`H_(3)A` is a weak triprotic acid `(K_(a1)=10^(-5),K_(a2)=10^(-9),K_(a3)=10^(-13)`
What is the value of pX of 0.1 M `H_(3)A` (aq.) solution ? Where pX=-log X and X=`[[A^(3-)]]/[[HA^(2-)]]`

Options

A. 7
B. 8
C. 9
D. 10

Answer

To solve the problem, we need to determine the value of \( pX \) for a 0.1 M solution of the weak triprotic acid \( H_3A \) with given dissociation constants \( K_{a1} = 10^{-5} \), \( K_{a2} = 10^{-9} \), and \( K_{a3} = 10^{-13} \).

### Step-by-Step Solution:

1. **Understanding the Dissociation of the Acid:**
The triprotic acid \( H_3A \) can dissociate in three steps:
- \( H_3A \rightleftharpoons H_2A^- + H^+ \) (with \( K_{a1} \))
- \( H_2A^- \rightleftharpoons HA^{2-} + H^+ \) (with \( K_{a2} \))
- \( HA^{2-} \rightleftharpoons A^{3-} + H^+ \) (with \( K_{a3} \))

2. **Identify Dominant Dissociation:**
Since \( K_{a1} \) is significantly larger than \( K_{a2} \) and \( K_{a3} \), the majority of \( H^+ \) ions will come from the first dissociation step. Therefore, we can assume that the concentration of \( H^+ \) is primarily determined by the first dissociation.

3. **Calculate \( [H^+] \):**
The concentration of \( H^+ \) can be approximated using the formula:
\[ [H^+] \approx \sqrt{K_{a1} \cdot C} \]
where \( C = 0.1 \, \text{M} \) (the concentration of \( H_3A \)).
\[ [H^+] \approx \sqrt{10^{-5} \cdot 0.1} = \sqrt{10^{-6}} = 10^{-3} \, \text{M} \]

4. **Calculate \( x = \frac{[A^{3-}]}{[HA^{2-}]} \):**
From the third dissociation step, we can use the expression for \( K_{a3} \):
\[ K_{a3} = \frac{[A^{3-}][H^+]}{[HA^{2-}]} \]
Plugging in the known values:
\[ 10^{-13} = \frac{[A^{3-}] \cdot 10^{-3}}{[HA^{2-}]} \]
Rearranging gives:
\[ [A^{3-}] = 10^{-13} \cdot [HA^{2-}] \cdot 10^3 \]
Thus,
\[ [A^{3-}] = 10^{-10} \cdot [HA^{2-}] \]
Therefore,
\[ x = \frac{[A^{3-}]}{[HA^{2-}]} = 10^{-10} \]

5. **Calculate \( pX \):**
Finally, we compute \( pX \):
\[ pX = -\log(x) = -\log(10^{-10}) = 10 \]

### Final Answer:
The value of \( pX \) for the 0.1 M solution of \( H_3A \) is **10**.