Problem 112 - Chemistry The Central Science

Question

The following observations are made about a diprotic acid \(\mathrm{H}_{2} \mathrm{A} :(\mathrm{i}) \mathrm{A} 0.10 \mathrm{M}\) solution of \(\mathrm{H}_{2} \mathrm{A}\) has \(\mathrm{pH}=3.30 .\) (ii) \(\mathrm{A} 0.10 \mathrm{M}\) solution of the salt NaHA is acidic. Which of the following could be the value of \(\mathrm{p} K_{a 2}\) for \(\mathrm{H}_{2} \mathrm{A} :\) (i) \(3.22,\) (ii) 5.30 , (iii) \(7.47,\) or (iv) 9.82\(?\)

Step-by-Step Solution

Answer

Based on the given observations, the possible value of \(pK_{a2}\) for H2A is 9.82.

Step 1: Calculate the concentration of H3O+ from the first observation

Given the pH of the 0.10 M solution of H2A is 3.30, first, we need to find the concentration of H3O+ ions in the solution using: \[ [\ce{H3O+}] = 10^{-pH} \] \[ [\ce{H3O+}] = 10^{-3.30} \approx 5.01 \times 10^{-4} \: M \]

Step 2: Determine the value of pKa1

Next, we will use the concentration of H3O+ to determine the value of \(pK_{a1}\) using the following equation with the initial concentration of H2A: \[ K_{a1} = \frac{[\ce{H3O+}]^2}{[\ce{H2A}] - [\ce{H3O+}]} \] Let X be the equilibrium concentration of H3O+ ions, then we have: \[ K_{a1} = \frac{(5.01 \times 10^{-4})^2}{0.10 - 5.01 \times 10^{-4}} \] Solving for \(K_{a1}\), we obtain: \[ K_{a1} \approx 2.52 \times 10^{-5} \] Now, we can calculate the value of \(pK_{a1}\): \[ pK_{a1} = -\log{K_{a1}} \approx 4.60 \]

Step 3: Check which value of pKa2 makes NaHA acidic

We are given 4 candidate values for \(pK_{a2}\). We can use the given salt NaHA and the possible values of \(pK_{a2}\) and check which one of them results in an acidic solution (pH less than 7).

Step 4: Candidate 1: pKa2 = 3.22

If \(pK_{a2}\) = 3.22, then since this value is less than that of \(pK_{a1}\), it would mean that the first ionization is weaker than the second, which is not possible for diprotic acids. Therefore, 3.22 cannot be the value for \(pK_{a2}\).

Step 5: Candidate 2: pKa2 = 5.30

If \(pK_{a2}\) = 5.30, it means that the second ionization is stronger than the first ionization (pKa1 = 4.60). The salt NaHA will not hydrolyze significant amounts of H2A (or H3O+), and the resulting solution will not be acidic. So, 5.30 is not a correct value for \(pK_{a2}\).

Step 6: Candidate 3: pKa2 = 7.47

If \(pK_{a2}\) = 7.47, then the second ionization is even weaker than the scenario with a value of 5.30. The salt NaHA will ionize H2A, but the solution's overall acidity will be very weak, and it will not be considered acidic. So, 7.47 is not a correct value for \(pK_{a2}\).

Step 7: Candidate 4: pKa2 = 9.82

If \(pK_{a2}\) = 9.82, then the second ionization of H2A is very weak. In such a situation, the presence of NaHA will still ionize some H2A, but only a small fraction. The concentration of NaHA being closer to the equivalence point will make the solution more acidic than in previous cases. Based on this analysis, 9.82 could be the value of \(pK_{a2}\).

Step 8: Final Answer

Based on the given observations, the possible value of \(pK_{a2}\) for H2A is 9.82.

Key Concepts

pH Calculation

Understanding the calculation of pH is crucial when working with solutions in chemistry, particularly for acidic or basic ones. pH is a scale used to specify the acidity or basicity of an aqueous solution. It is defined as the negative logarithm (base 10) of the concentration of hydronium ions (\text{H}_{3}\text{O}^+)\text{ in moles per liter. The equation used to calculate pH is:}
\[ \text{pH} = -\log([\text{H}_{3}\text{O}^+]) \]
For example, if the concentration of \text{H}_{3}\text{O}^+ in a solution is 5.01 \times 10^{-4} M, the pH is calculated as 3.30. This simple relationship can help us to understand the degree of acidity or basicity in any given solution.

Acid Dissociation Constant

The acid dissociation constant (Ka) is a quantitative measure of the strength of an acid in solution. It is the equilibrium constant for the dissociation reaction of the acid into anions and hydrogen ions in water. The higher the Ka value, the stronger the acid, as it tends to donate protons more completely.
For a generic acid dissociation reaction:\[ \text{HA} \rightleftharpoons \text{A}^- + \text{H}^+ \]
The acid dissociation constant can be calculated as follows:\[ K_a = \frac{[\text{A}^-][\text{H}^+]}{[\text{HA}]} \]
The related term pKa is the negative logarithm of Ka and is used for its simplicity:
\[ \text{pKa} = -\log(K_a) \]
This concept is essential when solving problems related to acid strength and the predictability of their behavior in solution.

Weak Acid Equilibria

Weak acids partially dissociate in solution, establishing an equilibrium between the undissociated acid and the ions produced. The equilibrium concentrations of the products and reactants are governed by the Ka value of the weak acid.
For a diprotic acid (H2A), the first deprotonation can be represented as:\[ \text{H}_2\text{A} \rightleftharpoons \text{H}^+ + \text{HA}^- \]
For the second deprotonation:\[ \text{HA}^- \rightleftharpoons \text{A}^{2-} + \text{H}^+ \]
Each deprotonation step has its Ka value, denoted as Ka1 and Ka2, respectively. The degree to which each step proceeds impacts the pH of the solution and the overall ionic composition. Equilibrium calculations often require assumptions and logical reasoning to simplify the process and yield a solution, a process which can be crucial in solving complex equilibria.

Salt Hydrolysis

Salt hydrolysis occurs when anion or cation components of a salt react with water, potentially affecting the pH of the solution. For instance, the salt of a weak acid and strong base will typically undergo anion hydrolysis, making the solution basic. Conversely, salts formed from strong acids and weak bases often result in acidic solutions due to cation hydrolysis.
For a generic salt such as NaHA, derived from a weak acid (H2A), the anion (HA-) may react with water:\[ \text{HA}^- + \text{H}_2\text{O} \rightleftharpoons \text{H}_2\text{A} + \text{OH}^- \]
If the second dissociation constant (pKa2) of the weak acid is high, indicating a very weakly acidic second hydrogen, the amount of OH- produced can increase, creating a more acidic solution. Therefore, an understanding of salt hydrolysis is necessary to predict and control the pH of salt solutions.