For the equilibrium system
`2HX(g) hArr H_(2)(g)+X_(2)(g)`
the equilibrium constant is `1.0xx10^(-5)`. What is the concentration of HX if the equilibrium concentration of `H_(2)` and `X_(2)` are `1.2xx10^(-3)` M, and `1.2xx10^(-4)` M respectively?

Options

A. `12xx10^(-4) M`
B. `12xx10^(-3) M`
C. `12xx10^(-2) M`
D. `12xx10^(-1) M`

Explanation

To solve the problem, we need to find the concentration of \( HX \) at equilibrium given the equilibrium constant \( K_c \) and the concentrations of the products \( H_2 \) and \( X_2 \).

### Step-by-Step Solution:

1. **Write the Expression for the Equilibrium Constant**:
The equilibrium constant \( K_c \) for the reaction
\[ 2HX(g) \rightleftharpoons H_2(g) + X_2(g) \]
is given by the formula:
\[ K_c = \frac{[H_2][X_2]}{[HX]^2} \]

2. **Substitute the Known Values**:
We know:
- \( K_c = 1.0 \times 10^{-5} \)
- \( [H_2] = 1.2 \times 10^{-3} \, M \)
- \( [X_2] = 1.2 \times 10^{-4} \, M \)

Substitute these values into the equilibrium expression:
\[ 1.0 \times 10^{-5} = \frac{(1.2 \times 10^{-3})(1.2 \times 10^{-4})}{[HX]^2} \]

3. **Calculate the Product of the Concentrations**:
First, calculate the product of the concentrations of \( H_2 \) and \( X_2 \):
\[ (1.2 \times 10^{-3})(1.2 \times 10^{-4}) = 1.44 \times 10^{-7} \]

4. **Set Up the Equation**:
Now, we can set up the equation:
\[ 1.0 \times 10^{-5} = \frac{1.44 \times 10^{-7}}{[HX]^2} \]

5. **Cross Multiply to Solve for \([HX]^2\)**:
Rearranging gives:
\[ [HX]^2 = \frac{1.44 \times 10^{-7}}{1.0 \times 10^{-5}} \]

6. **Perform the Division**:
Calculate the right side:
\[ [HX]^2 = 1.44 \times 10^{-2} \]

7. **Take the Square Root**:
Now, take the square root to find \([HX]\):
\[ [HX] = \sqrt{1.44 \times 10^{-2}} = 1.2 \times 10^{-1} \, M \]

### Final Answer:
The concentration of \( HX \) at equilibrium is:
\[ [HX] = 1.2 \times 10^{-1} \, M \]