Problem 112 - Chemistry: The Central Science in SI Units, Global Edition

Question

The following observations are made about a diprotic acid \(\mathrm{H}_{2} \mathrm{~A}:\) (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

The correct value for pKa2 of H2A is \(5.30\). This is the only option among the given values that would result in an acidic NaHA solution, as stated in the problem.

Step 1: Analyzing the given information

We are given that the pH of a 0.10 M solution of H2A is 3.30. Additionally, we know that a 0.10 M solution of the salt NaHA is acidic, which means the pH is less than 7. However, we don't have a specific pH value for the NaHA solution.

Step 2: Writing the ionization equations

The ionization of H2A in water can be written in two steps: Equation 1: H2A → HA- + H+ Equation 2: HA- → A2- + H+

Step 3: Using the definition of pKa

The definition of pKa can be written as: pKa = -log(Ka) Where Ka is the ionization constant of the acid. We can also write the relation between pH, pKa, and buffer ratio (Henderson-Hasselbalch equation): pH = pKa1 + log([HA-]/[H2A]) Since the pH of the H2A solution is given as 3.30, we can write: 3.30 = pKa1 + log(0.10 / [H2A])

Step 4: Analyzing the four pKa2 options

Now, let's analyze the four options for pKa2 values and see if they correspond to an acidic pH for NaHA. Option (i): If pKa2 = 3.22, it means that the second ionization is almost complete (pKa2 is very close to pH). This will make the salt NaHA almost neutral, which contradicts the given information that NaHA is acidic. So, pKa2 cannot be 3.22. Option (ii): If pKa2 = 5.30, it means that there is still significant ionization occurring in the second step (Equation 2), which will lower the pH more. As a result, the solution of NaHA will be acidic, which corresponds to the given information. Option (iii): If pKa2 = 7.47, the second ionization will be very weak since it will be attempting to ionize into a more alkaline solution. This will not lower the pH, and as a result, NaHA will not be acidic enough. So, pKa2 cannot be 7.47. Option (iv): If pKa2 = 9.82, the same argument as option (iii) applies here. pKa2 is too high for NaHA to be acidic, so pKa2 cannot be 9.82.

Step 5: Concluding the pKa2 value for H2A

Based on our analysis, option (ii) with pKa2 = 5.30 is the only possible value that corresponds to an acidic NaHA solution. Therefore, the correct value for pKa2 of H2A is 5.30.

Key Concepts

Ionization Equations

For a diprotic acid like \(\text{H}_2\text{A}\), ionization occurs in two distinct steps. Each step represents the loss of one proton \((\text{H}^+)\). To understand these steps, it's important to visualize the process:\[\text{Equation 1: } \text{H}_2\text{A} \rightleftharpoons \text{HA}^- + \text{H}^+\]This equation shows the first ionization of the diprotic acid, where \(\text{H}_2\text{A}\) loses one hydrogen ion to form \(\text{HA}^-\), a monobasic acid ion.\[\text{Equation 2: } \text{HA}^- \rightleftharpoons \text{A}^{2-} + \text{H}^+\]In this second step, the \(\text{HA}^-\) ion loses another hydrogen ion, resulting in \(\text{A}^{2-}\). The process of ionization in steps showcases how strong or weak each step of ionization is, ultimately affecting the pH and composition of the solution.Understanding these two equations is crucial for analyzing diprotic acids, as each ionization step impacts calculations for pH and pKa, which are central to many chemical reactions.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a handy tool when working with acid-base chemistry and buffer solutions. It's commonly used to calculate the pH of a solution based on the concentration of acid and its conjugate base. For a diprotic acid like \(\text{H}_2\text{A}\), the equation is expressed as:\[\text{pH} = \text{pKa}_1 + \log\left( \frac{[\text{HA}^-]}{[\text{H}_2\text{A}]} \right)\]This takes into account the concentration ratio of the acid (\(\text{H}_2\text{A}\)) and its conjugate base (\(\text{HA}^-\)). Using the equation, we can understand how the pH changes in relation to the relative quantities of the two species.- **pH**: Measure of the acidity of a solution.- **pKa**: A measure of the acid's strength, indicating how easily the acid donates protons.- **Logarithmic Ratio**: Reflects the relative concentrations of the conjugate base and acid.This equation proves extremely useful for calculations involving buffer systems and helps predict how small changes in concentration affect pH. Overall, it's an essential formula for anyone studying acid-base equilibria.

pKa Calculation

Calculating the pKa is an important step in determining the strength of an acid. It's crucial for understanding how diprotic acids like \(\text{H}_2\text{A}\) behave in solution. The pKa is defined as the negative logarithm of the acid dissociation constant \((K_a)\):\[\text{pKa} = -\log(K_a)\]Knowing the pKa helps chemists predict the degree of ionization at a given pH. For diprotic acids, both the first \((\text{pKa}_1)\) and second \((\text{pKa}_2)\) values are considered. These values allow us to:

• Understand the ionization process more deeply.
• Predict the behavior of the acid in different solutions.
• Distinguish between the strength of the two ionization stages.

In the context of the exercise, both the observation that \(\text{NaHA}\) solution is acidic and the given pH inform us about the reasonable pKa values. Here, recognizing that option (ii) with \(\text{pKa}_2 = 5.30\) fits the acidity of \(\text{NaHA}\) guides us to the correct conclusion regarding the strength of the second ionization step. With a lower pKa, the second hydrogen is more easily donated, confirming the acidic character of the solution.