Selenious acid `(H_(2)SeO_(3))`, a diprotic acid has `K_(a1)=3.0xx10^(-3)` and `K_(a2)=5.0xx10^(-8)`. What is the `[OH^(-)]` of a 0.30 M solution of selenious acid?

Options

A. `2.85xx10^(-3)`
B. `5.0xx10^(-6)`
C. `3.5xx10^(-12)`
D. `3.5xx10^(-13)`

Answer

To find the concentration of hydroxide ions \([OH^-]\) in a 0.30 M solution of selenious acid \((H_2SeO_3)\), we will follow these steps:

### Step 1: Understand the dissociation of selenious acid
Selenious acid is a diprotic acid, meaning it can donate two protons (H\(^+\)). The dissociation can be represented in two steps:
1. \( H_2SeO_3 \rightleftharpoons H^+ + HSeO_3^- \) (with \( K_{a1} = 3.0 \times 10^{-3} \))
2. \( HSeO_3^- \rightleftharpoons H^+ + SeO_3^{2-} \) (with \( K_{a2} = 5.0 \times 10^{-8} \))

### Step 2: Set up the equilibrium expressions
Let \( C = 0.30 \) M be the initial concentration of \( H_2SeO_3 \).
- After the first dissociation, let \( x \) be the concentration of \( H^+ \) produced.
- At equilibrium for the first dissociation:
- \([H_2SeO_3] = C - x\)
- \([HSeO_3^-] = x\)
- \([H^+] = x\)

For the second dissociation, let \( y \) be the concentration of \( H^+ \) produced from this step:
- At equilibrium for the second dissociation:
- \([HSeO_3^-] = x - y\)
- \([SeO_3^{2-}] = y\)
- \([H^+] = x + y\)

### Step 3: Apply the equilibrium constant expressions
Using the first dissociation:
\[ K_{a1} = \frac{[H^+][HSeO_3^-]}{[H_2SeO_3]} = \frac{x \cdot x}{C - x} = \frac{x^2}{0.30 - x} \]
Given \( K_{a1} = 3.0 \times 10^{-3} \):
\[ 3.0 \times 10^{-3} = \frac{x^2}{0.30 - x} \]

### Step 4: Simplify the equation
Assuming \( x \) is small compared to 0.30, we can neglect \( x \) in the denominator:
\[ 3.0 \times 10^{-3} \approx \frac{x^2}{0.30} \]
Thus,
\[ x^2 = 3.0 \times 10^{-3} \times 0.30 \]
\[ x^2 = 9.0 \times 10^{-4} \]
\[ x = \sqrt{9.0 \times 10^{-4}} = 3.0 \times 10^{-2} \text{ M} \]

### Step 5: Calculate the total \([H^+]\)
Since \( y \) is negligible compared to \( x \) (because \( K_{a2} \) is much smaller than \( K_{a1} \)), we can consider:
\[ [H^+] \approx x = 3.0 \times 10^{-2} \text{ M} \]

### Step 6: Calculate \([OH^-]\)
Using the relation \( [H^+][OH^-] = 1.0 \times 10^{-14} \):
\[ [OH^-] = \frac{1.0 \times 10^{-14}}{[H^+]} = \frac{1.0 \times 10^{-14}}{3.0 \times 10^{-2}} = 3.33 \times 10^{-13} \text{ M} \]

### Conclusion
The concentration of hydroxide ions \([OH^-]\) in a 0.30 M solution of selenious acid is approximately \( 3.33 \times 10^{-13} \text{ M} \).