundational understanding also applies in broader contexts, such as buffer solutions where weak acids and their salts maintain pH by neutralizing added acids or bases, illustrating the practical applications of acid-base chemistry.