10 mL of 0.1 M tribasic acid `H_(3)A` is titrated with 0.1 M NaOH solution. What is the ratio of `[[H_(3)A]]/[[A^(3-)]]" at " 2^(nd)` equivalence points? `[Given : K_(a1)=10^(-3),K_(a2)=10^(-8),K_(a3)=10^(-12)]`
(a)`cong10^(-4)`
(b)`cong10^(+4)`
(c)`cong10^(-7)`
(d)`cong10^(+6)`

Options

A. `cong10^(-4)`
B. `cong10^(+4)`
C. `cong10^(-7)`
D. `cong10^(+6)`

Answer

To solve the problem, we need to find the ratio of the concentrations of the tribasic acid \( H_3A \) and its fully deprotonated form \( A^{3-} \) at the second equivalence point during the titration with NaOH.

### Step-by-Step Solution:

1. **Understanding the Titration Process**:
- We have a tribasic acid \( H_3A \) which can donate three protons.
- The titration with 0.1 M NaOH will neutralize the protons in a stepwise manner.
- At the second equivalence point, two protons have been neutralized, resulting in the formation of \( HA^{2-} \).

2. **Determine the Volume of NaOH Required**:
- The initial volume of \( H_3A \) is 10 mL with a concentration of 0.1 M.
- The number of moles of \( H_3A \) is:
\[ \text{Moles of } H_3A = \text{Volume (L)} \times \text{Concentration (M)} = 0.01 \, \text{L} \times 0.1 \, \text{mol/L} = 0.001 \, \text{mol} \]
- At the second equivalence point, 2 moles of NaOH will react with 1 mole of \( H_3A \).
- Therefore, the moles of NaOH required will be:
\[ \text{Moles of NaOH} = 2 \times 0.001 \, \text{mol} = 0.002 \, \text{mol} \]
- The volume of 0.1 M NaOH needed:
\[ \text{Volume of NaOH} = \frac{\text{Moles}}{\text{Concentration}} = \frac{0.002 \, \text{mol}}{0.1 \, \text{mol/L}} = 20 \, \text{mL} \]

3. **Total Volume at the Second Equivalence Point**:
- The total volume after adding NaOH is:
\[ \text{Total Volume} = 10 \, \text{mL} + 20 \, \text{mL} = 30 \, \text{mL} \]

4. **Concentration of \( A^{2-} \) at the Second Equivalence Point**:
- At the second equivalence point, the concentration of \( A^{2-} \) will be:
\[ [A^{2-}] = \frac{0.002 \, \text{mol}}{0.030 \, \text{L}} = \frac{0.002}{0.030} \approx 0.0667 \, \text{M} \]

5. **Calculating \( [H^+] \) at the Second Equivalence Point**:
- The pH at the second equivalence point can be calculated using the average of \( pK_{a2} \) and \( pK_{a3} \):
\[ pH = \frac{pK_{a2} + pK_{a3}}{2} = \frac{8 + 12}{2} = 10 \]
- Therefore, the concentration of \( H^+ \) ions is:
\[ [H^+] = 10^{-pH} = 10^{-10} \, \text{M} \]

6. **Finding the Ratio \( \frac{[H_3A]}{[A^{3-}]} \)**:
- Using the ionization constant expression:
\[ K_{a2} = \frac{[H^+]^2 \cdot [A^{3-}]}{[H_3A]} \]
- Rearranging gives:
\[ [H_3A] = \frac{[H^+]^2 \cdot [A^{3-}]}{K_{a2}} \]
- Substituting the values:
\[ K_{a2} = 10^{-8}, \quad [H^+] = 10^{-10}, \quad [A^{3-}] = 0.0667 \]
- Thus,
\[ [H_3A] = \frac{(10^{-10})^2 \cdot 0.0667}{10^{-8}} = \frac{10^{-20} \cdot 0.0667}{10^{-8}} = 6.67 \times 10^{-13} \, \text{M} \]

7. **Calculating the Final Ratio**:
- Now we can find the ratio:
\[ \frac{[H_3A]}{[A^{3-}]} = \frac{6.67 \times 10^{-13}}{0.0667} \approx 10^{-6} \]

### Final Answer:
The ratio \( \frac{[H_3A]}{[A^{3-}]} \) at the second equivalence point is approximately \( 10^{-6} \), which corresponds to option (d) \( \cong 10^{+6} \).