The equilibrium composition for the reaction is
`{:(PCl_(3),+,Cl_(2),hArr,PCl_(5)),(0.20,,0.10,,0.40 mol L^(-1)):}`
What will be the equilibrium concentration of `PCl_(5)` on adding `0.10 mol` of `Cl_(2)` at the same temperature?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ \text{PCl}_3 + \text{Cl}_2 \rightleftharpoons \text{PCl}_5 \]

### Step 2: Identify the initial equilibrium concentrations
From the problem, we have the equilibrium concentrations:
- \([\text{PCl}_3] = 0.20 \, \text{mol L}^{-1}\)
- \([\text{Cl}_2] = 0.10 \, \text{mol L}^{-1}\)
- \([\text{PCl}_5] = 0.40 \, \text{mol L}^{-1}\)

### Step 3: Calculate the equilibrium constant \(K_c\)
The equilibrium constant \(K_c\) for the reaction can be calculated using the formula:
\[ K_c = \frac{[\text{PCl}_5]}{[\text{PCl}_3][\text{Cl}_2]} \]
Substituting the equilibrium concentrations:
\[ K_c = \frac{0.40}{(0.20)(0.10)} = \frac{0.40}{0.02} = 20 \]

### Step 4: Determine the change in concentration upon adding \(0.10 \, \text{mol}\) of \(\text{Cl}_2\)
When we add \(0.10 \, \text{mol}\) of \(\text{Cl}_2\), the new concentration of \(\text{Cl}_2\) becomes:
\[ [\text{Cl}_2] = 0.10 + 0.10 = 0.20 \, \text{mol L}^{-1} \]

### Step 5: Set up the expression for the new equilibrium
Let \(x\) be the change in concentration of \(\text{PCl}_5\) at the new equilibrium. The changes in concentrations will be:
- \([\text{PCl}_3] = 0.20 - x\)
- \([\text{Cl}_2] = 0.20 - x\)
- \([\text{PCl}_5] = 0.40 + x\)

### Step 6: Write the equilibrium expression with the new concentrations
Using the equilibrium constant \(K_c\):
\[ 20 = \frac{0.40 + x}{(0.20 - x)(0.20 - x)} \]

### Step 7: Solve the equation
Cross-multiplying gives:
\[ 20(0.20 - x)^2 = 0.40 + x \]
Expanding and rearranging:
\[ 20(0.04 - 0.4x + x^2) = 0.40 + x \]
\[ 0.80 - 8x + 20x^2 = 0.40 + x \]
\[ 20x^2 - 9x + 0.40 = 0 \]

### Step 8: Use the quadratic formula to solve for \(x\)
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 20\), \(b = -9\), and \(c = 0.40\):
\[ x = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 20 \cdot 0.40}}{2 \cdot 20} \]
\[ x = \frac{9 \pm \sqrt{81 - 32}}{40} \]
\[ x = \frac{9 \pm \sqrt{49}}{40} \]
\[ x = \frac{9 \pm 7}{40} \]
Calculating the two possible values for \(x\):
1. \(x = \frac{16}{40} = 0.40\) (not valid as it makes \([\text{PCl}_3]\) negative)
2. \(x = \frac{2}{40} = 0.05\) (valid)

### Step 9: Calculate the new equilibrium concentration of \(\text{PCl}_5\)
Now substituting \(x = 0.05\) back to find the new concentration of \(\text{PCl}_5\):
\[ [\text{PCl}_5] = 0.40 + 0.05 = 0.45 \, \text{mol L}^{-1} \]

### Final Answer
The equilibrium concentration of \(\text{PCl}_5\) after adding \(0.10 \, \text{mol}\) of \(\text{Cl}_2\) is:
\[ \boxed{0.45 \, \text{mol L}^{-1}} \]