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) \(9.82 ?\)

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.

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

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])

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.

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.